INSTRUCTOR’S

SOLUTION MANUAL

KEYING YE AND SHARON MYERS

for

PROBABILITY & STATISTICS

FOR ENGINEERS & SCIENTISTS

EIGHTH EDITION

WALPOLE, MYERS, MYERS, YE

Contents

1 Introduction to Statistics and Data Analysis 1

2 Probability 11

3 Random Variables and Probability Distributions 29

4 Mathematical Expectation 45

5 Some Discrete Probability Distributions 59

6 Some Continuous Probability Distributions 71

7 Functions of Random Variables 85

8 Fundamental Sampling Distributions and Data Descriptions 91

9 One- and Two-Sample Estimation Problems 103

10 One- and Two-Sample Tests of Hypotheses 121

11 Simple Linear Regression and Correlation 149

12 Multiple Linear Regression and Certain Nonlinear Regression Models 171

13 One-Factor Experiments: General 185

14 Factorial Experiments (Two or More Factors) 213

15 2k Factorial Experiments and Fractions 237

16 Nonparametric Statistics 257

iii

iv CONTENTS

17 Statistical Quality Control 273

18 Bayesian Statistics 277

Chapter 1

Introduction to Statistics and Data

Analysis

1.1 (a) 15.

(b) ¯x = 1

15 (3.4 + 2.5 + 4.8 + · · · + 4.8) = 3.787.

(c) Sample median is the 8th value, after the data is sorted from smallest to largest:

3.6.

(d) A dot plot is shown below.

2.5 3.0 3.5 4.0 4.5 5.0 5.5

(e) After trimming total 40% of the data (20% highest and 20% lowest), the data

becomes:

2.9 3.0 3.3 3.4 3.6

3.7 4.0 4.4 4.8

So. the trimmed mean is

¯xtr20 =

1

9

(2.9 + 3.0 + · · · + 4.8) = 3.678.

1.2 (a) Mean=20.768 and Median=20.610.

(b) ¯xtr10 = 20.743.

(c) A dot plot is shown below.

18 19 20 21 22 23

1

2 Chapter 1 Introduction to Statistics and Data Analysis

1.3 (a) A dot plot is shown below.

200 205 210 215 220 225 230

In the figure, “×” represents the “No aging” group and “◦” represents the “Aging”

group.

(b) Yes; tensile strength is greatly reduced due to the aging process.

(c) MeanAging = 209.90, and MeanNo aging = 222.10.

(d) MedianAging = 210.00, and MedianNo aging = 221.50. The means and medians for

each group are similar to each other.

1.4 (a) ¯XA = 7.950 and ˜XA = 8.250;

¯X

B = 10.260 and ˜XB = 10.150.

(b) A dot plot is shown below.

6.5 7.5 8.5 9.5 10.5 11.5

In the figure, “×” represents company A and “◦” represents company B. The

steel rods made by company B show more flexibility.

1.5 (a) A dot plot is shown below.

−10 0 10 20 30 40

In the figure, “×” represents the control group and “◦” represents the treatment

group.

(b) ¯XControl = 5.60, ˜XControl = 5.00, and ¯Xtr(10);Control = 5.13;

¯X

Treatment = 7.60, ˜XTreatment = 4.50, and ¯Xtr(10);Treatment = 5.63.

(c) The difference of the means is 2.0 and the differences of the medians and the

trimmed means are 0.5, which are much smaller. The possible cause of this might

be due to the extreme values (outliers) in the samples, especially the value of 37.

1.6 (a) A dot plot is shown below.

1.95 2.05 2.15 2.25 2.35 2.45 2.55

In the figure, “×” represents the 20◦C group and “◦” represents the 45◦C group.

(b) ¯X20◦C = 2.1075, and ¯X45◦C = 2.2350.

(c) Based on the plot, it seems that high temperature yields more high values of

tensile strength, along with a few low values of tensile strength. Overall, the

temperature does have an influence on the tensile strength.

Solutions for Exercises in Chapter 1 3

(d) It also seems that the variation of the tensile strength gets larger when the cure

temperature is increased.

1.7 s2 = 1

15−1 [(3.4−3.787)2+(2.5−3.787)2+(4.8−3.787)2+· · ·+(4.8−3.787)2] = 0.94284;

s = √s2 = √0.9428 = 0.971.

1.8 s2 = 1

20−1 [(18.71 − 20.768)2 + (21.41 − 20.768)2 + · · · + (21.12 − 20.768)2] = 2.5345;

s = √2.5345 = 1.592.

1.9 s2

No Aging = 1

10−1 [(227 − 222.10)2 + (222 − 222.10)2 + · · · + (221 − 222.10)2] = 42.12;

sNo Aging = √42.12 = 6.49.

s2

Aging = 1

10−1 [(219 − 209.90)2 + (214 − 209.90)2 + · · · + (205 − 209.90)2] = 23.62;

sAging = √23.62 = 4.86.

1.10 For company A: s2

A = 1.2078 and sA = √1.2078 = 1.099.

For company B: s2

B = 0.3249 and sB = √0.3249 = 0.570.

1.11 For the control group: s2

Control = 69.39 and sControl = 8.33.

For the treatment group: s2

Treatment = 128.14 and sTreatment = 11.32.

1.12 For the cure temperature at 20◦C: s2

20◦C = 0.005 and s20◦C = 0.071.

For the cure temperature at 45◦C: s2

45◦C = 0.0413 and s45◦C = 0.2032.

The variation of the tensile strength is influenced by the increase of cure temperature.

1.13 (a) Mean = ¯X = 124.3 and median = ˜X = 120;

(b) 175 is an extreme observation.

1.14 (a) Mean = ¯X = 570.5 and median = ˜X = 571;

(b) Variance = s2 = 10; standard deviation= s = 3.162; range=10;

(c) Variation of the diameters seems too big.

1.15 Yes. The value 0.03125 is actually a P-value and a small value of this quantity means

that the outcome (i.e., HHHHH) is very unlikely to happen with a fair coin.

1.16 The term on the left side can be manipulated to

Xn

i=1

xi − n¯x =

Xn

i=1

xi −

Xn

i=1

xi = 0,

which is the term on the right side.

1.17 (a) ¯Xsmokers = 43.70 and ¯Xnonsmokers = 30.32;

(b) ssmokers = 16.93 and snonsmokers = 7.13;

4 Chapter 1 Introduction to Statistics and Data Analysis

(c) A dot plot is shown below.

10 20 30 40 50 60 70

In the figure, “×” represents the nonsmoker group and “◦” represents the smoker

group.

(d) Smokers appear to take longer time to fall asleep and the time to fall asleep for

smoker group is more variable.

1.18 (a) A stem-and-leaf plot is shown below.

Stem Leaf Frequency

1 057 3

2 35 2

3 246 3

4 1138 4

5 22457 5

6 00123445779 11

7 01244456678899 14

8 00011223445589 14

9 0258 4

(b) The following is the relative frequency distribution table.

Relative Frequency Distribution of Grades

Class Interval Class Midpoint Frequency, f Relative Frequency

10 − 19

20 − 29

30 − 39

40 − 49

50 − 59

60 − 69

70 − 79

80 − 89

90 − 99

14.5

24.5

34.5

44.5

54.5

64.5

74.5

84.5

94.5

3

2

3

4

5

11

14

14

4

0.05

0.03

0.05

0.07

0.08

0.18

0.23

0.23

0.07

(c) A histogram plot is given below.

14.5 24.5 34.5 44.5 54.5 64.5 74.5 84.5 94.5

Final Exam Grades

Relative Frequency

Solutions for Exercises in Chapter 1 5

The distribution skews to the left.

(d) ¯X = 65.48, ˜X = 71.50 and s = 21.13.

1.19 (a) A stem-and-leaf plot is shown below.

Stem Leaf Frequency

0 22233457 8

1 023558 6

2 035 3

3 03 2

4 057 3

5 0569 4

6 0005 4

(b) The following is the relative frequency distribution table.

Relative Frequency Distribution of Years

Class Interval Class Midpoint Frequency, f Relative Frequency

0.0 − 0.9

1.0 − 1.9

2.0 − 2.9

3.0 − 3.9

4.0 − 4.9

5.0 − 5.9

6.0 − 6.9

0.45

1.45

2.45

3.45

4.45

5.45

6.45

8

6

3

2

3

4

4

0.267

0.200

0.100

0.067

0.100

0.133

0.133

(c) ¯X = 2.797, s = 2.227 and Sample range is 6.5 − 0.2 = 6.3.

1.20 (a) A stem-and-leaf plot is shown next.

Stem Leaf Frequency

0* 34 2

0 56667777777889999 17

1* 0000001223333344 16

1 5566788899 10

2* 034 3

2 7 1

3* 2 1

(b) The relative frequency distribution table is shown next.

6 Chapter 1 Introduction to Statistics and Data Analysis

Relative Frequency Distribution of Fruit Fly Lives

Class Interval Class Midpoint Frequency, f Relative Frequency

0 − 4

5 − 9

10 − 14

15 − 19

20 − 24

25 − 29

30 − 34

2

7

12

17

22

27

32

2

17

16

10

3

1

1

0.04

0.34

0.32

0.20

0.06

0.02

0.02

(c) A histogram plot is shown next.

2 7 12 17 22 27 32

Fruit fly lives (seconds)

Relative Frequency

(d) ˜X = 10.50.

1.21 (a) ¯X = 1.7743 and ˜X = 1.7700;

(b) s = 0.3905.

1.22 (a) ¯X = 6.7261 and ˜X = 0.0536.

(b) A histogram plot is shown next.

6.62 6.66 6.7 6.74 6.78 6.82

Relative Frequency Histogram for Diameter

(c) The data appear to be skewed to the left.

1.23 (a) A dot plot is shown next.

0 100 200 300 400 500 600 700 800 900 1000

160.15 395.10

(b) ¯X1980 = 395.1 and ¯X1990 = 160.2.

Solutions for Exercises in Chapter 1 7

(c) The sample mean for 1980 is over twice as large as that of 1990. The variability

for 1990 decreased also as seen by looking at the picture in (a). The gap represents

an increase of over 400 ppm. It appears from the data that hydrocarbon emissions

decreased considerably between 1980 and 1990 and that the extreme large emission

(over 500 ppm) were no longer in evidence.

1.24 (a) ¯X = 2.8973 and s = 0.5415.

(b) A histogram plot is shown next.

1.8 2.1 2.4 2.7 3 3.3 3.6 3.9

Salaries

Relative Frequency

(c) Use the double-stem-and-leaf plot, we have the following.

Stem Leaf Frequency

1 (84) 1

2* (05)(10)(14)(37)(44)(45) 6

2 (52)(52)(67)(68)(71)(75)(77)(83)(89)(91)(99) 11

3* (10)(13)(14)(22)(36)(37) 6

3 (51)(54)(57)(71)(79)(85) 6

1.25 (a) ¯X = 33.31;

(b) ˜X = 26.35;

(c) A histogram plot is shown next.

10 20 30 40 50 60 70 80 90

Percentage of the families

Relative Frequency

8 Chapter 1 Introduction to Statistics and Data Analysis

(d) ¯Xtr(10) = 30.97. This trimmed mean is in the middle of the mean and median

using the full amount of data. Due to the skewness of the data to the right (see

plot in (c)), it is common to use trimmed data to have a more robust result.

1.26 If a model using the function of percent of families to predict staff salaries, it is likely

that the model would be wrong due to several extreme values of the data. Actually if

a scatter plot of these two data sets is made, it is easy to see that some outlier would

influence the trend.

1.27 (a) The averages of the wear are plotted here.

700 800 900 1000 1100 1200 1300

250 300 350

load

wear

(b) When the load value increases, the wear value also increases. It does show certain

relationship.

(c) A plot of wears is shown next.

700 800 900 1000 1100 1200 1300

100 300 500 700

load

wear

(d) The relationship between load and wear in (c) is not as strong as the case in (a),

especially for the load at 1300. One reason is that there is an extreme value (750)

which influence the mean value at the load 1300.

1.28 (a) A dot plot is shown next.

71.45 71.65 71.85 72.05 72.25 72.45 72.65 72.85

High Low

In the figure, “×” represents the low-injection-velocity group and “◦” represents

the high-injection-velocity group.

Solutions for Exercises in Chapter 1 9

(b) It appears that shrinkage values for the low-injection-velocity group is higher than

those for the high-injection-velocity group. Also, the variation of the shrinkage

is a little larger for the low injection velocity than that for the high injection

velocity.

1.29 (a) A dot plot is shown next.

76 79 82 85 88 91 94

Low High

In the figure, “×” represents the low-injection-velocity group and “◦” represents

the high-injection-velocity group.

(b) In this time, the shrinkage values are much higher for the high-injection-velocity

group than those for the low-injection-velocity group. Also, the variation for the

former group is much higher as well.

(c) Since the shrinkage effects change in different direction between low mode temperature

and high mold temperature, the apparent interactions between the mold

temperature and injection velocity are significant.

1.30 An interaction plot is shown next.

Low high

injection velocity

low mold temp

high mold temp

mean shrinkage value

It is quite obvious to find the interaction between the two variables. Since in this experimental

data, those two variables can be controlled each at two levels, the interaction

can be investigated. However, if the data are from an observational studies, in which

the variable values cannot be controlled, it would be difficult to study the interactions

among these variables.

Chapter 2

Probability

2.1 (a) S = {8, 16, 24, 32, 40, 48}.

(b) For x2 + 4x − 5 = (x + 5)(x − 1) = 0, the only solutions are x = −5 and x = 1.

S = {−5, 1}.

(c) S = {T,HT,HHT,HHH}.

(d) S = {N. America, S. America, Europe,Asia,Africa,Australia,Antarctica}.

(e) Solving 2x − 4 ≥ 0 gives x ≥ 2. Since we must also have x < 1, it follows that

S = φ.

2.2 S = {(x, y) | x2 + y2 < 9; x ≥ 0, y ≥ 0}.

2.3 (a) A = {1, 3}.

(b) B = {1, 2, 3, 4, 5, 6}.

(c) C = {x | x2 − 4x + 3 = 0} = {x | (x − 1)(x − 3) = 0} = {1, 3}.

(d) D = {0, 1, 2, 3, 4, 5, 6}. Clearly, A = C.

2.4 (a) S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6),

(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6),

(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}.

(b) S = {(x, y) | 1 ≤ x, y ≤ 6}.

2.5 S = {1HH, 1HT, 1TH, 1TT, 2H, 2T, 3HH, 3HT, 3TH, 3TT, 4H, 4T, 5HH, 5HT, 5TH,

5TT, 6H, 6T}.

2.6 S = {A1A2,A1A3,A1A4,A2A3,A2A4,A3A4}.

2.7 S1 = {MMMM,MMMF,MMFM,MFMM, FMMM,MMFF,MFMF,MFFM,

FMFM, FFMM, FMMF,MFFF, FMFF, FFMF, FFFM, FFFF}.

S2 = {0, 1, 2, 3, 4}.

2.8 (a) A = {(3, 6), (4, 5), (4, 6), (5, 4), (5, 5), (5, 6), (6, 3), (6, 4), (6, 5), (6, 6)}.

11

12 Chapter 2 Probability

(b) B = {(1, 2), (2, 2), (3, 2), (4, 2), (5, 2), (6, 2), (2, 1), (2, 3), (2, 4),

(2, 5), (2, 6)}.

(c) C = {(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}.

(d) A ∩ C = {(5, 4), (5, 5), (5, 6), (6, 3), (6, 4), (6, 5), (6, 6)}.

(e) A ∩ B = φ.

(f) B ∩ C = {(5, 2), (6, 2)}.

(g) A Venn diagram is shown next.

A

A C

B

B C

C

S

Ç

Ç

2.9 (a) A = {1HH, 1HT, 1TH, 1TT, 2H, 2T}.

(b) B = {1TT, 3TT, 5TT}.

(c) A′ = {3HH, 3HT, 3TH, 3TT, 4H, 4T, 5HH, 5HT, 5TH, 5TT, 6H, 6T}.

(d) A′ ∩ B = {3TT, 5TT}.

(e) A ∪ B = {1HH, 1HT, 1TH, 1TT, 2H, 2T, 3TT, 5TT}.

2.10 (a) S = {FFF, FFN, FNF,NFF, FNN,NFN,NNF,NNN}.

(b) E = {FFF, FFN, FNF,NFF}.

(c) The second river was safe for fishing.

2.11 (a) S = {M1M2,M1F1,M1F2,M2M1,M2F1,M2F2, F1M1, F1M2, F1F2, F2M1, F2M2,

F2F1}.

(b) A = {M1M2,M1F1,M1F2,M2M1,M2F1,M2F2}.

(c) B = {M1F1,M1F2,M2F1,M2F2, F1M1, F1M2, F2M1, F2M2}.

(d) C = {F1F2, F2F1}.

(e) A ∩ B = {M1F1,M1F2,M2F1,M2F2}.

(f) A ∪ C = {M1M2,M1F1,M1F2,M2M1,M2F1,M2F2, F1F2, F2F1}.

Solutions for Exercises in Chapter 2 13

(g)

A

A B

B

C

S

Ç

2.12 (a) S = {ZY F, ZNF,WY F,WNF, SY F, SNF, ZYM}.

(b) A ∪ B = {ZY F, ZNF,WY F,WNF, SY F, SNF} = A.

(c) A ∩ B = {WY F, SY F}.

2.13 A Venn diagram is shown next.

S

P

F

S

2.14 (a) A ∪ C = {0, 2, 3, 4, 5, 6, 8}.

(b) A ∩ B = φ.

(c) C′ = {0, 1, 6, 7, 8, 9}.

(d) C′ ∩ D = {1, 6, 7}, so (C′ ∩ D) ∪ B = {1, 3, 5, 6, 7, 9}.

(e) (S ∩ C)′ = C′ = {0, 1, 6, 7, 8, 9}.

(f) A ∩ C = {2, 4}, so A ∩ C ∩ D′ = {2, 4}.

2.15 (a) A′ = {nitrogen, potassium, uranium, oxygen}.

(b) A ∪ C = {copper, sodium, zinc, oxygen}.

(c) A ∩ B′ = {copper, zinc} and

C′ = {copper, sodium, nitrogen, potassium, uranium, zinc};

so (A ∩ B′) ∪ C′ = {copper, sodium, nitrogen, potassium, uranium, zinc}.

14 Chapter 2 Probability

(d) B′ ∩ C′ = {copper, uranium, zinc}.

(e) A ∩ B ∩ C = φ.

(f) A′ ∪ B′ = {copper, nitrogen, potassium, uranium, oxygen, zinc} and

A′ ∩ C = {oxygen}; so, (A′ ∪ B′) ∩ (A′ ∩ C) = {oxygen}.

2.16 (a) M ∪ N = {x | 0 < x < 9}.

(b) M ∩ N = {x | 1 < x < 5}.

(c) M′ ∩ N′ = {x | 9 < x < 12}.

2.17 A Venn diagram is shown next.

A B

S

1 2 3 4

(a) From the above Venn diagram, (A ∩ B)′ contains the regions of 1, 2 and 4.

(b) (A ∪ B)′ contains region 1.

(c) A Venn diagram is shown next.

A

B

C

S

1

2

3

4

5

6

7

8

(A ∩ C) ∪ B contains the regions of 3, 4, 5, 7 and 8.

2.18 (a) Not mutually exclusive.

(b) Mutually exclusive.

(c) Not mutually exclusive.

(d) Mutually exclusive.

2.19 (a) The family will experience mechanical problems but will receive no ticket for

traffic violation and will not arrive at a campsite that has no vacancies.

(b) The family will receive a traffic ticket and arrive at a campsite that has no vacancies

but will not experience mechanical problems.

Solutions for Exercises in Chapter 2 15

(c) The family will experience mechanical problems and will arrive at a campsite that

has no vacancies.

(d) The family will receive a traffic ticket but will not arrive at a campsite that has

no vacancies.

(e) The family will not experience mechanical problems.

2.20 (a) 6;

(b) 2;

(c) 2, 5, 6;

(d) 4, 5, 6, 8.

2.21 With n1 = 6 sightseeing tours each available on n2 = 3 different days, the multiplication

rule gives n1n2 = (6)(3) = 18 ways for a person to arrange a tour.

2.22 With n1 = 8 blood types and n2 = 3 classifications of blood pressure, the multiplication

rule gives n1n2 = (8)(3) = 24 classifications.

2.23 Since the die can land in n1 = 6 ways and a letter can be selected in n2 = 26 ways, the

multiplication rule gives n1n2 = (6)(26) = 156 points in S.

2.24 Since a student may be classified according to n1 = 4 class standing and n2 = 2 gender

classifications, the multiplication rule gives n1n2 = (4)(2) = 8 possible classifications

for the students.

2.25 With n1 = 5 different shoe styles in n2 = 4 different colors, the multiplication rule

gives n1n2 = (5)(4) = 20 different pairs of shoes.

2.26 Using Theorem 2.8, we obtain the followings.

(a) There are

7

5

= 21 ways.

(b) There are

5

3

= 10 ways.

2.27 Using the generalized multiplication rule, there are n1×n2×n3×n4 = (4)(3)(2)(2) = 48

different house plans available.

2.28 With n1 = 5 different manufacturers, n2 = 3 different preparations, and n3 = 2

different strengths, the generalized multiplication rule yields n1n2n3 = (5)(3)(2) = 30

different ways to prescribe a drug for asthma.

2.29 With n1 = 3 race cars, n2 = 5 brands of gasoline, n3 = 7 test sites, and n4 = 2 drivers,

the generalized multiplication rule yields (3)(5)(7)(2) = 210 test runs.

2.30 With n1 = 2 choices for the first question, n2 = 2 choices for the second question,

and so forth, the generalized multiplication rule yields n1n2 · · ·n9 = 29 = 512 ways to

answer the test.

16 Chapter 2 Probability

2.31 (a) With n1 = 4 possible answers for the first question, n2 = 4 possible answers

for the second question, and so forth, the generalized multiplication rule yields

45 = 1024 ways to answer the test.

(b) With n1 = 3 wrong answers for the first question, n2 = 3 wrong answers for the

second question, and so forth, the generalized multiplication rule yields

n1n2n3n4n5 = (3)(3)(3)(3)(3) = 35 = 243

ways to answer the test and get all questions wrong.

2.32 (a) By Theorem 2.3, 7! = 5040.

(b) Since the first letter must be m, the remaining 6 letters can be arranged in 6! = 720

ways.

2.33 Since the first digit is a 5, there are n1 = 9 possibilities for the second digit and then

n2 = 8 possibilities for the third digit. Therefore, by the multiplication rule there are

n1n2 = (9)(8) = 72 registrations to be checked.

2.34 (a) By Theorem 2.3, there are 6! = 720 ways.

(b) A certain 3 persons can follow each other in a line of 6 people in a specified order is

4 ways or in (4)(3!) = 24 ways with regard to order. The other 3 persons can then

be placed in line in 3! = 6 ways. By Theorem 2.1, there are total (24)(6) = 144

ways to line up 6 people with a certain 3 following each other.

(c) Similar as in (b), the number of ways that a specified 2 persons can follow each

other in a line of 6 people is (5)(2!)(4!) = 240 ways. Therefore, there are 720 − 240 = 480 ways if a certain 2 persons refuse to follow each other.

2.35 The first house can be placed on any of the n1 = 9 lots, the second house on any of the

remaining n2 = 8 lots, and so forth. Therefore, there are 9! = 362, 880 ways to place

the 9 homes on the 9 lots.

2.36 (a) Any of the 6 nonzero digits can be chosen for the hundreds position, and of the

remaining 6 digits for the tens position, leaving 5 digits for the units position. So,

there are (6)(5)(5) = 150 three digit numbers.

(b) The units position can be filled using any of the 3 odd digits. Any of the remaining

5 nonzero digits can be chosen for the hundreds position, leaving a choice of 5

digits for the tens position. By Theorem 2.2, there are (3)(5)(5) = 75 three digit

odd numbers.

(c) If a 4, 5, or 6 is used in the hundreds position there remain 6 and 5 choices,

respectively, for the tens and units positions. This gives (3)(6)(5) = 90 three

digit numbers beginning with a 4, 5, or 6. If a 3 is used in the hundreds position,

then a 4, 5, or 6 must be used in the tens position leaving 5 choices for the units

position. In this case, there are (1)(3)(5) = 15 three digit number begin with

a 3. So, the total number of three digit numbers that are greater than 330 is

90 + 15 = 105.

Solutions for Exercises in Chapter 2 17

2.37 The first seat must be filled by any of 5 girls and the second seat by any of 4 boys.

Continuing in this manner, the total number of ways to seat the 5 girls and 4 boys is

(5)(4)(4)(3)(3)(2)(2)(1)(1) = 2880.

2.38 (a) 8! = 40320.

(b) There are 4! ways to seat 4 couples and then each member of a couple can be

interchanged resulting in 24(4!) = 384 ways.

(c) By Theorem 2.3, the members of each gender can be seated in 4! ways. Then

using Theorem 2.1, both men and women can be seated in (4!)(4!) = 576 ways.

2.39 (a) Any of the n1 = 8 finalists may come in first, and of the n2 = 7 remaining finalists

can then come in second, and so forth. By Theorem 2.3, there 8! = 40320 possible

orders in which 8 finalists may finish the spelling bee.

(b) The possible orders for the first three positions are 8P3 = 8!

5! = 336.

2.40 By Theorem 2.4, 8P5 = 8!

3! = 6720.

2.41 By Theorem 2.4, 6P4 = 6!

2! = 360.

2.42 By Theorem 2.4, 40P3 = 40!

37! = 59, 280.

2.43 By Theorem 2.5, there are 4! = 24 ways.

2.44 By Theorem 2.5, there are 7! = 5040 arrangements.

2.45 By Theorem 2.6, there are 8!

3!2! = 3360.

2.46 By Theorem 2.6, there are 9!

3!4!2! = 1260 ways.

2.47 By Theorem 2.7, there are

12

7,3,2

= 7920 ways.

2.48

9

1,4,4

+

9

2,4,3

+

9

1,3,5

+

9

2,3,4

+

9

2,2,5

= 4410.

2.49 By Theorem 2.8, there are

8

3

= 56 ways.

2.50 Assume February 29th as March 1st for the leap year. There are total 365 days in a

year. The number of ways that all these 60 students will have different birth dates (i.e,

arranging 60 from 365) is 365P60. This is a very large number.

2.51 (a) Sum of the probabilities exceeds 1.

(b) Sum of the probabilities is less than 1.

(c) A negative probability.

(d) Probability of both a heart and a black card is zero.

2.52 Assuming equal weights

18 Chapter 2 Probability

(a) P(A) = 5

18 ;

(b) P(C) = 1

3 ;

(c) P(A ∩ C) = 7

36 .

2.53 S = {$10, $25, $100} with weights 275/500 = 11/20, 150/500 = 3/10, and 75/500 =

3/20, respectively. The probability that the first envelope purchased contains less than

$100 is equal to 11/20 + 3/10 = 17/20.

2.54 (a) P(S ∩ D′) = 88/500 = 22/125.

(b) P(E ∩ D ∩ S′) = 31/500.

(c) P(S′ ∩ E′) = 171/500.

2.55 Consider the events

S: industry will locate in Shanghai,

B: industry will locate in Beijing.

(a) P(S ∩ B) = P(S) + P(B) − P(S ∪ B) = 0.7 + 0.4 − 0.8 = 0.3.

(b) P(S′ ∩ B′) = 1 − P(S ∪ B) = 1 − 0.8 = 0.2.

2.56 Consider the events

B: customer invests in tax-free bonds,

M: customer invests in mutual funds.

(a) P(B ∪M) = P(B) + P(M) − P(B ∩M) = 0.6 + 0.3 − 0.15 = 0.75.

(b) P(B′ ∩M′) = 1 − P(B ∪M) = 1 − 0.75 = 0.25.

2.57 (a) Since 5 of the 26 letters are vowels, we get a probability of 5/26.

(b) Since 9 of the 26 letters precede j, we get a probability of 9/26.

(c) Since 19 of the 26 letters follow g, we get a probability of 19/26.

2.58 (a) Let A = Defect in brake system; B = Defect in fuel system; P(A ∪ B) = P(A) +

P(B) − P(A ∩ B) = 0.25 + 0.17 − 0.15 = 0.27.

(b) P(No defect) = 1 − P(A ∪ B) = 1 − 0.27 = 0.73.

2.59 By Theorem 2.2, there are N = (26)(25)(24)(9)(8)(7)(6) = 47, 174, 400 possible ways

to code the items of which n = (5)(25)(24)(8)(7)(6)(4) = 4, 032, 000 begin with a vowel

and end with an even digit. Therefore, n

N = 10

117 .

2.60 (a) Of the (6)(6) = 36 elements in the sample space, only 5 elements (2,6), (3,5),

(4,4), (5,3), and (6,2) add to 8. Hence the probability of obtaining a total of 8 is

then 5/36.

(b) Ten of the 36 elements total at most 5. Hence the probability of obtaining a total

of at most is 10/36=5/18.

Solutions for Exercises in Chapter 2 19

2.61 Since there are 20 cards greater than 2 and less than 8, the probability of selecting two

of these in succession is

20

52

19

51

=

95

663

.

2.62 (a)

(1

1)(8

2)

(9

3)

= 1

3 .

(b)

(5

2)(3

1)

(9

3)

= 5

14 .

2.63 (a)

(4

3)(48

2 )

(52

5 )

= 94

54145 .

(b)

(13

4 )(13

1 )

(52

5 )

= 143

39984 .

2.64 Any four of a kind, say four 2’s and one 5 occur in

5

1

= 5 ways each with probability

(1/6)(1/6)(1/6)(1/6)(1/6) = (1/6)5. Since there are 6P2 = 30 ways to choose various

pairs of numbers to constitute four of one kind and one of the other (we use permutation

instead of combination is because that four 2’s and one 5, and four 5’s and one 2 are

two different ways), the probability is (5)(30)(1/6)5 = 25/1296.

2.65 (a) P(M ∪ H) = 88/100 = 22/25;

(b) P(M′ ∩ H′) = 12/100 = 3/25;

(c) P(H ∩M′) = 34/100 = 17/50.

2.66 (a) 9;

(b) 1/9.

2.67 (a) 0.32;

(b) 0.68;

(c) office or den.

2.68 (a) 1 − 0.42 = 0.58;

(b) 1 − 0.04 = 0.96.

2.69 P(A) = 0.2 and P(B) = 0.35

(a) P(A′) = 1 − 0.2 = 0.8;

(b) P(A′ ∩ B′) = 1 − P(A ∪ B) = 1 − 0.2 − 0.35 = 0.45;

(c) P(A ∪ B) = 0.2 + 0.35 = 0.55.

2.70 (a) 0.02 + 0.30 = 0.32 = 32%;

(b) 0.32 + 0.25 + 0.30 = 0.87 = 87%;

20 Chapter 2 Probability

(c) 0.05 + 0.06 + 0.02 = 0.13 = 13%;

(d) 1 − 0.05 − 0.32 = 0.63 = 63%.

2.71 (a) 0.12 + 0.19 = 0.31;

(b) 1 − 0.07 = 0.93;

(c) 0.12 + 0.19 = 0.31.

2.72 (a) 1 − 0.40 = 0.60.

(b) The probability that all six purchasing the electric oven or all six purchasing the

gas oven is 0.007 + 0.104 = 0.111. So the probability that at least one of each

type is purchased is 1 − 0.111 = 0.889.

2.73 (a) P(C) = 1 − P(A) − P(B) = 1 − 0.990 − 0.001 = 0.009;

(b) P(B′) = 1 − P(B) = 1 − 0.001 = 0.999;

(c) P(B) + P(C) = 0.01.

2.74 (a) ($4.50 − $4.00) × 50, 000 = $25, 000;

(b) Since the probability of underfilling is 0.001, we would expect 50, 000×0.001 = 50

boxes to be underfilled. So, instead of having ($4.50 − $4.00) × 50 = $25 profit

for those 50 boxes, there are a loss of $4.00 × 50 = $200 due to the cost. So, the

loss in profit expected due to underfilling is $25 + $200 = $250.

2.75 (a) 1 − 0.95 − 0.002 = 0.048;

(b) ($25.00 − $20.00) × 10, 000 = $50, 000;

(c) (0.05)(10, 000) × $5.00 + (0.05)(10, 000) × $20 = $12, 500.

2.76 P(A′∩B′) = 1−P(A∪B) = 1−(P(A)+P(B)−P(A∩B) = 1+P(A∩B)−P(A)−P(B).

2.77 (a) The probability that a convict who pushed dope, also committed armed robbery.

(b) The probability that a convict who committed armed robbery, did not push dope.

(c) The probability that a convict who did not push dope also did not commit armed

robbery.

2.78 P(S | A) = 10/18 = 5/9.

2.79 Consider the events:

M: a person is a male;

S: a person has a secondary education;

C: a person has a college degree.

(a) P(M | S) = 28/78 = 14/39;

(b) P(C′ | M′) = 95/112.

Solutions for Exercises in Chapter 2 21

2.80 Consider the events:

A: a person is experiencing hypertension,

B: a person is a heavy smoker,

C: a person is a nonsmoker.

(a) P(A | B) = 30/49;

(b) P(C | A′) = 48/93 = 16/31.

2.81 (a) P(M ∩ P ∩ H) = 10

68 = 5

34 ;

(b) P(H ∩M | P′) = P(H∩M∩P′)

P(P′) = 22−10

100−68 = 12

32 = 3

8 .

2.82 (a) (0.90)(0.08) = 0.072;

(b) (0.90)(0.92)(0.12) = 0.099.

2.83 (a) 0.018;

(b) 0.22 + 0.002 + 0.160 + 0.102 + 0.046 + 0.084 = 0.614;

(c) 0.102/0.614 = 0.166;

(d) 0.102+0.046

0.175+0.134 = 0.479.

2.84 Consider the events:

C: an oil change is needed,

F: an oil filter is needed.

(a) P(F | C) = P(F∩C)

P(C) = 0.14

0.25 = 0.56.

(b) P(C | F) = P(C∩F)

P(F) = 0.14

0.40 = 0.35.

2.85 Consider the events:

H: husband watches a certain show,

W: wife watches the same show.

(a) P(W ∩ H) = P(W)P(H | W) = (0.5)(0.7) = 0.35.

(b) P(W | H) = P(W∩H)

P(H) = 0.35

0.4 = 0.875.

(c) P(W ∪ H) = P(W) + P(H) − P(W ∩ H) = 0.5 + 0.4 − 0.35 = 0.55.

2.86 Consider the events:

H: the husband will vote on the bond referendum,

W: the wife will vote on the bond referendum.

Then P(H) = 0.21, P(W) = 0.28, and P(H ∩W) = 0.15.

(a) P(H ∪W) = P(H) + P(W) − P(H ∩W) = 0.21 + 0.28 − 0.15 = 0.34.

(b) P(W | H) = P(H∩W)

P(H) = 0.15

0.21 = 5

7 .

(c) P(H | W′) = P(H∩W′)

P(W′) = 0.06

0.72 = 1

12 .

22 Chapter 2 Probability

2.87 Consider the events:

A: the vehicle is a camper,

B: the vehicle has Canadian license plates.

(a) P(B | A) = P(A∩B)

P(A) = 0.09

0.28 = 9

28 .

(b) P(A | B) = P(A∩B)

P(B) = 0.09

0.12 = 3

4 .

(c) P(B′ ∪ A′) = 1 − P(A ∩ B) = 1 − 0.09 = 0.91.

2.88 Define

H: head of household is home,

C: a change is made in long distance carriers.

P(H ∩ C) = P(H)P(C | H) = (0.4)(0.3) = 0.12.

2.89 Consider the events:

A: the doctor makes a correct diagnosis,

B: the patient sues.

P(A′ ∩ B) = P(A′)P(B | A′) = (0.3)(0.9) = 0.27.

2.90 (a) 0.43;

(b) (0.53)(0.22) = 0.12;

(c) 1 − (0.47)(0.22) = 0.90.

2.91 Consider the events:

A: the house is open,

B: the correct key is selected.

P(A) = 0.4, P(A′) = 0.6, and P(B) =

(1

1)(7

2)

(8

3)

= 3

8 = 0.375.

So, P[A ∪ (A′ ∩ B)] = P(A) + P(A′)P(B) = 0.4 + (0.6)(0.375) = 0.625.

2.92 Consider the events:

F: failed the test,

P: passed the test.

(a) P(failed at least one tests) = 1 − P(P1P2P3P4) = 1 − (0.99)(0.97)(0.98)(0.99) =

1 − 0.93 = 0.07,

(b) P(failed 2 or 3) = P(P1)P(P4)(1 − P(P2P3)) = (0.99)(0.99)(1 − (0.97)(0.98)) =

0.0484.

(c) 100 × 0.07 = 7.

(d) 0.25.

2.93 Let A and B represent the availability of each fire engine.

(a) P(A′ ∩ B′) = P(A′)P(B′) = (0.04)(0.04) = 0.0016.

(b) P(A ∪ B) = 1 − P(A′ ∩ B′) = 1 − 0.0016 = 0.9984.

Solutions for Exercises in Chapter 2 23

2.94 P(T′ ∩ N′) = P(T′)P(N′) = (1 − P(T))(1 − P(N)) = (0.3)(0.1) = 0.03.

2.95 Consider the events:

A1: aspirin tablets are selected from the overnight case,

A2: aspirin tablets are selected from the tote bag,

L2: laxative tablets are selected from the tote bag,

T1: thyroid tablets are selected from the overnight case,

T2: thyroid tablets are selected from the tote bag.

(a) P(T1 ∩ T2) = P(T1)P(T2) = (3/5)(2/6) = 1/5.

(b) P(T′

1 ∩ T′

2) = P(T′

1)P(T′

2) = (2/5)(4/6) = 4/15.

(c) 1−P(A1 ∩A2)−P(T1 ∩T2) = 1−P(A1)P(A2)−P(T1)P(T2) = 1−(2/5)(3/6)−

(3/5)(2/6) = 3/5.

2.96 Consider the events:

X: a person has an X-ray,

C: a cavity is filled,

T: a tooth is extracted.

P(X ∩ C ∩ T) = P(X)P(C | X)P(T | X ∩ C) = (0.6)(0.3)(0.1) = 0.018.

2.97 (a) P(Q1 ∩Q2 ∩Q3 ∩Q4) = P(Q1)P(Q2 | Q1)P(Q3 | Q1 ∩Q2)P(Q4 | Q1 ∩Q2 ∩Q3) =

(15/20)(14/19)(13/18)(12/17) = 91/323.

(b) Let A be the event that 4 good quarts of milk are selected. Then

P(A) =

15

4

20

4

=

91

323

.

2.98 P = (0.95)[1 − (1 − 0.7)(1 − 0.8)](0.9) = 0.8037.

2.99 This is a parallel system of two series subsystems.

(a) P = 1 − [1 − (0.7)(0.7)][1 − (0.8)(0.8)(0.8)] = 0.75112.

(b) P = P(A′∩C∩D∩E)

Psystem works = (0.3)(0.8)(0.8)(0.8)

0.75112 = 0.2045.

2.100 Define S: the system works.

P(A′ | S′) = P(A′∩S′)

P(S′) = P(A′)(1−P(C∩D∩E))

1−P(S) = (0.3)[1−(0.8)(0.8)(0.8)]

1−0.75112 = 0.588.

2.101 Consider the events:

C: an adult selected has cancer,

D: the adult is diagnosed as having cancer.

P(C) = 0.05, P(D | C) = 0.78, P(C′) = 0.95 and P(D | C′) = 0.06. So, P(D) =

P(C ∩ D) + P(C′ ∩ D) = (0.05)(0.78) + (0.95)(0.06) = 0.096.

24 Chapter 2 Probability

2.102 Let S1, S2, S3, and S4 represent the events that a person is speeding as he passes through

the respective locations and let R represent the event that the radar traps is operating

resulting in a speeding ticket. Then the probability that he receives a speeding ticket:

P(R) =

P4

i=1

P(R | Si)P(Si) = (0.4)(0.2) + (0.3)(0.1) + (0.2)(0.5) + (0.3)(0.2) = 0.27.

2.103 P(C | D) = P(C∩D)

P(D) = 0.039

0.096 = 0.40625.

2.104 P(S2 | R) = P(R∩ S2)

P(R) = 0.03

0.27 = 1/9.

2.105 Consider the events:

A: no expiration date,

B1: John is the inspector, P(B1) = 0.20 and P(A | B1) = 0.005,

B2: Tom is the inspector, P(B2) = 0.60 and P(A | B2) = 0.010,

B3: Jeff is the inspector, P(B3) = 0.15 and P(A | B3) = 0.011,

B4: Pat is the inspector, P(B4) = 0.05 and P(A | B4) = 0.005,

P(B1 | A) = (0.005)(0.20)

(0.005)(0.20)+(0.010)(0.60)+(0.011)(0.15)+(0.005)(0.05) = 0.1124.

2.106 Consider the events

E: a malfunction by other human errors,

A: station A, B: station B, and C: station C.

P(C | E) = P(E | C)P(C)

P(E | A)P(A)+P(E | B)P(B)+P(E | C)P(C) = (5/10)(10/43)

(7/18)(18/43)+(7/15)(15/43)+(5/10)(10/43) =

0.1163

0.4419 = 0.2632.

2.107 (a) P(A ∩ B ∩ C) = P(C | A ∩ B)P(B | A)P(A) = (0.20)(0.75)(0.3) = 0.045.

(b) P(B′ ∩ C) = P(A ∩ B′ ∩ C) + P(A′ ∩ B′ ∩ C) = P(C | A ∩ B′)P(B′ | A)P(A) +

P(C | A′∩B′)P(B′ | A′)P(A′) = (0.80)(1−0.75)(0.3)+(0.90)(1−0.20)(1−0.3) =

0.564.

(c) Use similar argument as in (a) and (b), P(C) = P(A∩B ∩C)+P(A∩B′ ∩ C)+

P(A′ ∩ B ∩ C) + P(A′ ∩ B′ ∩ C) = 0.045 + 0.060 + 0.021 + 0.504 = 0.630.

(d) P(A | B′ ∩ C) = P(A ∩ B′ ∩ C)/P(B′ ∩ C) = (0.06)(0.564) = 0.1064.

2.108 Consider the events:

A: a customer purchases latex paint,

A′: a customer purchases semigloss paint,

B: a customer purchases rollers.

P(A | B) = P(B | A)P(A)

P(B | A)P(A)+P(B | A′)P(A′) = (0.60)(0.75)

(0.60)(0.75)+(0.25)(0.30) = 0.857.

2.109 Consider the events:

G: guilty of committing a crime,

I: innocent of the crime,

i: judged innocent of the crime,

g: judged guilty of the crime.

P(I | g) = P(g | I)P(I)

P(g | G)P(G)+P(g | I)P(I) = (0.01)(0.95)

(0.05)(0.90)+(0.01)(0.95) = 0.1743.

Solutions for Exercises in Chapter 2 25

2.110 Let Ai be the event that the ith patient is allergic to some type of week.

(a) P(A1 ∩ A2 ∩ A3 ∩ A′

4) + P(A1 ∩ A2 ∩ A′

3 ∩ A4) + P(A1 ∩ A′

2 ∩ A3 ∩ A4) +

P(A′

1 ∩ A2 ∩ A3 ∩ A4) = P(A1)P(A2)P(A3)P(A′

4) + P(A1)P(A2)P(A′

3)P(A4) +

P(A1)P(A′

2)P(A3)P(A4) + P(A′

1)P(A2)P(A3)P(A4) = (4)(1/2)4 = 1/4.

(b) P(A′

1 ∩ A′

2 ∩ A′

3 ∩ A′

4) = P(A′

1)P(A′

2)P(A′

3)P(A′

4) = (1/2)4 = 1/16.

2.111 No solution necessary.

2.112 (a) 0.28 + 0.10 + 0.17 = 0.55.

(b) 1 − 0.17 = 0.83.

(c) 0.10 + 0.17 = 0.27.

2.113 P =

(13

4 )(13

6 )(13

1 )(13

2 )

(52

13)

.

2.114 (a) P(M1 ∩M2 ∩M3 ∩M4) = (0.1)4 = 0.0001, where Mi represents that ith person

make a mistake.

(b) P(J ∩ C ∩ R′ ∩W′) = (0.1)(0.1)(0.9)(0.9) = 0.0081.

2.115 Let R, S, and L represent the events that a client is assigned a room at the Ramada

Inn, Sheraton, and Lakeview Motor Lodge, respectively, and let F represents the event

that the plumbing is faulty.

(a) P(F) = P(F | R)P(R) + P(F | S)P(S) + P(F | L)P(L) = (0.05)(0.2) +

(0.04)(0.4) + (0.08)(0.3) = 0.054.

(b) P(L | F) = (0.08)(0.3)

0.054 = 4

9 .

2.116 (a) There are

9

3

= 84 possible committees.

(b) There are

4

1

5

2

= 40 possible committees.

(c) There are

3

1

1

1

5

1

= 15 possible committees.

2.117 Denote by R the event that a patient survives. Then P(R) = 0.8.

(a) P(R1 ∩ R2 ∩ R′

3) + P(R1 ∩ R′

2 ∩ R3)P(R′

1 ∩ R2 ∩ R3) = P(R1)P(R2)P(R′

3) +

P(R1)P(R′

2)P(R3) + P(R′

1)P(R2)P(R3) = (3)(0.8)(0.8)(0.2) = 0.384.

(b) P(R1 ∩ R2 ∩ R3) = P(R1)P(R2)P(R3) = (0.8)3 = 0.512.

2.118 Consider events

M: an inmate is a male,

N: an inmate is under 25 years of age.

P(M′ ∩ N′) = P(M′) + P(N′) − P(M′ ∪ N′) = 2/5 + 1/3 − 5/8 = 13/120.

2.119 There are

4

3

5

3

6

3

= 800 possible selections.

26 Chapter 2 Probability

2.120 Consider the events:

Bi: a black ball is drawn on the ith drawl,

Gi: a green ball is drawn on the ith drawl.

(a) P(B1 ∩B2 ∩B3)+P(G1 ∩G2 ∩G3) = (6/10)(6/10)(6/10)+(4/10)(4/10)(4/10) =

7/25.

(b) The probability that each color is represented is 1 − 7/25 = 18/25.

2.121 The total number of ways to receive 2 or 3 defective sets among 5 that are purchased

is

3

2

9

3

+

3

3

9

2

= 288.

2.122 A Venn diagram is shown next.

A

B

C

S

1

2

3

4

5

6

7

8

(a) (A ∩ B)′: 1, 2, 3, 6, 7, 8.

(b) (A ∪ B)′: 1, 6.

(c) (A ∩ C) ∪ B: 3, 4, 5, 7, 8.

2.123 Consider the events:

O: overrun,

A: consulting firm A,

B: consulting firm B,

C: consulting firm C.

(a) P(C | O) = P(O | C)P(C)

P(O | A)P(A)+P(O | B)P(B)+P(O | C)P(C) = (0.15)(0.25)

(0.05)(0.40)+(0.03)(0.35)+(0.15)(0.25) =

0.0375

0.0680 = 0.5515.

(b) P(A | O) = (0.05)(0.40)

0.0680 = 0.2941.

2.124 (a) 36;

(b) 12;

(c) order is not important.

2.125 (a) 1

(36

2 )

= 0.0016;

(b)

(12

1 )(24

1 )

(36

2 )

= 288

630 = 0.4571.

Solutions for Exercises in Chapter 2 27

2.126 Consider the events:

C: a woman over 60 has the cancer,

P: the test gives a positive result.

So, P(C) = 0.07, P(P′ | C) = 0.1 and P(P | C′) = 0.05.

P(C | P′) = P(P′ | C)P(C)

P(P′ | C)P(C)+P(P′ | C′)P(C′) = (0.1)(0.07)

(0.1)(0.07)+(1−0.05)(1−0.07) = 0.007

0.8905 = 0.00786.

2.127 Consider the events:

A: two nondefective components are selected,

N: a lot does not contain defective components, P(N) = 0.6, P(A | N) = 1,

O: a lot contains one defective component, P(O) = 0.3, P(A | O) =

(19

2 )

(20

2 )

= 9

10 ,

T: a lot contains two defective components,P(T) = 0.1, P(A | T) =

(18

2 )

(20

2 )

= 153

190 .

(a) P(N | A) = P(A | N)P(N)

P(A | N)P(N)+P(A | O)P(O)+P(A | T)P(T) = (1)(0.6)

(1)(0.6)+(9/10)(0.3)+(153/190)(0.1)

= 0.6

0.9505 = 0.6312;

(b) P(O | A) = (9/10)(0.3)

0.9505 = 0.2841;

(c) P(T | A) = 1 − 0.6312 − 0.2841 = 0.0847.

2.128 Consider events:

D: a person has the rare disease, P(D) = 1/500,

P: the test shows a positive result, P(P | D) = 0.95 and P(P | D′) = 0.01.

P(D | P) = P(P | D)P(D)

P(P | D)P(D)+P(P | D′)P(D′) = (0.95)(1/500)

(0.95)(1/500)+(0.01)(1−1/500) = 0.1599.

2.129 Consider the events:

1: engineer 1, P(1) = 0.7, and 2: engineer 2, P(2) = 0.3,

E: an error has occurred in estimating cost, P(E | 1) = 0.02 and P(E | 2) = 0.04.

P(1 | E) = P(E | 1)P(1)

P(E | 1)P(1)+P(E | 2)P(2) = (0.02)(0.7)

(0.02)(0.7)+(0.04)(0.3) = 0.5385, and

P(2 | E) = 1 − 0.5385 = 0.4615. So, more likely engineer 1 did the job.

2.130 Consider the events: D: an item is defective

(a) P(D1D2D3) = P(D1)P(D2)P(D3) = (0.2)3 = 0.008.

(b) P(three out of four are defectives) =

4

3

(0.2)3(1 − 0.2) = 0.0256.

2.131 Let A be the event that an injured worker is admitted to the hospital and N be the event

that an injured worker is back to work the next day. P(A) = 0.10, P(N) = 0.15 and

P(A∩N) = 0.02. So, P(A∪N) = P(A)+P(N)−P(A∩N) = 0.1+0.15−0.02 = 0.23.

2.132 Consider the events:

T: an operator is trained, P(T) = 0.5,

M an operator meets quota, P(M | T) = 0.9 and P(M | T′) = 0.65.

P(T | M) = P(M | T)P(T)

P(M | T)P(T)+P(M | T′)P(T′) = (0.9)(0.5)

(0.9)(0.5)+(0.65)(0.5) = 0.5807.

28 Chapter 2 Probability

2.133 Consider the events:

A: purchased from vendor A,

D: a customer is dissatisfied.

Then P(A) = 0.2, P(A | D) = 0.5, and P(D) = 0.1.

So, P(D | A) = P(A | D)P(D)

P(A) = (0.5)(0.1)

0.2 = 0.25.

2.134 (a) P(Union member | New company (same field)) = 13

13+10 = 13

23 = 0.5652.

(b) P(Unemployed | Union member) = 2

40+13+4+2 = 2

59 = 0.034.

2.135 Consider the events:

C: the queen is a carrier, P(C) = 0.5,

D: a prince has the disease, P(D | C) = 0.5.

P(C | D′

1D′

2D′

3) = P(D

′

1D

′

2D

′

3 | C)P(C)

P(D

′

1D

′

2D

′

3 | C)P(C)+P(D

′

1D

′

2D

′

3 | C′)P(C′)

= (0.5)3(0.5)

(0.5)3(0.5)+1(0.5) = 1

9 .

2.136 Using the solutions to Exercise 2.50, we know that there are total 365P60 ways that no

two students have the same birth date. Since the total number of ways of the birth

dates that 60 students can have is 36560, the probability that at least two students

will have the same birth date in a class of 60 is P = 1 − 365P60

36560 . To compute this

number, regular calculator may not be able to handle it. Using approximation (such

as Stirling’s approximation formula), we obtain P = 0.9941, which is quite high.

Chapter 3

Random Variables and Probability

Distributions

3.1 Discrete; continuous; continuous; discrete; discrete; continuous.

3.2 A table of sample space and assigned values of the random variable is shown next.

Sample Space x

NNN

NNB

NBN

BNN

NBB

BNB

BBN

BBB

0

1

1

1

2

2

2

3

3.3 A table of sample space and assigned values of the random variable is shown next.

Sample Space w

HHH

HHT

HTH

THH

HTT

THT

TTH

TTT

3

1

1

1

−1

−1

−1

−3

3.4 S = {HHH, THHH,HTHHH, TTHHH, TTTHHH,HTTHHH, THTHHH,

HHTHHH, . . . }; The sample space is discrete containing as many elements as there

are positive integers.

29

30 Chapter 3 Random Variables and Probability Distributions

3.5 (a) c = 1/30 since 1 =

P3

x=0

c(x2 + 4) = 30c.

(b) c = 1/10 since

1 =

X2

x=0

c

2

x

3

3 − x

= c

2

0

3

3

+

2

1

3

2

+

2

2

3

1

= 10c.

3.6 (a) P(X > 200) =

R

∞

200

20000

(x+100)3 dx = − 10000

(x+100)2

∞

200

= 1

9 .

(b) P(80 < X < 200) =

R 120

80

20000

(x+100)3 dx = − 10000

(x+100)2

120

80

= 1000

9801 = 0.1020.

3.7 (a) P(X < 1.2) =

R 1

0 x dx +

R 1.2

1 (2 − x) dx = x2

2

1

0

+

2x − x2

2

1.2

1

= 0.68.

(b) P(0.5 < X < 1) =

R 1

0.5 x dx = x2

2

1

0.5

= 0.375.

3.8 Referring to the sample space in Exercise 3.3 and making use of the fact that P(H) =

2/3 and P(T) = 1/3, we have

P(W = −3) = P(TTT) = (1/3)3 = 1/27;

P(W = −1) = P(HTT) + P(THT) + P(TTH) = 3(2/3)(1/3)2 = 2/9;

P(W = 1) = P(HHT) + P(HTH) + P(THH) = 3(2/3)2(1/3) = 2/9;

P(W = 3) = P(HHH) = (2/3)3 = 8/27;

The probability distribution for W is then

w −3 −1 1 3

P(W = w) 1/27 2/9 2/9 8/27

3.9 (a) P(0 < X < 1) =

R 1

0

2(x+2)

5 dx = (x+2)2

5

1

0

= 1.

(b) P(1/4 < X < 1/2) =

R 1/2

1/4

2(x+2)

5 dx = (x+2)2

5

1/2

1/4

= 19/80.

3.10 The die can land in 6 different ways each with probability 1/6. Therefore, f(x) = 1

6 ,

for x = 1, 2, . . . , 6.

3.11 We can select x defective sets from 2, and 3 − x good sets from 5 in

2

x

5

3−x

ways. A

random selection of 3 from 7 sets can be made in

7

3

ways. Therefore,

f(x) =

2

x

5

3−x

7

3

, x = 0, 1, 2.

In tabular form

x 0 1 2

f(x) 2/7 4/7 1/7

Solutions for Exercises in Chapter 3 31

The following is a probability histogram:

1 2 3

x

f(x)

1/7

2/7

3/7

4/7

3.12 (a) P(T = 5) = F(5) − F(4) = 3/4 − 1/2 = 1/4.

(b) P(T > 3) = 1 − F(3) = 1 − 1/2 = 1/2.

(c) P(1.4 < T < 6) = F(6) − F(1.4) = 3/4 − 1/4 = 1/2.

3.13 The c.d.f. of X is

F(x) =

0, for x < 0,

0.41, for 0 ≤ x < 1,

0.78, for 1 ≤ x < 2,

0.94, for 2 ≤ x < 3,

0.99, for 3 ≤ x < 4,

1, for x ≥ 4.

3.14 (a) P(X < 0.2) = F(0.2) = 1 − e−1.6 = 0.7981;

(b) f(x) = F′(x) = 8e−8x. Therefore, P(X < 0.2) = 8

R 0.2

0 e−8x dx = −e−8x|0.2

0 =

0.7981.

3.15 The c.d.f. of X is

F(x) =

0, for x < 0,

2/7, for 0 ≤ x < 1,

6/7, for 1 ≤ x < 2,

1, for x ≥ 2.

(a) P(X = 1) = P(X ≤ 1) − P(X ≤ 0) = 6/7 − 2/7 = 4/7;

(b) P(0 < X ≤ 2) = P(X ≤ 2) − P(X ≤ 0) = 1 − 2/7 = 5/7.

32 Chapter 3 Random Variables and Probability Distributions

3.16 A graph of the c.d.f. is shown next.

F(x)

x

1/7

2/7

3/7

4/7

5/7

6/7

1

0 1 2

3.17 (a) Area =

R 3

1 (1/2) dx = x

2

3

1 = 1.

(b) P(2 < X < 2.5)

R 2.5

2 (1/2) dx = x

2

2.5

2 = 1

4 .

(c) P(X ≤ 1.6) =

R 1.6

1 (1/2) dx = x

2

1

.

6

1 = 0.3.

3.18 (a) P(X < 4) =

R 4

2

2(1+x)

27 dx = (1+x)2

27

4

2

= 16/27.

(b) P(3 ≤ X < 4) =

R 4

3

2(1+x)

27 dx = (1+x)2

27

4

3

= 1/3.

3.19 F(x) =

R x

1 (1/2) dt = x−1

2 ,

P(2 < X < 2.5) = F(2.5) − F(2) = 1.5

2 − 1

2 = 1

4 .

3.20 F(x) = 2

27

R x

2 (1 + t) dt = 2

27

t + t2

2

x

2

= (x+4)(x−2)

27 ,

P(3 ≤ X < 4) = F(4) − F(3) = (8)(2)

27 − (7)(1)

27 = 1

3 .

3.21 (a) 1 = k

R 1

0 √x dx = 2k

3 x3/2

1

0 = 2k

3 . Therefore, k = 3

2 .

(b) F(x) = 3

2

R x

0

√t dt = t3/2

x

0 = x3/2.

P(0.3 < X < 0.6) = F(0.6) − F(0.3) = (0.6)3/2 − (0.3)3/2 = 0.3004.

3.22 Denote by X the number of spades int he three draws. Let S and N stand for a spade

and not a spade, respectively. Then

P(X = 0) = P(NNN) = (39/52)(38/51)(37/50) = 703/1700,

P(X = 1) = P(SNN) + P(NSN) + P(NNS) = 3(13/52)(39/51)(38/50) = 741/1700,

P(X = 3) = P(SSS) = (13/52)(12/51)(11/50) = 11/850, and

P(X = 2) = 1 − 703/1700 − 741/1700 − 11/850 = 117/850.

The probability mass function for X is then

x 0 1 2 3

f(x) 703/1700 741/1700 117/850 11/850

Solutions for Exercises in Chapter 3 33

3.23 The c.d.f. of X is

F(x) =

0, for w < −3,

1/27, for − 3 ≤ w < −1,

7/27, for − 1 ≤ w < 1,

19/27, for 1 ≤ w < 3,

1, for w ≥ 3,

(a) P(W > 0 = 1 − P(W ≤ 0) = 1 − 7/27 = 20/27.

(b) P(−1 ≤ W < 3) = F(2) − F(−3) = 19/27 − 1/27 = 2/3.

3.24 There are

10

4

ways of selecting any 4 CDs from 10. We can select x jazz CDs from 5

and 4 − x from the remaining CDs in

5

x

5

4−x

ways. Hence

f(x) =

5

x

5

4−x

10

4

, x = 0, 1, 2, 3, 4.

3.25 Let T be the total value of the three coins. Let D and N stand for a dime and nickel,

respectively. Since we are selecting without replacement, the sample space containing

elements for which t = 20, 25, and 30 cents corresponding to the selecting of 2 nickels

and 1 dime, 1 nickel and 2 dimes, and 3 dimes. Therefore, P(T = 20) =

(2

2)(4

1)

(6

3)

= 1

5 ,

P(T = 25) =

(2

1)(4

2)

(6

3)

= 3

5 ,

P(T = 30) =

(4

3)

(6

3)

= 1

5 ,

and the probability distribution in tabular form is

t 20 25 30

P(T = t) 1/5 3/5 1/5

As a probability histogram

20 25 30

x

f(x)

1/5

2/5

3/5

34 Chapter 3 Random Variables and Probability Distributions

3.26 Denote by X the number of green balls in the three draws. Let G and B stand for the

colors of green and black, respectively.

Simple Event x P(X = x)

BBB

GBB

BGB

BBG

BGG

GBG

GGB

GGG

0

1

1

1

2

2

2

3

(2/3)3 = 8/27

(1/3)(2/3)2 = 4/27

(1/3)(2/3)2 = 4/27

(1/3)(2/3)2 = 4/27

(1/3)2(2/3) = 2/27

(1/3)2(2/3) = 2/27

(1/3)2(2/3) = 2/27

(1/3)3 = 1/27

The probability mass function for X is then

x 0 1 2 3

P(X = x) 8/27 4/9 2/9 1/27

3.27 (a) For x ≥ 0, F(x) =

R x

0

1

2000 exp(−t/2000) dt = − exp(−t/2000)|x

0

= 1 − exp(−x/2000). So

F(x) =

(

0, x < 0,

1 − exp(−x/2000), x ≥ 0.

(b) P(X > 1000) = 1 − F(1000) = 1 − [1 − exp(−1000/2000)] = 0.6065.

(c) P(X < 2000) = F(2000) = 1 − exp(−2000/2000) = 0.6321.

3.28 (a) f(x) ≥ 0 and

R 26.25

23.75

2

5 dx = 2

5 t

26.25

23.75 = 2.5

2.5 = 1.

(b) P(X < 24) =

R 24

23.75

2

5 dx = 2

5 (24 − 23.75) = 0.1.

(c) P(X > 26) =

R 26.25

26

2

5 dx = 2

5 (26.25 − 26) = 0.1. It is not extremely rare.

3.29 (a) f(x) ≥ 0 and

R

∞

1 3x−4 dx = −3 x−3

3

∞

1

= 1. So, this is a density function.

(b) For x ≥ 1, F(x) =

R x

1 3t−4 dt = 1 − x−3. So,

F(x) =

(

0, x < 1,

1 − x−3, x ≥ 1.

(c) P(X > 4) = 1 − F(4) = 4−3 = 0.0156.

3.30 (a) 1 = k

R 1

−1(3 − x2) dx = k

3x − x3

3

1

−1

= 16

3 k. So, k = 3

16 .

Solutions for Exercises in Chapter 3 35

(b) For −1 ≤ x < 1, F(x) = 3

16

R x

−1(3 − t2) dt =

3t − 1

3 t3

x

−1 = 1

2 + 9

16x − x3

16 .

So, P

X < 1

2

= 1

2 −

9

16

1

2

− 1

16

1

2

3

= 99

128 .

(c) P(|X| < 0.8) = P(X < −0.8) + P(X > 0.8) = F(−0.8) + 1 − F(0.8)

= 1 +

1

2 − 9

160.8 + 1

160.83

−

1

2 + 9

160.8 − 1

160.83

= 0.164.

3.31 (a) For y ≥ 0, F(y) = 1

4

R y

0 e−t/4 dy = 1 − ey/4. So, P(Y > 6) = e−6/4 = 0.2231. This

probability certainly cannot be considered as “unlikely.”

(b) P(Y ≤ 1) = 1 − e−1/4 = 0.2212, which is not so small either.

3.32 (a) f(y) ≥ 0 and

R 1

0 5(1 − y)4 dy = − (1 − y)5|1

0 = 1. So, this is a density function.

(b) P(Y < 0.1) = − (1 − y)5|0.1

0 = 1 − (1 − 0.1)5 = 0.4095.

(c) P(Y > 0.5) = (1 − 0.5)5 = 0.03125.

3.33 (a) Using integral by parts and setting 1 = k

R 1

0 y4(1 − y)3 dy, we obtain k = 280.

(b) For 0 ≤ y < 1, F(y) = 56y5(1 − Y )3 + 28y6(1 − y)2 + 8y7(1 − y) + y8. So,

P(Y ≤ 0.5) = 0.3633.

(c) Using the cdf in (b), P(Y > 0.8) = 0.0563.

3.34 (a) The event Y = y means that among 5 selected, exactly y tubes meet the specification

(M) and 5 − y (M′) does not. The probability for one combination of

such a situation is (0.99)y(1 − 0.99)5−y if we assume independence among the

tubes. Since there are 5!

y!(5−y)! permutations of getting y Ms and 5 − y M′s, the

probability of this event (Y = y) would be what it is specified in the problem.

(b) Three out of 5 is outside of specification means that Y = 2. P(Y = 2) = 9.8×10−6

which is extremely small. So, the conjecture is false.

3.35 (a) P(X > 8) = 1 − P(X ≤ 8) =

P8

x=0

e−6 6x

x! = e−6

60

0! + 61

1! + · · · + 68

8!

= 0.1528.

(b) P(X = 2) = e−6 62

2! = 0.0446.

3.36 For 0 < x < 1, F(x) = 2

R x

0 (1 − t) dt = − (1 − t)2|x

0 = 1 − (1 − x)2.

(a) P(X ≤ 1/3) = 1 − (1 − 1/3)2 = 5/9.

(b) P(X > 0.5) = (1 − 1/2)2 = 1/4.

(c) P(X < 0.75 | X ≥ 0.5) = P(0.5≤X<0.75)

P(X≥0.5) = (1−0.5)2−(1−0.75)2

(1−0.5)2 = 3

4 .

3.37 (a)

P3

x=0

P3

y=0

f(x, y) = c

P3

x=0

P3

y=0

xy = 36c = 1. Hence c = 1/36.

(b)

P

x

P

y

f(x, y) = c

P

x

P

y |x − y| = 15c = 1. Hence c = 1/15.

3.38 The joint probability distribution of (X, Y ) is

36 Chapter 3 Random Variables and Probability Distributions

x

f(x, y) 0 1 2 3

0 0 1/30 2/30 3/30

y 1 1/30 2/30 3/30 4/30

2 2/30 3/30 4/30 5/30

(a) P(X ≤ 2, Y = 1) = f(0, 1) + f(1, 1) + f(2, 1) = 1/30 + 2/30 + 3/30 = 1/5.

(b) P(X > 2, Y ≤ 1) = f(3, 0) + f(3, 1) = 3/30 + 4/30 = 7/30.

(c) P(X > Y ) = f(1, 0) + f(2, 0) + f(3, 0) + f(2, 1) + f(3, 1) + f(3, 2)

= 1/30 + 2/30 + 3/30 + 3/30 + 4/30 + 5/30 = 3/5.

(d) P(X + Y = 4) = f(2, 2) + f(3, 1) = 4/30 + 4/30 = 4/15.

3.39 (a) We can select x oranges from 3, y apples from 2, and 4 − x − y bananas from 3

in

3

x

2

y

3

4−x−y

ways. A random selection of 4 pieces of fruit can be made in

8

4

ways. Therefore,

f(x, y) =

3

x

2

y

3

4−x−y

8

4

, x = 0, 1, 2, 3; y = 0, 1, 2; 1 ≤ x + y ≤ 4.

(b) P[(X, Y ) ∈ A] = P(X + Y ≤ 2) = f(1, 0) + f(2, 0) + f(0, 1) + f(1, 1) + f(0, 2)

= 3/70 + 9/70 + 2/70 + 18/70 + 3/70 = 1/2.

3.40 (a) g(x) = 2

3

R 1

0 (x + 2y) dy = 2

3 (x + 1), for 0 ≤ x ≤ 1.

(b) h(y) = 2

3

R 1

0 (x + 2y) dy = 1

3 (1 + 4y), for 0 ≤ y ≤ 1.

(c) P(X < 1/2) = 2

3

R 1/2

0 (x + 1) dx = 5

12 .

3.41 (a) P(X + Y ≤ 1/2) =

R 1/2

0

R 1/2−y

0 24xy dx dy = 12

R 1/2

0

1

2 − y

2

y dy = 1

16 .

(b) g(x) =

R 1−x

0 24xy dy = 12x(1 − x)2, for 0 ≤ x < 1.

(c) f(y|x) = 24xy

12x(1−x)2 = 2y

(1−x)2 , for 0 ≤ y ≤ 1 − x.

Therefore, P(Y < 1/8 | X = 3/4) = 32

R 1/8

0 y dy = 1/4.

3.42 Since h(y) = e−y

R

∞

0 e−x dx = e−y, for y > 0, then f(x|y) = f(x, y)/h(y) = e−x, for

x > 0. So, P(0 < X < 1 | Y = 2) =

R 1

0 e−x dx = 0.6321.

3.43 (a) P(0 ≤ X ≤ 1/2, 1/4 ≤ Y ≤ 1/2) =

R 1/2

0

R 1/2

1/4 4xy dy dx = 3/8

R 1/2

0 x dx = 3/64.

(b) P(X < Y ) =

R 1

0

R y

0 4xy dx dy = 2

R 1

0 y3 dy = 1/2.

3.44 (a) 1 = k

R 50

30

R 50

30 (x2 + y2) dx dy = k(50 − 30)

R 50

30 x2 dx +

R 50

30 y2 dy

= 392k

3 · 104.

So, k = 3

392 · 10−4.

Solutions for Exercises in Chapter 3 37

(b) P(30 ≤ X ≤ 40, 40 ≤ Y ≤ 50) = 3

392 · 10−4

R 40

30

R 50

40 (x2 + y2) dy dx

= 3

392 · 10−3(

R 40

30 x2 dx +

R 50

40 y2 dy) = 3

392 · 10−3

403−303

3 + 503−403

3

= 49

196 .

(c) P(30 ≤ X ≤ 40, 30 ≤ Y ≤ 40) = 3

392 · 10−4

R 40

30

R 40

30 (x2 + y2) dx dy

= 2 3

392 · 10−4(40 − 30)

R 40

30 x2 dx = 3

196 · 10−3 403−303

3 = 37

196 .

3.45 P(X + Y > 1/2) = 1 − P(X + Y < 1/2) = 1 −

R 1/4

0

R 1/2−x

x

1

y dy dx

= 1 −

R 1/4

0

ln

1

2 − x

− ln x

dx = 1 +

1

2 − x

ln

1

2 − x

− x ln x

1/4

0

= 1 + 1

4 ln

1

4

= 0.6534.

3.46 (a) From the column totals of Exercise 3.38, we have

x 0 1 2 3

g(x) 1/10 1/5 3/10 2/5

(b) From the row totals of Exercise 3.38, we have

y 0 1 2

h(y) 1/5 1/3 7/15

3.47 (a) g(x) = 2

R 1

x dy = 2(1 − x) for 0 < x < 1;

h(y) = 2

R y

0 dx = 2y, for 0 < y < 1.

Since f(x, y) 6= g(x)h(y), X and Y are not independent.

(b) f(x|y) = f(x, y)/h(y) = 1/y, for 0 < x < y.

Therefore, P(1/4 < X < 1/2 | Y = 3/4) = 4

3

R 1/2

1/4 dx = 1

3 .

3.48 (a) g(2) =

P2

y=0

f(2, y) = f(2, 0) + f(2, 1) + f(2, 2) = 9/70 + 18/70 + 3/70 = 3/7. So,

f(y|2) = f(2, y)/g(2) = (7/3)f(2, y).

f(0|2) = (7/3)f(2, 0) = (7/3)(9/70) = 3/10, f(1|2) = 3/5 and f(2|2) = 1/10. In

tabular form,

y 0 1 2

f(y|2) 3/10 3/5 1/10

(b) P(Y = 0 | X = 2) = f(0|2) = 3/10.

3.49 (a)

x 1 2 3

g(x) 0.10 0.35 0.55

(b)

y 1 2 3

h(y) 0.20 0.50 0.30

(c) P(Y = 3 | X = 2) = 0.2

0.05+0.10+0.20 = 0.5714.

38 Chapter 3 Random Variables and Probability Distributions

3.50

x

f(x, y) 2 4 h(y)

1 0.10 0.15 0.25

y 3 0.20 0.30 0.50

5 0.10 0.15 0.25

g(x) 0.40 0.60

(a)

x 2 4

g(x) 0.40 0.60

(b)

y 1 3 5

h(y) 0.25 0.50 0.25

3.51 (a) Let X be the number of 4’s and Y be the number of 5’s. The sample space

consists of 36 elements each with probability 1/36 of the form (m, n) where

m is the outcome of the first roll of the die and n is the value obtained on

the second roll. The joint probability distribution f(x, y) is defined for x =

0, 1, 2 and y = 0, 1, 2 with 0 ≤ x + y ≤ 2. To find f(0, 1), for example,

consider the event A of obtaining zero 4’s and one 5 in the 2 rolls. Then

A = {(1, 5), (2, 5), (3, 5), (6, 5), (5, 1), (5, 2), (5, 3), (5, 6)}, so f(0, 1) = 8/36 = 2/9.

In a like manner we find f(0, 0) = 16/36 = 4/9, f(0, 2) = 1/36, f(1, 0) = 2/9,

f(2, 0) = 1/36, and f(1, 1) = 1/18.

(b) P[(X, Y ) ∈ A] = P(2X + Y < 3) = f(0, 0) + f(0, 1) + f(0, 2) + f(1, 0) =

4/9 + 1/9 + 1/36 + 2/9 = 11/12.

3.52 A tabular form of the experiment can be established as,

Sample Space x y

HHH

HHT

HTH

THH

HTT

THT

TTH

TTT

3

2

2

2

1

1

1

0

3

1

1

1

−1

−1

−1

−3

So, the joint probability distribution is,

x

f(x, y) 0 1 2 3

−3 1/8

y −1 3/8

1 3/8

3 1/8

Solutions for Exercises in Chapter 3 39

3.53 (a) If (x, y) represents the selection of x kings and y jacks in 3 draws, we must have

x = 0, 1, 2, 3; y = 0, 1, 2, 3; and 0 ≤ x + y ≤ 3. Therefore, (1, 2) represents the

selection of 1 king and 2 jacks which will occur with probability

f(1, 2) =

4

1

4

2

12

3

=

6

55

.

Proceeding in a similar fashion for the other possibilities, we arrive at the following

joint probability distribution:

x

f(x, y) 0 1 2 3

0 1/55 6/55 6/55 1/55

y 1 6/55 16/55 6/55

2 6/55 6/55

3 1/55

(b) P[(X, Y ) ∈ A] = P(X +Y ≥ 2) = 1−P(X +Y < 2) = 1−1/55−6/55−6/55 =

42/55.

3.54 (a) P(H) = 0.4, P(T) = 0.6, and S = {HH,HT, TH, TT}. Let (W, Z) represent

a typical outcome of the experiment. The particular outcome (1, 0) indicating a

total of 1 head and no heads on the first toss corresponds to the event TH. Therefore,

f(1, 0) = P(W = 1, Z = 0) = P(TH) = P(T)P(H) = (0.6)(0.4) = 0.24.

Similar calculations for the outcomes (0, 0), (1, 1), and (2, 1) lead to the following

joint probability distribution:

w

f(w, z) 0 1 2

z 0 0.36 0.24

1 0.24 0.16

(b) Summing the columns, the marginal distribution of W is

w 0 1 2

g(w) 0.36 0.48 0.16

(c) Summing the rows, the marginal distribution of Z is

z 0 1

h(z) 0.60 0.40

(d) P(W ≥ 1) = f(1, 0) + f(1, 1) + f(2, 1) = 0.24 + 0.24 + 0.16 = 0.64.

40 Chapter 3 Random Variables and Probability Distributions

3.55 g(x) = 1

8

R 4

2 (6 − x − y) dy = 3−x

4 , for 0 < x < 2.

So, f(y|x) = f(x,y)

g(x) = 6−x−y

2(3−x) , for 2 < y < 4,

and P(1 < Y < 3 | X = 1) = 1

4

R 3

2 (5 − y) dy = 5

8 .

3.56 Since f(1, 1) 6= g(1)h(1), the variables are not independent.

3.57 X and Y are independent since f(x, y) = g(x)h(y) for all (x, y).

3.58 (a) h(y) = 6

R 1−y

0 x dx = 3(1 − y)2, for 0 < y < 1. Since f(x|y) = f(x,y)

h(y) = 2x

(1−y)2 , for

0 < x < 1 − y, involves the variable y, X and Y are not independent.

(b) P(X > 0.3 | Y = 0.5) = 8

R 0.5

0.3 x dx = 0.64.

3.59 (a) 1 = k

R 1

0

R 1

0

R 2

0 xy2z dx dy dz = 2k

R 1

0

R 1

0 y2z dy dz = 2k

3

R 1

0 z dz = k

3 . So, k = 3.

(b) P

X < 1

4 , Y > 1

2 , 1 < Z < 2

= 3

R 1/4

0

R 1

1/2

R 2

1 xy2z dx dy dz = 9

2

R 1/4

0

R 1

1/2 y2z dy dz

= 21

16

R 1/4

0 z dz = 21

512 .

3.60 g(x) = 4

R 1

0 xy dy = 2x, for 0 < x < 1; h(y) = 4

R 1

0 xy dx = 2y, for 0 < y < 1. Since

f(x, y) = g(x)h(y) for all (x, y), X and Y are independent.

3.61 g(x) = k

R 50

30 (x2 + y2) dy = k

x2y + y3

3

50

30

= k

20x2 + 98,000

3

, and

h(y) = k

20y2 + 98,000

3

.

Since f(x, y) 6= g(x)h(y), X and Y are not independent.

3.62 (a) g(y, z) = 4

9

R 1

0 xyz2 dx = 2

9yz2, for 0 < y < 1 and 0 < z < 3.

(b) h(y) = 2

9

R 3

0 yz2 dz = 2y, for 0 < y < 1.

(c) P

1

4 < X < 1

2 , Y > 1

3 , Z < 2

= 4

9

R 2

1

R 1

1/3

R 1/2

1/4 xyz2 dx dy dz = 7

162 .

(d) Since f(x|y, z) = f(x,y,z)

g(y,z) = 2x, for 0 < x < 1, P

0 < X < 1

2 | Y = 1

4 , Z = 2

=

2

R 1/2

0 x dx = 1

4 .

3.63 g(x) = 24

R 1−x

0 xy dy = 12x(1 − x)2, for 0 < x < 1.

(a) P(X ≥ 0.5) = 12

R 1

0.5 x(1 − x)2 dx =

R 1

0.5(12x − 24x2 + 12x3) dx = 5

16 = 0.3125.

(b) h(y) = 24

R 1−y

0 xy dx = 12y(1 − y)2, for 0 < y < 1.

(c) f(x|y) = f(x,y)

h(y) = 24xy

12y(1−y)2 = 2x

(1−y)2 , for 0 < x < 1 − y.

So, P

X < 1

8 | Y = 3

4

=

R 1/8

0

2x

1/16 dx = 32

R 1/8

0 = 0.25.

3.64 (a)

x 1 3 5 7

f(x) 0.4 0.2 0.2 0.2

(b) P(4 < X ≤ 7) = P(X ≤ 7) − P(X ≤ 4) = F(7) − F(4) = 1 − 0.6 = 0.4.

Solutions for Exercises in Chapter 3 41

3.65 (a) g(x) =

R

∞

0 ye−y(1+x) dy = − 1

1+xye−y(1+x)

∞0

+ 1

1+x

R

∞

0 e−y(1+x) dy

= − 1

(1+x)2 e−y(1+x)

∞0

= 1

(1+x)2 , for x > 0.

h(y) = ye−y

R

∞

0 e−yx dx = −e−y e−yx|∞0 = e−y, for y > 0.

(b) P(X ≥ 2, Y ≥ 2) =

R

∞

2

R

∞

2 ye−y(1+x) dx dy = −

R

∞

2 e−y e−yx|∞2 dy =

R

∞

2 e−3ydy

= − 1

3e−3y

∞2

= 1

3e6 .

3.66 (a) P

X ≤ 1

2 , Y ≤ 1

2

= 3

2

R 1/2

0

R 1/2

0 (x2 + y2) dxdy = 3

2

R 1/2

0

x2y + y3

3

1/2

0

dx

= 3

4

R 1/2

0

x2 + 1

12

dx = 1

16 .

(b) P

X ≥ 3

4

= 3

2

R 1

3/4

x2 + 1

3

dx = 53

128 .

3.67 (a)

x 0 1 2 3 4 5 6

f(x) 0.1353 0.2707 0.2707 0.1804 0.0902 0.0361 0.0120

(b) A histogram is shown next.

1 2 3 4 5 6 7

x

f(x)

0.0

0.1

0.2

0.3

(c)

x 0 1 2 3 4 5 6

F(x) 0.1353 0.4060 0.6767 0.8571 0.9473 0.9834 0.9954

3.68 (a) g(x) =

R 1

0 (x + y) dy = x + 1

2 , for 0 < x < 1, and h(y) = y + 1

2 for 0 < y < 1.

(b) P(X > 0.5, Y > 0.5) =

R 1

0.5

R 1

0.5(x + y) dx dy =

R 1

0.5

x2

2 + xy

1

0.5

dy

=

R 1

0.5

1

2 + y

−

1

8 + y

2

dy = 3

8 .

3.69 f(x) =

5

x

(0.1)x(1 − 0.1)5−x, for x = 0, 1, 2, 3, 4, 5.

3.70 (a) g(x) =

R 2

1

3x−y

9

dy = 3xy−y2/2

9

2

1

= x

3 − 1

6 , for 1 < x < 3, and

h(y) =

R 3

1

3x−y

9

dx = 4

3 − 2

9y, for 1 < y < 2.

(b) No, since g(x)h(y) 6= f(x, y).

(c) P(X > 2) =

R 3

2

x

3 − 1

6

dx =

x2

6 − x

6

3

2

= 2

3 .

42 Chapter 3 Random Variables and Probability Distributions

3.71 (a) f(x) = d

dxF(x) = 1

50e−x/50, for x > 0.

(b) P(X > 70) = 1 − P(X ≤ 70) = 1 − F(70) = 1 − (1 − e−70/50) = 0.2466.

3.72 (a) f(x) = 1

10 , for x = 1, 2, . . . , 10.

(b) A c.d.f. plot is shown next.

F(x)

x

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

1 2 3 4 5 6 7 8 9 10

3.73 P(X ≥ 3) = 1

2

R

∞

3 e−y/2 = e−3/2 = 0.2231.

3.74 (a) f(x) ≥ 0 and

R 10

0

1

10 dx = 1. This is a continuous uniform distribution.

(b) P(X ≤ 7) = 1

10

R 7

0 dx = 0.7.

3.75 (a) f(y) ≥ 0 and

R 1

0 f(y) dy = 10

R 1

0 (1 − y)9 dy = − 10

10 (1 − y)10

1

0 = 1.

(b) P(Y > 0.6) =

R 1

0.6 f(y) dy = − (1 − y)10|1

0.6 = (1 − 0.6)10 = 0.0001.

3.76 (a) P(Z > 20) = 1

10

R

∞

20 e−z/10 dz = − e−z/10

∞ 2

0

=

e−2

0

/

1

0

=

0.1353.

(b) P(Z ≤ 10) = − e−z/10

1

0

0 = 1 − e−10/10 = 0.6321.

3.77 (a) g(x1) =

R 1

x1

2 dx2 = 2(1 − x1), for 0 < x1 < 1.

(b) h(x2) =

R x2

0 2 dx1 = 2x2, for 0 < x2 < 1.

(c) P(X1 < 0.2,X2 > 0, 5) =

R 1

0.5

R 0.2

0 2 dx1 dx2 = 2(1 − 0.5)(0.2 − 0) = 0.2.

(d) fX1|X2(x1|x2) = f(x1,x2)

h(x2) = 2

2x2

= 1

x2

, for 0 < x1 < x2.

3.78 (a) fX1(x1) =

R x1

0 6x2 dx2 = 3x2

1, for 0 < x1 < 1. Apparently, fX1R (x1) ≥ 0 and 1

0 fX1(x1) dx1 =

R 1

0 3x2

1 dx1 = 1. So, fX1(x1) is a density function.

(b) fX2|X1(x2|x1) = f(x1,x2)

fX1 (x1) = 6x2

3x21

= 2x2

x21

, for 0 < x2 < x1.

So, P(X2 < 0.5 | X1 = 0.7) = 2

0.72

R 0.5

0 x2 dx2 = 25

49 .

Solutions for Exercises in Chapter 3 43

3.79 (a) g(x) = 9

(16)4y

∞P

x=0

1

4x = 9

(16)4y

1

1−1/4 = 3

4 · 1

4x , for x = 0, 1, 2, . . . ; similarly, h(y) = 3

4 · 1

4y ,

for y = 0, 1, 2, . . . . Since f(x, y) = g(x)h(y), X and Y are independent.

(b) P(X +Y < 4) = f(0, 0)+f(0, 1)+f(0, 2)+f(0, 3)+f(1, 0)+f(1, 1)+f(1, 2)+

f(2, 0) + f(2, 1) + f(3, 0) = 9

16

1 + 1

4 + 1

42 + 1

43 + 1

4 + 1

42 + 1

43 + 1

r2 + 1

43 + 1

43

=

9

16

1 + 2

4 + 3

42 + 4

43

= 63

64 .

3.80 P(the system works) = P(all components work) = (0.95)(0.99)(0.92) = 0.86526.

3.81 P(the system does not fail) = P(at least one of the components works)

= 1−P(all components fail) = 1−(1−0.95)(1−0.94)(1−0.90)(1−0.97) = 0.999991.

3.82 Denote by X the number of components (out of 5) work.

Then, P(the system is operational) = P(X ≥ 3) = P(X = 3) + P(X = 4) + P(X =

5) =

5

3

(0.92)3(1 − 0.92)2 +

5

4

(0.92)4(1 − 0.92) +

5

5

(0.92)5 = 0.9955.

Chapter 4

Mathematical Expectation

4.1 E(X) = 1

a2

R a

−a

R√a2−y2

−√a2−y2

x dx dy = 1

a2

h

a2−y2

2

−

a2−y2

2

i

dy = 0.

4.2 E(X) =

P3

x=0

x f(x) = (0)(27/64) + (1)(27/64) + (2)(9/64) + (3)(1/64) = 3/4.

4.3 μ = E(X) = (20)(1/5) + (25)(3/5) + (30)(1/5) = 25 cents.

4.4 Assigning wrights of 3w and w for a head and tail, respectively. We obtain P(H) = 3/4

and P(T) = 1/4. The sample space for the experiment is S = {HH,HT, TH, TT}.

Now if X represents the number of tails that occur in two tosses of the coin, we have

P(X = 0) = P(HH) = (3/4)(3/4) = 9/16,

P(X = 1) = P(HT) + P(TH) = (2)(3/4)(1/4) = 3/8,

P(X = 2) = P(TT) = (1/4)(1/4) = 1/16.

The probability distribution for X is then

x 0 1 2

f(x) 9/16 3/8 1/16

from which we get μ = E(X) = (0)(9/16) + (1)(3/8) + (2)(1/16) = 1/2.

4.5 μ = E(X) = (0)(0.41) + (1)(0.37) + (2)(0.16) + (3)(0.05) + (4)(0.01) = 0.88.

4.6 μ = E(X) = ($7)(1/12)+($9)(1/12)+($11)(1/4)+($13)(1/4)+($15)(1/6)+($17)(1/6)

= $12.67.

4.7 Expected gain = E(X) = (4000)(0.3) + (−1000)(0.7) = $500.

4.8 Let X = profit. Then

μ = E(X) = (250)(0.22) + (150)(0.36) + (0)(0.28) + (−150)(0.14) = $88.

45

46 Chapter 4 Mathematical Expectation

4.9 Let c = amount to play the game and Y = amount won.

y 5 − c 3 − c −c

f(y) 2/13 2/13 9/13

E(Y ) = (5 − c)(2/13) + (3 − c)(2/13) + (−c)(9/13) = 0. So, 13c = 16 which implies

c = $1.23.

4.10 μX =

P

xg(x) = (1)(0.17) + (2)(0.5) + (3)(0.33) = 2.16,

μY =

P

yh(y) = (1)(0.23) + (2)(0.5) + (3)(0.27) = 2.04.

4.11 For the insurance of $200,000 pilot, the distribution of the claim the insurance company

would have is as follows:

Claim Amount $200,000 $100,000 $50,000 0

f(x) 0.002 0.01 0.1 0.888

So, the expected claim would be

($200, 000)(0.002) + ($100, 000)(0.01) + ($50, 000)(0.1) + ($0)(0.888) = $6, 400.

Hence the insurance company should charge a premium of $6, 400 + $500 = $6, 900.

4.12 E(X) =

R 1

0 2x(1 − x) dx = 1/3. So, (1/3)($5, 000) = $1, 667.67.

4.13 E(X) = 4

R 1

0

x

1+x2 dx = ln 4

.

4.14 E(X) =

R 1

0

2x(x+2)

5 dx = 8

15 .

4.15 E(X) =

R 1

0 x2 dx +

R 2

1 x(2 − x) dx = 1. Therefore, the average number of hours per

year is (1)(100) = 100 hours.

4.16 P(X1 + X2 = 1) = P(X1 = 1,X2 = 0) + P(X1 = 0,X2 = 1)

=

(980

1 )(20

1 )

(1000

2 )

+

(980

1 )(20

1 )

(1000

2 )

= (2)(0.0392) = 0.0784.

4.17 The probability density function is,

x −3 6 9

f(x) 1/6 1/2 1/3

g(x) 25 169 361

μg(X) = E[(2X + 1)2] = (25)(1/6) + (169)(1/2) + (361)(1/3) = 209.

4.18 E(X2) = (0)(27/64) + (1)(27/64) + (4)(9/64) + (9)(1/64) = 9/8.

4.19 Let Y = 1200X − 50X2 be the amount spent.

Solutions for Exercises in Chapter 4 47

x 0 1 2 3

f(x) 1/10 3/10 2/5 1/5

y = g(x) 0 1150 2200 3150

μY = E(1200X − 50X2) = (0)(1/10) + (1150)(3/10) + (2200)(2/5) + (3150)(1/5)

= $1, 855.

4.20 E[g(X)] = E(e2X/3) =

R

∞

0 e2x/3e−x dx =

R

∞

0 e−x/3 dx = 3.

4.21 E(X2) =

R 1

0 2x2(1 − x) dx = 1

6 . Therefore, the average profit per new automobile is

(1/6)($5000.00) = $833.33.

4.22 E(Y ) = E(X + 4) =

R

∞

0 32(x + 4) 1

(x+4)3 dx = 8 days.

4.23 (a) E[g(X, Y )] = E(XY 2) =

P

x

P

y

xy2f(x, y)

= (2)(1)2(0.10) + (2)(3)2(0.20) + (2)(5)2(0.10) + (4)(1)2(0.15) + (4)(3)2(0.30)

+ (4)(5)2(0.15) = 35.2.

(b) μX = E(X) = (2)(0.40) + (4)(0.60) = 3.20,

μY = E(Y ) = (1)(0.25) + (3)(0.50) + (5)(0.25) = 3.00.

4.24 (a) E(X2Y − 2XY ) =

P3

x=0

P2

y=0

(x2y − 2xy)f(x, y) = (1− 2)(18/70)+ (4− 4)(18/70)+

· · · + (8 − 8)(3/70) = −3/7.

(b)

x 0 1 2 3

g(x) 5/70 30/70 30/70 5/70

y 0 1 2

h(y) 15/70 40/70 15/70

μX = E(X) = (0)(5/70) + (1)(30/70) + (2)(30/70) + (3)(5/70) = 3/2,

μY = E(Y ) = (0)(15/70) + (1)(40/70) + (2)(15/70) = 1.

4.25 μX+Y = E(X +Y ) =

P3

x=0

P3

y=0

(x+y)f(x, y) = (0+0)(1/55)+(1+0)(6/55)+· · ·+(0+

3)(1/55) = 2.

4.26 E(Z) = E(√X2 + Y 2) =

R 1

0

R 1

0 4xy

p

x2 + y2 dx dy = 4

3

R 1

0 [y(1 + y2)3/2 − y4] dy

= 8(23/2 − 1)/15 = 0.9752.

4.27 E(X) = 1

2000

R

∞

0 x exp(−x/2000) dx = 2000

R

∞

0 y exp(−y) dy = 2000.

4.28 (a) The density function is shown next.

f(x)

23.75 26.25

2/5

48 Chapter 4 Mathematical Expectation

(b) E(X) = 2

5

R 26.25

23.75 x dx = 1

5 (26.252 − 23.752) = 25.

(c) The mean is exactly in the middle of the interval. This should not be surprised

due to the symmetry of the density at 25.

4.29 (a) The density function is shown next

f(x)

0 1 1.5 2 2.5 3 3.5 4

1

2

3

(b) μ = E(X) =

R

∞

1 3x−3 dx = 3

2 .

4.30 E(Y ) = 1

4

R

∞

0 ye−y/4 dy = 4.

4.31 (a) μ = E(Y ) = 5

R 1

0 y(1 − y)4 dy = −

R 1

0 y d(1 − y)5 =

R

∞

0 (1 − y)5 dy = 1

6 .

(b) P(Y > 1/6) =

R 1

1/6 5(1 − y)4 dy = − (1 − y)5|1

1/6 = (1 − 1/6)5 = 0.4019.

4.32 (a) A histogram is shown next.

1 2 3 4 5

x

f(x)

0

0.1

0.2

0.3

0.4

(b) μ = (0)(0.41) + (1)(0.37) + (2)(0.16) + (3)(0.05) + (4)(0.01) = 0.88.

(c) E(X2) = (0)2(0.41) + (1)2(0.37) + (2)2(0.16) + (3)2(0.05) + (4)2(0.01) = 1.62.

(d) V ar(X) = 1.62 − 0.882 = 0.8456.

4.33 μ = $500. So, σ2 = E[(X − μ)2] =

P

x

(x − μ)2f(x) = (−1500)2(0.7) + (3500)2(0.3) =

$5, 250, 000.

Solutions for Exercises in Chapter 4 49

4.34 μ = (−2)(0.3) + (3)(0.2) + (5)(0.5) = 2.5 and

E(X2) = (−2)2(0.3) + (3)2(0.2) + (5)2(0.5) = 15.5.

So, σ2 = E(X2) − μ2 = 9.25 and σ = 3.041.

4.35 μ = (2)(0.01) + (3)(0.25) + (4)(0.4) + (5)(0.3) + (6)(0.04) = 4.11,

E(X2) = (2)2(0.01) + (3)2(0.25) + (4)2(0.4) + (5)2(0.3) + (6)2(0.04) = 17.63.

So, σ2 = 17.63 − 4.112 = 0.74.

4.36 μ = (0)(0.4) + (1)(0.3) + (2)(0.2) + (3)(0.1) = 1.0,

and E(X2) = (0)2(0.4) + (1)2(0.3) + (2)2(0.2) + (3)2(0.1) = 2.0.

So, σ2 = 2.0 − 1.02 = 1.0.

4.37 It is know μ = 1/3.

So, E(X2) =

R 1

0 2x2(1 − x) dx = 1/6 and σ2 = 1/6 − (1/3)2 = 1/18. So, in the actual

profit, the variance is 1

18 (5000)2.

4.38 It is known μ = 8/15.

Since E(X2) =

R 1

0

2

5x2(x + 2) dx = 11

30 , then σ2 = 11/30 − (8/15)2 = 37/450.

4.39 It is known μ = 1.

Since E(X2) =

R 1

0 x2 dx +

R 2

1 x2(2 − x) dx = 7/6, then σ2 = 7/6 − (1)2 = 1/6.

4.40 μg(X) = E[g(X)] =

R 1

0 (3x2 + 4)

2x+4

5

dx = 1

5

R 1

0 (6x3 + 12x2 + 8x + 16) dx = 5.1.

So, σ2 = E[g(X) − μ]2 =

R 1

0 (3x2 + 4 − 5.1)2

2x+4

5

dx

=

R 1

0 (9x4 − 6.6x2 + 1.21)

2x+4

5

dx = 0.83.

4.41 It is known μg(X) = E[(2X + 1)2] = 209. Hence

σ2

g(X) =

P

x

[(2X + 1)2 − 209]2g(x)

= (25 − 209)2(1/6) + (169 − 209)2(1/2) + (361 − 209)2(1/3) = 14, 144.

So, σg(X) = √14, 144 = 118.9.

4.42 It is known μg(X) = E(X2) = 1/6. Hence

σ2

g(X) =

R 1

0 2

x2 − 1

6

2

(1 − x) dx = 7/180.

4.43 μY = E(3X − 2) = 1

4

R

∞

0 (3x − 2)e−x/4 dx = 10. So

σ2

Y = E{[(3X − 2) − 10]2} = 9

4

R

∞

0 (x − 4)2e−x/4 dx = 144.

4.44 E(XY ) =

P

x

P

y

xyf(x, y) = (1)(1)(18/70) + (2)(1)(18/70)

+ (3)(1)(2/70) + (1)(2)(9/70) + (2)(2)(3/70) = 9/7;

μX =

P

x

P

y

xf(x, y) = (0)f(0, 1) + (0)f(0, 2) + (1)f(1, 0) + · · · + (3)f(3, 1) = 3/2,

and μY = 1.

So, σXY = E(XY ) − μXμY = 9/7 − (3/2)(1) = −3/14.

50 Chapter 4 Mathematical Expectation

4.45 μX =

P

x

xg(x) = 2.45, μY =

P

y

yh(y) = 2.10, and

E(XY ) =

P

x

P

x

xyf(x, y) = (1)(0.05) + (2)(0.05) + (3)(0.10) + (2)(0.05)

+ (4)(0.10) + (6)(0.35) + (3)(0) + (6)(0.20) + (9)(0.10) = 5.15.

So, σXY = 5.15 − (2.45)(2.10) = 0.005.

4.46 From previous exercise, k =

3

392

10−4, and g(x) = k

20x2 + 98000

3

, with

μX = E(X) =

R 50

30 xg(x) dx = k

R 50

30

20x3 + 98000

3 x

dx = 40.8163.

Similarly, μY = 40.8163. On the other hand,

E(XY ) = k

R 50

30

R 50

30 xy(x2 + y2) dy dx = 1665.3061.

Hence, σXY = E(XY ) − μXμY = 1665.3061 − (40.8163)2 = −0.6642.

4.47 g(x) = 2

3

R 1

0 (x + 2y) dy = 2

3 (x + 1, for 0 < x < 1, so μX = 2

3

R 1

0 x(x + 1) dx = 5

9 ;

h(y) = 2

3

R 1

0 (x + 2y) dx = 2

3

1

2 + 2y

, so μY = 2

3

R 1

0 y

1

2 + 2y

dy = 11

18 ; and

E(XY ) = 2

3

R 1

0

R 1

0 xy(x + 2y) dy dx = 1

3 .

So, σXY = E(XY ) − μXμY = 1

3 −

5

9

11

18

= −0.0062.

4.48 Since σXY = Cov(a + bX,X) = bσ2

X and σ2

Y = b2σ2

X , then

ρ = XY

X

σY = b 2

X √ 2

X b2 2

X

= b

|b|

= sign of b.

Hence ρ = 1 if b > 0 and ρ = −1 if b < 0.

4.49 E(X) = (0)(0.41) + (1)(0.37) + (2)(0.16) + (3)(0.05) + (4)(0.01) = 0.88

and E(X2) = (0)2(0.41) + (1)2(0.37) + (2)2(0.16) + (3)2(0.05) + (4)2(0.01) = 1.62.

So, V ar(X) = 1.62 − 0.882 = 0.8456 and σ = √0.8456 = 0.9196.

4.50 E(X) = 2

R 1

0 x(1 − x) dx = 2

x2

2 − x3

3

1

0

= 1

3 and

E(X2) = 2

R 1

0 x2(1 − x) dx = 2

x3

3 − x4

4

1

0

= 1

6 . Hence,

V ar(X) = 1

6 −

1

3

2 = 1

18 , and σ =

p

1/18 = 0.2357.

4.51 Previously we found μ = 4.11 and σ2 = 0.74, Therefore,

μg(X) = E(3X − 2) = 3μ − 2 = (3)(4.11) − 2 = 10.33 and σg(X) = 9σ2 = 6.66.

4.52 Previously we found μ = 1 and σ2 = 1. Therefore,

μg(X) = E(5X + 3) = 5μ + 3 = (5)(1) + 3 = 8 and σg(X) = 25σ2 = 25.

4.53 Let X = number of cartons sold and Y = profit.

We can write Y = 1.65X + (0.90)(5 − X) − 6 = 0.75X − 1.50. Now

E(X) = (0)(1/15)+(1)(2/15)+(2)(2/15)+(3)(3/15)+(4)(4/15)+(5)(3/15) = 46/15,

and E(Y ) = (0.75)E(X) − 1.50 = (0.75)(46/15) − 1.50 = $0.80.

4.54 μX = E(X) = 1

4

R

∞

0 xe−x/4 dx = 4.

Therefore, μY = E(3X − 2) = 3E(X) − 2 = (3)(4) − 2 = 10.

Since E(X2) = 1

4

R

∞

0 x2e−x/4 dx = 32, therefore, σ2

X = E(X2) − μ2

X = 32 − 16 = 16.

Hence σ2

Y = 9σ2

X = (9)(16) = 144.

Solutions for Exercises in Chapter 4 51

4.55 E(X) = (−3)(1/6) + (6)(1/2) + (9)(1/3) = 11/2,

E(X2) = (−3)2(1/6) + (6)2(1/2) + (9)2(1/3) = 93/2. So,

E[(2X + 1)2] = 4E(X2) + 4E(X) + 1 = (4)(93/2) + (4)(11/2) + 1 = 209.

4.56 Since E(X) =

R 1

0 x2 dx +

R 2

1 x(2 − x) dx = 1, and

E(X2) =

R 1

0 x32 dx +

R 2

1 x2(2 − x) dx = 7/6,then

E(Y ) = 60E(X2) + 39E(X) = (60)(7/6) + (39)(1) = 109 kilowatt hours.

4.57 The equations E[(X − 1)2] = 10 and E[(X − 2)2] = 6 may be written in the form:

E(X2) − 2E(X) = 9, E(X2) − 4E(X) = 2.

Solving these two equations simultaneously we obtain

E(X) = 7/2, and E(X2) = 16.

Hence μ = 7/2 and σ2 = 16 − (7/2)2 = 15/4.

4.58 E(X) = (2)(0.40) + (4)(0.60) = 3.20, and

E(Y ) = (1)(0.25) + (3)(0.50) + (5)(0.25) = 3. So,

(a) E(2X − 3Y ) = 2E(X) − 3E(Y ) = (2)(3.20) − (3)(3.00) = −2.60.

(b) E(XY ) = E(X)E(Y ) = (3.20)(3.00) = 9.60.

4.59 E(2XY 2 − X2Y ) = 2E(XY 2) − E(X2Y ). Now,

E(XY 2) =

P2

x=0

P2

y=0

xy2f(x, y) = (1)(1)2(3/14) = 3.14, and

E(X2Y ) =

P2

x=0

P2

y=0

x2yf(x, y) = (1)2(1)(3/14) = 3.14.

Therefore, E(2XY 2 − X2Y ) = (2)(3/14) − (3/14) = 3/14.

4.60 Using μ = 60 and σ = 6 and Chebyshev’s theorem

P(μ − kσ < X < μ + kσ) ≥ 1 −

1

k2 ,

since from μ + kσ = 84 we obtain k = 4.

So, P(X < 84) ≥ P(36 < X < 84) ≥ 1 − 1

42 = 0.9375. Therefore,

P(X ≥ 84) ≤ 1 − 0.9375 = 0.0625.

Since 1000(0.0625) = 62.5, we claim that at most 63 applicants would have a score as

84 or higher. Since there will be 70 positions, the applicant will have the job.

4.61 μ = 900 hours and σ = 50 hours. Solving μ − kσ = 700 we obtain k = 4.

So, using Chebyshev’s theorem with P(μ − 4σ < X < μ + 4σ) ≥ 1 − 1/42 = 0.9375,

we obtain P(700 < X < 1100) ≥ 0.9375. Therefore, P(X ≤ 700) ≤ 0.03125.

52 Chapter 4 Mathematical Expectation

4.62 μ = 52 and σ = 6.5. Solving μ + kσ = 71.5 we obtain k = 3. So,

P(μ − 3σ < X < μ + 3σ) ≥ 1 −

1

32 = 0.8889,

which is

P(32.5 < X < 71.5) ≥ 0.8889.

we obtain P(X > 71.5) < 1−0.8889

2 = 0.0556 using the symmetry.

4.63 n = 500, μ = 4.5 and σ = 2.8733. Solving μ + k(σ/√500) = 5 we obtain

k =

5 − 4.5

2.87333/√500

=

0.5

0.1284

= 3.8924.

So, P(4 ≤ ¯X ≤ 5) ≥ 1 − 1

k2 = 0.9340.

4.64 σ2

Z = σ2

−2X+4Y −3 = 4σ2

X + 16σ2

Y = (4)(5) + (16)(3) = 68.

4.65 σ2

Z = σ2

−2X+4Y −3 = 4σ2

X + 16σ2

Y − 16σXY = (4)(5) + (16)(3) − (16)(1) = 52.

4.66 (a) P(6 < X < 18) = P[12 − (2)(3) < X < 12 + (2)(3)] ≥ 1 − 1

22 = 3

4 .

(b) P(3 < X < 21) = P[12 − (3)(3) < X < 12 + (3)(3)] ≥ 1 − 1

32 = 8

9 .

4.67 (a) P(|X − 10| ≥ 3) = 1 − P(|X − 10| < 3)

= 1 − P[10 − (3/2)(2) < X < 10 + (3/2)(2)] ≤ 1 −

h

1 − 1

(3/2)2

i

= 4

9 .

(b) P(|X − 10| < 3) = 1 − P(|X − 10| ≥ 3) ≥ 1 − 4

9 = 5

9 .

(c) P(5 < X < 15) = P[10 − (5/2)(2) < X < 10 + (5/2)(2)] ≥ 1 − 1

(5/2)2 = 21

25 .

(d) P(|X − 10| ≥ c) ≤ 0.04 implies that P(|X − 10| < c) ≥ 1 − 0.04 = 0.96.

Solving 0.96 = 1 − 1

k2 we obtain k = 5. So, c = kσ = (5)(2) = 10.

4.68 μ = E(X) = 6

R 1

0 x2(1 − x) dx = 0.5, E(X2) = 6

R 1

0 x3(1 − x) dx = 0.3, which imply

σ2 = 0.3 − (0.5)2 = 0.05 and σ = 0.2236. Hence,

P(μ − 2σ < X < μ + 2σ) = P(0.5 − 0.4472 < X < 0.5 + 0.4472)

= P(0.0528 < X < 0.9472) = 6

Z 0.9472

0.0528

x(1 − x) dx = 0.9839,

compared to a probability of at least 0.75 given by Chebyshev’s theorem.

4.69 It is easy to see that the expectations of X and Y are both 3.5. So,

(a) E(X + Y ) = E(X) + E(Y ) = 3.5 + 3.5 = 7.0.

(b) E(X − Y ) = E(X) − E(Y ) = 0.

Solutions for Exercises in Chapter 4 53

(c) E(XY ) = E(X)E(Y ) = (3.5)(3.5) = 12.25.

4.70 E(Z) = E(XY ) = E(X)E(Y ) =

R 1

0

R

∞

2 16xy(y/x3) dx dy = 8/3.

4.71 E[g(X, Y )] = E(X/Y 3 + X2Y ) = E(X/Y 3) + E(X2Y ).

E(X/Y 3) =

R 2

1

R 1

0

2x(x+2y)

7y3 dx dy = 2

7

R 2

1

1

3y3 + 1

y2

dy = 15

84 ;

E(X2Y ) =

R 2

1

R 1

0

2x2y(x+2y)

7 dx dy = 2

7

R 2

1 y

1

4 + 2y

3

dy = 139

252 .

Hence, E[g(X, Y )] = 15

84 + 139

252 = 46

63 .

4.72 μX = μY = 3.5. σ2

X = σ2

Y = [(1)2 + (2)2 + · · · + (6)2](1/6) − (3.5)2 = 35

12 .

(a) σ2X−Y = 4σ2

X + σ2

Y = 175

12 ;

(b) σX+3Y −5 = σ2

X + 9σ2

Y = 175

6 .

4.73 (a) μ = 1

5

R 5

0 x dx = 2.5, σ2 = E(X2) − μ2 = 1

5x2

R 5

0 x2 dx − 2.52 = 2.08.

So, σ = √σ2 = 1.44.

(b) By Chebyshev’s theorem,

P[2.5 − (2)(1.44) < X < 2.5 + (2)(1.44)] = P(−0.38 < X < 5.38) ≥ 0.75.

Using integration, P(−0.38 < X < 5.38) = 1 ≥ 0.75;

P[2.5 − (3)(1.44) < X < 2.5 + (3)(1.44)] = P(−1.82 < X < 6.82) ≥ 0.89.

Using integration, P(−1.82 < X < 6.82) = 1 ≥ 0.89.

4.74 P = I2R with R = 50, μI = E(I) = 15 and σ2

I = V ar(I) = 0.03.

E(P) = E(I2R) = 50E(I2) = 50[V ar(I) + μ2

I ] = 50(0.03 + 152) = 11251.5. If we use

the approximation formula, with g(I) = I2, g′(I) = 2I and g′′(I) = 2, we obtain,

E(P) ≈ 50

g(μI) + 2

σ2

I

2

= 50(152 + 0.03) = 11251.5.

Since V ar[g(I)] ≈

h

@g(i)

@i

i2

i=μI

σ2

I , we obtain

V ar(P) = 502V ar(I2) = 502(2μI)2σ2

I = 502(30)2(0.03) = 67500.

4.75 For 0 < a < 1, since g(a) =

∞P

x=0

ax = 1

1−a , g′(a) =

∞P

x=1

xax−1 = 1

(1−a)2 and

g′′(a) =

∞P

x=2

x(x − 1)ax−2 = 2

(1−a)3 .

54 Chapter 4 Mathematical Expectation

(a) E(X) = (3/4)

∞P

x=1

x(1/4)x = (3/4)(1/4)

∞P

x=1

x(1/4)x−1 = (3/16)[1/(1 − 1/4)2]

= 1/3, and E(Y ) = E(X) = 1/3.

E(X2) − E(X) = E[X(X − 1)] = (3/4)

∞P

x=2

x(x − 1)(1/4)x

= (3/4)(1/4)2 ∞P

x=2

x(x − 1)(1/4)x−2 = (3/43)[2/(1 − 1/4)3] = 2/9.

So, V ar(X) = E(X2) − [E(X)]2 = [E(X2) − E(X)] + E(X) − [E(X)]2

2/9 + 1/3 − (1/3)2 = 4/9, and V ar(Y ) = 4/9.

(b) E(Z) = E(X) + E(Y ) = (1/3) + (1/3) = 2/3, and

V ar(Z) = V ar(X +Y ) = V ar(X)+V ar(Y ) = (4/9)+ (4/9) = 8/9, since X and

Y are independent (from Exercise 3.79).

4.76 (a) g(x) = 3

2

R 1

0 (x2 + y2) dy = 1

2 (3x2 + 1) for 0 < x < 1 and

h(y) = 1

2 (3y2 + 1) for 0 < y < 1.

Since f(x, y) 6= g(x)h(y), X and Y are not independent.

(b) E(X + Y ) = E(X) + E(Y ) = 2E(X) =

R 1

0 x(3x2 + 1) dx = 3/4 + 1/2 = 5/4.

E(XY ) = 3

2

R 1

0

R 1

0 xy(x2 + y2) dx dy = 3

2

R 1

0 y

1

4 + y2

2

dy

= 3

2

1

4

1

2

+

1

2

1

4

= 3

8 .

(c) V ar(X) = E(X2) − [E(X)]2 = 1

2

R 1

0 x2(3x2 + 1) dx −

5

8

2 = 7

15 − 25

64 = 73

960 , and

V ar(Y ) = 73

960 . Also, Cov(X, Y ) = E(XY ) − E(X)E(Y ) = 3

8 −

5

8

2

= − 1

64 .

(d) V ar(X + Y ) = V ar(X) + V ar(Y ) + 2Cov(X, Y ) = 2 73

960 − 2 1

64 = 29

240 .

4.77 (a) E(Y ) =

R

∞

0 ye−y/4 dy = 4.

(b) E(Y 2) =

R

∞

0 y2e−y/4 dy = 32 and V ar(Y ) = 32 − 42 = 16.

4.78 (a) The density function is shown next.

x

f(x)

0 7 8

1

(b) E(Y ) =

R 8

7 y dy = 1

2 [82 − 72] = 15

2 = 7.5,

E(Y 2) =

R 8

7 y2 dy = 1

3 [83 − 73] = 169

3 , and V ar(Y ) = 169

3 −

15

2

2

= 1

12 .

4.79 Using the exact formula, we have

E(eY ) =

Z 8

7

ey dy = ey|8

7 = 1884.32.

Solutions for Exercises in Chapter 4 55

Using the approximation, since g(y) = ey, so g′′(y) = ey. Hence, using the approximation

formula,

E(eY ) ≈ eμY + eμY σ2

Y

2

=

1 +

1

24

e7.5 = 1883.38.

The approximation is very close to the true value.

4.80 Using the exact formula, E(Z2) =

R 8

7 e2y dy = 1

2 e2y|8

7 = 3841753.12. Hence,

V ar(Z) = E(Z2) − [E(Z)]2 = 291091.3.

Using the approximation formula, we have

V ar(eY ) = (eμY )2V ar(Y ) =

e(2)(7.5)

12

= 272418.11.

The approximation is not so close to each other. One reason is that the first order

approximation may not always be good enough.

4.81 Define I1 = {xi| |xi − μ| < kσ} and I2 = {xi| |xi − μ| ≥ kσ}. Then

σ2 = E[(X − μ)2] =

X

x

(x − μ)2f(x) =

X

xi∈I1

(xi − μ)2f(xi) +

X

xi∈I2

(xi − μ)2f(xi)

≥

X

xi∈I2

(xi − μ)2f(xi) ≥ k2σ2

X

xi∈I2

f(xi) = k2σ2P(|X − μ| ≥ kσ),

which implies

P(|X − μ| ≥ kσ) ≤

1

k2 .

Hence, P(|X − μ| < kσ) ≥ 1 − 1

k2 .

4.82 E(XY ) =

R 1

0

R 1

0 xy(x+y) dx dy = 1

3 , E(X) =

R 1

0

R 1

0 x(x+y) dx dy = 7

12 and E(Y ) = 7

12 .

Therefore, σXY = E(XY ) − μXμY = 1

3 −

7

12

2

= − 1

144 .

4.83 E(Y − X) =

R 1

0

R y

0 2(y − x) dx dy =

R 1

0 y2 dy = 1

3 . Therefore, the average amount of

kerosene left in the tank at the end of each day is (1/3)(1000) = 333 liters.

4.84 (a) E(X) =

R

∞

0

x

5 e−x/5 dx = 5.

(b) E(X2) =

R

∞

0

x2

5 e−x/5 dx = 50, so V ar(X) = 50 − 52 = 25, and σ = 5.

(c) E[(X + 5)2] = E{[(X − 5) + 10]2} = E[(X − 5)2] + 102 + 20E(X − 5)

= V ar(X) + 100 = 125.

4.85 E(XY ) = 24

R 1

0

R 1−y

0 x2y2 dx dy = 8

R 1

0 y2(1 − y)3 dy = 2

15 ,

μX = 24

R 1

0

R 1−y

0 x2y dx dy = 2

5 and μY = 24

R 1

0

R 1−y

0 xy2 dx dy = 2

5 . Therefore,

σXY = E(XY ) − μXμY = 2

15 −

2

5

2

= − 2

75 .

56 Chapter 4 Mathematical Expectation

4.86 E(X + Y ) =

R 1

0

R 1−y

0 24(x + y)xy dx dy = 4

5 .

4.87 (a) E(X) =

R

∞

0

x

900e−x/900 dx = 900 hours.

(b) E(X2) =

R

∞

0

x2

900e−x/900 dx = 1620000 hours2.

(c) V ar(X) = E(X2) − [E(X)]2 = 810000 hours2 and σ = 900 hours.

4.88 It is known g(x) = 2

3 (x + 1), for 0 < x < 1, and h(y) = 1

3 (1 + 4y), for 0 < y < 1.

(a) μX =

R 1

0

2

3x(x + 1) dx = 5

9 and μY =

R 1

0

1

3y(1 + 4y) dy = 11

18 .

(b) E[(X + Y )/2] = 1

2 [E(X) + E(Y )] = 7

12 .

4.89 Cov(aX, bY ) = E[(aX − aμX)(bY − bμY )] = abE[(X − μX)(Y − μY )] = abCov(X, Y ).

4.90 It is known μ = 900 and σ = 900. For k = 2,

P(μ − 2σ < X < μ + 2σ) = P(−900 < X < 2700) ≥ 0.75

using Chebyshev’s theorem. On the other hand,

P(μ − 2σ < X < μ + 2σ) = P(−900 < X < 2700) = 1 − e−3 = 0.9502.

For k = 3, Chebyshev’s theorem yields

P(μ − 3σ < X < μ + 3σ) = P(−1800 < X < 3600) ≥ 0.8889,

while P(−1800 < X < 3600) = 1 − e−4 = 0.9817.

4.91 g(x) =

R 1

0

16y

x3 dy = 8

x3 , for x > 2, with μX =

R

∞

2

8

x2 dx = − 8

x

∞2

= 4,

h(y) =

R

∞

2

16y

x3 dx = − 8y

x2

∞2

= 2y, for 0 < y < 1, with μY =

R 1

0 2y2 = 2

3 , and

E(XY ) =

R

∞

2

R 1

0

16y2

x2 dy dx = 16

3

R

∞

2

1

x2 dx = 8

3 . Hence,

σXY = E(XY ) − μXμY = 8

3 − (4)

2

3

= 0.

4.92 Since σXY = 1, σ2

X = 5 and σ2

Y = 3, we have ρ = XY

X Y

= 1 √(5)(3)

= 0.2582.

4.93 (a) From Exercise 4.37, we have σ2 = 1/18, so σ = 0.2357.

(b) Also, μX = 1/3 from Exercise 4.12. So,

P(μ − 2σ < X < μ + 2σ) = P[1/3 − (2)(0.2357) < X < 1/3 + (2)(0.2357)]

= P(0 < X < 0.8047) =

Z 0.8047

0

2(1 − x) dx = 0.9619.

Using Chebyshev’s theorem, the probability of this event should be larger than

0.75, which is true.

(c) P(profit > $500) = P(X > 0.1) =

R 1

0.1 2(1 − x) = 0.81.

Solutions for Exercises in Chapter 4 57

4.94 Since g(0)h(0) = (0.17)(0.23) 6= 0.10 = f(0, 0), X and Y are not independent.

4.95 E(X) = (−5000)(0.2) + (10000)(0.5) + (30000)(0.3) = $13, 000.

4.96 (a) f(x) =

3

x

(0.15)x(0.85)3−x, for x = 0, 1, 2, 3.

x 0 1 2 3

f(x) 0.614125 0.325125 0.057375 0.003375

(b) E(X) = 0.45.

(c) E(X2) = 0.585, so V ar(X) = 0.585 − 0.452 = 0.3825.

(d) P(X ≤ 2) = 1 − P(X = 3) = 1 − 0.003375 = 0.996625.

(e) 0.003375.

(f) Yes.

4.97 (a) E(X) = (−$15k)(0.05)+($15k)(0.15)+($25k)(0.30)+($40k)(0.15)+($50k)(0.10)+

($100k)(0.05) + ($150k)(0.03) + ($200k)(0.02) = $33.5k.

(b) E(X2) = 2, 697, 500, 000 dollars2. So, σ =

p

E(X2) − [E(X)]2 = $39.689k.

4.98 (a) E(X) = 3

4×503

R 50

−50 x(502 − x2) dx = 0.

(b) E(X2) = 3

4×503

R 50

−50 x2(502 − x2) dx = 500.

(c) σ =

p

E(X2) − [E(X)]2 = √500 − 0 = 22.36.

4.99 (a) The marginal density of X is

x1 0 1 2 3 4

fX1(x1) 0.13 0.21 0.31 0.23 0.12

(b) The marginal density of Y is

x2 0 1 2 3 4

fX2(x2) 0.10 0.30 0.39 0.15 0.06

(c) Given X2 = 3, the conditional density function of X1 is f(x1, 3)/0.15. So

x1 0 1 2 3 4

fX2(x2) 7

15

1

5

1

15

1

5

1

15

(d) E(X1) = (0)(0.13) + (1)(0.21) + (2)(0.31) + (3)(0.23) + (4)(0.12) = 2.

(e) E(X2) = (0)(0.10) + (1)(0.30) + (2)(0.39) + (3)(0.15) + (4)(0.06) = 1.77.

(f) E(X1|X2 = 3) = (0)

7

15

+ (1)

1

5

+ (2)

1

15

+ (3)

1

5

+ (4)

1

15

= 18

15 = 6

5 = 1.2.

(g) E(X2

1 ) = (0)2(0.13) + (1)2(0.21) + (2)2(0.31) + (3)2(0.23) + (4)2(0.12) = 5.44.

So, σX1 =

p

E(X2

1 ) − [E(X1)]2 = √5.44 − 22 = √1.44 = 1.2.

4.100 (a) The marginal densities of X and Y are, respectively,

58 Chapter 4 Mathematical Expectation

x 0 1 2

g(x) 0.2 0.32 0.48

y 0 1 2

h(y) 0.26 0.35 0.39

The conditional density of X given Y = 2 is

x 0 1 2

fX|Y =2(x|2) 4

39

5

39

30

39

(b) E(X) = (0)(0.2) + (1)(0.32) + (2)(0.48) = 1.28,

E(X2) = (0)2(0.2) + (1)2(0.32) + (2)2(0.48) = 2.24, and

V ar(X) = 2.24 − 1.282 = 0.6016.

(c) E(X|Y = 2) = (1) 5

39 + (2) 30

39 = 65

39 and E(X2|Y = 2) = (1)2 5

39 + (2)2 30

39 = 125

39 . So,

V ar(X) = 125

39 −

65

39

2 = 650

1521 = 50

117 .

4.101 The profit is 8X + 3Y − 10 for each trip. So, we need to calculate the average of this

quantity. The marginal densities of X and Y are, respectively,

x 0 1 2

g(x) 0.34 0.32 0.34

y 0 1 2 3 4 5

h(y) 0.05 0.18 0.15 0.27 0.19 0.16

So, E(8X+3Y −10) = (8)[(1)(0.32)+(2)(0.34)]+(3)[(1)(0.18)+(2)(0.15)+(3)(0.27)+

(4)(0.19) + (5)(0.16)] − 10 = $6.55.

4.102 Using the approximation formula, V ar(Y ) ≈

Pk

i=1

h

@h(x1,x2,...,xk)

@xi

i2

xi=μi, 1≤i≤k

σ2

i , we

have

V ar( ˆ Y ) ≈

X2

i=0

∂eb0+b1k1+b2k2

∂bi

2

bi= i, 0≤i≤2

σ2

bi = e2( 0+k1 1+k2 2)(σ2

0 + k2

1σ2

1 + k2

2σ2

2).

4.103 (a) E(Y ) = 10

R 1

0 y(1 − y)9 dy = − y(1 − y)10|1

0 +

R 1

0 (1 − y)10 dy = 1

11 .

(b) E(1 − Y ) = 1 − E(Y ) = 10

11 .

(c) V ar(Z) = V ar(1 − Y ) = V ar(Y ) = E(Y 2) − [E(Y )]2 = 10

112×12 = 0.006887.

Chapter 5

Some Discrete Probability

Distributions

5.1 This is a uniform distribution: f(x) = 1

10 , for x = 1, 2, . . . , 10.

Therefore P(X < 4) =

P3

x=1

f(x) = 3

10 .

5.2 Binomial distribution with n = 12 and p = 0.5. Hence

P(X = 3) = P(X ≤ 3) − P(X ≤ 2) = 0.0730 − 0.0193 = 0.0537.

5.3 μ =

P10

x=1

x

10 = 5.5, and σ2 =

P10

x=1

(x−5.5)2

10 = 8.25.

5.4 For n = 5 and p = 3/4, we have

(a) P(X = 2) =

5

2

(3/4)2(1/4)3 = 0.0879,

(b) P(X ≤ 3) =

P3

x=0

b(x; 5, 3/4) = 1 − P(X = 4) − P(X = 5)

= 1 −

5

4

(3/4)4(1/4)1 −

5

5

(3/4)5(1/4)0 = 0.3672.

5.5 We are considering a b(x; 20, 0.3).

(a) P(X ≥ 10) = 1 − P(X ≤ 9) = 1 − 0.9520 = 0.0480.

(b) P(X ≤ 4) = 0.2375.

(c) P(X = 5) = 0.1789. This probability is not very small so this is not a rare event.

Therefore, P = 0.30 is reasonable.

5.6 For n = 6 and p = 1/2.

(a) P(2 ≤ X ≤ 5) = P(X ≤ 5) − P(X ≤ 1) = 0.9844 − 0.1094.

(b) P(X < 3) = P(X ≤ 2) = 0.3438.

5.7 p = 0.7.

59

60 Chapter 5 Some Discrete Probability Distributions

(a) For n = 10, P(X < 5) = P(X ≤ 4) = 0.0474.

(b) For n = 20, P(X < 10) = P(X ≤ 9) = 0.0171.

5.8 For n = 8 and p = 0.6, we have

(a) P(X = 3) = b(3; 8, 0.6) = P(X ≤ 3) − P(X ≤ 2) = 0.1737 − 0.0498 = 0.1239.

(b) P(X ≥ 5) = 1 − P(X ≤ 4) = 1 − 0.4059 = 0.5941.

5.9 For n = 15 and p = 0.25, we have

(a) P(3 ≤ X ≤ 6) = P(X ≤ 6) − P(X ≤ 2) = 0.9434 − 0.2361 = 0.7073.

(b) P(X < 4) = P(X ≤ 3) = 0.4613.

(c) P(X > 5) = 1 − P(X ≤ 5) = 1 − 0.8516 = 0.1484.

5.10 From Table A.1 with n = 12 and p = 0.7, we have

(a) P(7 ≤ X ≤ 9) = P(X ≤ 9) − P(X ≤ 6) = 0.7472 − 0.1178 = 0.6294.

(b) P(X ≤ 5) = 0.0386.

(c) P(X ≥ 8) = 1 − P(X ≤ 7) = 1 − 0.2763 = 0.7237.

5.11 From Table A.1 with n = 7 and p = 0.9, we have

P(X = 5) = P(X ≤ 5) − P(X ≤ 4) = 0.1497 − 0.0257 = 0.1240.

5.12 From Table A.1 with n = 9 and p = 0.25, we have P(X < 4) = 0.8343.

5.13 From Table A.1 with n = 5 and p = 0.7, we have

P(X ≥ 3) = 1 − P(X ≤ 2) = 1 − 0.1631 = 0.8369.

5.14 (a) n = 4, P(X = 4) = 1 − 0.3439 = 0.6561.

(b) Assuming the series went to the seventh game, the probability that the Bulls won

3 of the first 6 games and then the seventh game is given by

6

3

(0.9)3(0.1)3

(0.9) = 0.0131.

(c) The probability that the Bulls win is always 0.9.

5.15 p = 0.4 and n = 5.

(a) P(X = 0) = 0.0778.

(b) P(X < 2) = P(X ≤ 1) = 0.3370.

(c) P(X > 3) = 1 − P(X ≤ 3) = 1 − 0.9130 = 0.0870.

Solutions for Exercises in Chapter 5 61

5.16 Probability of 2 or more of 4 engines operating when p = 0.6 is

P(X ≥ 2) = 1 − P(X ≤ 1) = 0.8208,

and the probability of 1 or more of 2 engines operating when p = 0.6 is

P(X ≥ 1) = 1 − P(X = 0) = 0.8400.

The 2-engine plane has a slightly higher probability for a successful flight when p = 0.6.

5.17 Since μ = np = (5)(0.7) = 3.5 and σ2 = npq = (5)(0.7)(0.3) = 1.05 with σ = 1.025.

Then μ± 2σ = 3.5± (2)(1.025) = 3.5± 2.050 or from 1.45 to 5.55. Therefore, at least

3/4 of the time when 5 people are selected at random, anywhere from 2 to 5 are of the

opinion that tranquilizers do not cure but only cover up the real problem.

5.18 (a) μ = np = (15)(0.25) = 3.75.

(b) With k = 2 and σ = √npq =

p

(15)(0.25)(0.75) = 1.677, μ ± 2σ = 3.75 ± 3.354

or from 0.396 to 7.104.

5.19 Let X1 = number of times encountered green light with P(Green) = 0.35,

X2 = number of times encountered yellow light with P(Yellow) = 0.05, and

X3 = number of times encountered red light with P(Red) = 0.60. Then

f(x1, x2, x3) =

n

x1, x2, x3

(0.35)x1(0.05)x2(0.60)x3.

5.20 (a)

10

2,5,3

(0.225)2(0.544)5(0.231)3 = 0.0749.

(b)

10

10

(0.544)10(0.456)0 = 0.0023.

(c)

10

0

(0.225)0(0.775)10 = 0.0782.

5.21 Using the multinomial distribution with required probability is

7

0, 0, 1, 4, 2

(0.02)(0.82)4(0.1)2 = 0.0095.

5.22 Using the multinomial distribution, we have

8

5,2,1

(1/2)5(1/4)2(1/4) = 21/256.

5.23 Using the multinomial distribution, we have

9

3, 3, 1, 2

(0.4)3(0.2)3(0.3)(0.2)2 = 0.0077.

5.24 p = 0.40 and n = 6, so P(X = 4) = P(X ≤ 4)−P(X ≤ 3) = 0.9590−0.8208 = 0.1382.

5.25 n = 20 and the probability of a defective is p = 0.10. So, P(X ≤ 3) = 0.8670.

62 Chapter 5 Some Discrete Probability Distributions

5.26 n = 8 and p = 0.60;

(a) P(X = 6) =

8

6

(0.6)6(0.4)2 = 0.2090.

(b) P(X = 6) = P(X ≤ 6) − P(X ≤ 5) = 0.8936 − 0.6846 = 0.2090.

5.27 n = 20 and p = 0.90;

(a) P(X = 18) = P(X ≤ 18) − P(X ≤ 17) = 0.6083 − 0.3231 = 0.2852.

(b) P(X ≥ 15) = 1 − P(X ≤ 14) = 1 − 0.0113 = 0.9887.

(c) P(X ≤ 18) = 0.6083.

5.28 n = 20;

(a) p = 0.20, P(X ≥ x) ≤ 0.5 and P(X < x) > 0.5 yields x = 4.

(b) p = 0.80, P(Y ≥ y) ≥ 0.8 and P(Y < y) < 0.2 yields y = 14.

5.29 Using the hypergeometric distribution, we get

(a)

(12

2 )(40

5 )

(52

7 )

= 0.3246.

(b) 1 −

(48

7 )

(52

7 )

= 0.4496.

5.30 P(X ≥ 1) = 1 − P(X = 0) = 1 − h(0; 15, 3, 6) = 1 −

(6

0)(9

3)

(15

3 )

= 53

65 .

5.31 Using the hypergeometric distribution, we get h(2; 9, 6, 4) =

(4

2)(5

4)

(9

6)

= 5

14 .

5.32 (a) Probability that all 4 fire = h(4; 10, 4, 7) = 1

6 .

(b) Probability that at most 2 will not fire =

P2

x=0

h(x; 10, 4, 3) = 29

30 .

5.33 h(x; 6, 3, 4) =

(4

x)( 2

3−x)

(6

3)

, for x = 1, 2, 3.

P(2 ≤ X ≤ 3) = h(2; 6, 3, 4) + h(3; 6, 3, 4) = 4

5 .

5.34 h(2; 9, 5, 4) =

(4

2)(5

3)

(9

5)

= 10

21 .

5.35 P(X ≤ 2) =

P2

x=0

h(x; 50, 5, 10) = 0.9517.

5.36 (a) P(X = 0) = h(0; 25, 3, 3) = 77

115 .

(b) P(X = 1) = h(1; 25, 3, 1) = 3

25 .

5.37 (a) P(X = 0) = b(0; 3, 3/25) = 0.6815.

Solutions for Exercises in Chapter 5 63

(b) P(1 ≤ X ≤ 3) =

P3

x=1

b(x; 3, 1/25) = 0.1153.

5.38 Since μ = (4)(3/10) = 1.2 and σ2 = (4)(3/10)(7/10)(6/9) = 504/900 with σ = 0.7483,

at least 3/4 of the time the number of defectives will fall in the interval

μ ± 2σ = 1.2 ± (2)(0.7483), or from − 0.297 to 2.697,

and at least 8/9 of the time the number of defectives will fall in the interval

μ ± 3σ = 1.2 ± (3)(0.7483) or from − 1.045 to 3.445.

5.39 Since μ = (13)(13/52) = 3.25 and σ2 = (13)(1/4)(3/4)(39/51) = 1.864 with σ = 1.365,

at least 75% of the time the number of hearts lay between

μ ± 2σ = 3.25 ± (2)(1.365) or from 0.52 to 5.98.

5.40 The binomial approximation of the hypergeometric with p = 1 − 4000/10000 = 0.6

gives a probability of

P7

x=0

b(x; 15, 0.6) = 0.2131.

5.41 Using the binomial approximation of the hypergeometric with p = 0.5, the probability

is 1 −

P2

x=0

b(x; 10, 0.5) = 0.9453.

5.42 Using the binomial approximation of the hypergeometric distribution with p = 30/150 =

0.2, the probability is 1 −

P2

x=0

b(x; 10, 0.2) = 0.3222.

5.43 Using the binomial approximation of the hypergeometric distribution with 0.7, the

probability is 1 −

P13

x=10

b(x; 18, 0.7) = 0.6077.

5.44 Using the extension of the hypergeometric distribution the probability is

13

5

13

2

13

3

13

3

52

13

= 0.0129.

5.45 (a) The extension of the hypergeometric distribution gives a probability

2

1

3

1

5

1

2

1

12

4

=

4

33

.

(b) Using the extension of the hypergeometric distribution, we have

2

1

3

1

2

2

12

4

+

2

2

3

1

2

1

12

4

+

2

1

3

2

2

1

12

4

=

8

165

.

64 Chapter 5 Some Discrete Probability Distributions

5.46 Using the extension of the hypergeometric distribution the probability is

2

2

4

1

3

2

9

5

+

2

2

4

2

3

1

9

5

+

2

2

4

3

3

0

9

5

=

17

63

.

5.47 h(5; 25, 15, 10) =

(10

5 )(15

10)

(25

15)

= 0.2315.

5.48 (a)

(2

1)(13

4 )

(15

5 )

= 0.4762.

(b)

(2

2)(13

3 )

(15

5 )

= 0.0952.

5.49 (a)

(3

0)(17

5 )

(20

5 )

= 0.3991.

(b)

(3

2)(17

3 )

(20

5 )

= 0.1316.

5.50 N = 10000, n = 30 and k = 300. Using binomial approximation to the hypergeometric

distribution with p = 300/10000 = 0.03, the probability of {X ≥ 1} can be determined

by

1 − b(0; 30, 0.03) = 1 − (0.97)30 = 0.599.

5.51 Using the negative binomial distribution, the required probability is

b∗(10; 5, 0.3) =

9

4

(0.3)5(0.7)5 = 0.0515.

5.52 From the negative binomial distribution, we obtain

b∗(8; 2, 1/6) =

7

1

(1/6)2(5/6)6 = 0.0651.

5.53 (a) P(X > 5) =

∞P x

=

6

p(x; 5) = 1 −

P5

x=0

p(x; 5) = 0.3840.

(b) P(X = 0) = p(0; 5) = 0.0067.

5.54 (a) Using the negative binomial distribution, we get

b∗(7; 3, 1/2) =

6

2

(1/2)7 = 0.1172.

(b) From the geometric distribution, we have g(4; 1/2) = (1/2)(1/2)3 = 1/16.

Solutions for Exercises in Chapter 5 65

5.55 The probability that all coins turn up the same is 1/4. Using the geometric distribution

with p = 3/4 and q = 1/4, we have

P(X < 4) =

X3

x=1

g(x; 3/4) =

X3

x=1

(3/4)(1/4)x−1 =

63

64

.

5.56 (a) Using the geometric distribution, we have g(5; 2/3) = (2/3)(1/3)4 = 2/243.

(b) Using the negative binomial distribution, we have

b∗(5; 3, 2/3) =

4

2

(2/3)3(1/3)2 =

16

81

.

5.57 Using the geometric distribution

(a) P(X = 3) = g(3; 0.7) = (0.7)(0.3)2 = 0.0630.

(b) P(X < 4) =

P3

x=1

g(x; 0.7) =

P3

x=1

(0.7)(0.3)x−1 = 0.9730.

5.58 (a) Using the Poisson distribution with x = 5 and μ = 3, we find from Table A.2 that

p(5; 3) =

X5

x=0

p(x; 3) −

X4

x=0

p(x; 3) = 0.1008.

(b) P(X < 3) = P(X ≤ 2) = 0.4232.

(c) P(X ≥ 2) = 1 − P(X ≤ 1) = 0.8009.

5.59 (a) P(X ≥ 4) = 1 − P(X ≤ 3) = 0.1429.

(b) P(X = 0) = p(0; 2) = 0.1353.

5.60 (a) P(X < 4) = P(X ≤ 3) = 0.1512.

(b) P(6 ≤ X ≤ 8) = P(X ≤ 8) − P(X ≤ 5) = 0.4015.

5.61 (a) Using the negative binomial distribution, we obtain

b∗(6; 4, 0.8) =

5

3

(0.8)4(0.2)2 = 0.1638.

(b) From the geometric distribution, we have g(3; 0.8) = (0.8)(0.2)2 = 0.032.

5.62 (a) Using the Poisson distribution with μ = 12, we find from Table A.2 that

P(X < 7) = P(X ≤ 6) = 0.0458.

(b) Using the binomial distribution with p = 0.0458, we get

b(2; 3, 0.0458) =

3

2

(0.0458)2(0.9542) = 0.0060.

66 Chapter 5 Some Discrete Probability Distributions

5.63 (a) Using the Poisson distribution with μ = 5, we find

P(X > 5) = 1 − P(X ≤ 5) = 1 − 0.6160 = 0.3840.

(b) Using the binomial distribution with p = 0.3840, we get

b(3; 4, 0.384) =

4

3

(0.3840)3(0.6160) = 0.1395.

(c) Using the geometric distribution with p = 0.3840, we have

g(5; 0.384) = (0.394)(0.616)4 = 0.0553.

5.64 μ = np = (2000)(0.002) = 4, so P(X < 5) = P(X ≤ 4) ≈

P4

x=0

p(x; 4) = 0.6288.

5.65 μ = np = (10000)(0.001) = 10, so

P(6 ≤ X ≤ 8) = P(X ≤ 8) − P(X ≤ 5) ≈

X8

x=0

p(x; 10) −

X5

x=0

p(x; 10) = 0.2657.

5.66 (a) μ = np = (1875)(0.004) = 7.5, so P(X < 5) = P(X ≤ 4) ≈ 0.1321.

(b) P(8 ≤ X ≤ 10) = P(X ≤ 10) − P(X ≤ 7) ≈ 0.8622 − 0.5246 = 0.3376.

5.67 (a) μ = (2000)(0.002) = 4 and σ2 = 4.

(b) For k = 2, we have μ ± 2σ = 4 ± 4 or from 0 to 8.

5.68 (a) μ = (10000)(0.001) = 10 and σ2 = 10.

(b) For k = 3, we have μ ± 3σ = 10 ± 3√10 or from 0.51 to 19.49.

5.69 (a) P(X ≤ 3|λt = 5) = 0.2650.

(b) P(X > 1|λt = 5) = 1 − 0.0404 = 0.9596.

5.70 (a) P(X = 4|λt = 6) = 0.2851 − 0.1512 = 0.1339.

(b) P(X ≥ 4|λt = 6) = 1 − 0.1512 = 0.8488.

(c) P(X ≥ 75|λt = 72) = 1 −

P74

x=0

p(x; 74) = 0.3773.

5.71 (a) P(X > 10|λt = 14) = 1 − 0.1757 = 0.8243.

(b) λt = 14.

5.72 μ = np = (1875)(0.004) = 7.5.

5.73 μ = (4000)(0.001) = 4.

Solutions for Exercises in Chapter 5 67

5.74 μ = 1 and σ2 = 0.99.

5.75 μ = λt = (1.5)(5) = 7.5 and P(X = 0|λt = 7.5) = e−7.5 = 5.53 × 10−4.

5.76 (a) P(X ≤ 1|λt = 2) = 0.4060.

(b) μ = λt = (2)(5) = 10 and P(X ≤ 4|λt = 10) = 0.0293.

5.77 (a) P(X > 10|λt = 5) = 1 − P(X ≤ 10|λt = 5) = 1 − 0.9863 = 0.0137.

(b) μ = λt = (5)(3) = 15, so P(X > 20|λt = 15) = 1 − P(X ≤ 20|λ = 15) =

1 − 0.9170 = 0.0830.

5.78 p = 0.03 with a g(x; 0.03). So, P(X = 16) = (0.03)(1 − 0.03)15 = 0.0190 and

μ = 1

0.03 − 1 = 32.33.

5.79 So, Let Y = number of shifts until it fails. Then Y follows a geometric distribution

with p = 0.10. So,

P(Y ≤ 6) = g(1; 0.1) + g(2; 0.1) + · · · + g(6; 0.1)

= (0.1)[1 + (0.9) + (0.9)2 + · · · + (0.9)5] = 0.4686.

5.80 (a) The number of people interviewed before the first refusal follows a geometric

distribution with p = 0.2. So.

P(X ≥ 51) =

∞X

x=51

(0.2)(1 − 0.2)x = (0.2)

(1 − 0.2)50

1 − (1 − 0.2)

= 0.00001,

which is a very rare event.

(b) μ = 1

0.2 − 1 = 4.

5.81 n = 15 and p = 0.05.

(a) P(X ≥ 2) = 1 − P(X ≤ 1) = 1 −

P1

x=0

15

x

(0.05)x(1 − 0.05)15−x = 1 − 0.8290 =

0.1710.

(b) p = 0.07. So, P(X ≤ 1) =

P1

x=0

15

x

(0.07)x(1 − 0.07)15−x = 1 − 0.7168 = 0.2832.

5.82 n = 100 and p = 0.01.

(a) P(X > 3) = 1 − P(X ≤ 3) = 1 −

P3

x=0

100

x

(0.01)x(1 − 0.01)100−x = 1 − 0.9816 =

0.0184.

(b) For p = 0.05, P(X ≤ 3) =

P3

x=0

100

x

(0.05)x(1 − 0.05)100−x = 0.2578.

68 Chapter 5 Some Discrete Probability Distributions

5.83 Using the extension of the hypergeometric distribution, the probability is

5

2

7

3

4

1

3

1

4

2

5

2

= 0.0308.

5.84 λ = 2.7 call/min.

(a) P(X ≤ 4) =

P4

x=0

e−2.7(2.7)x

x! = 0.8629.

(b) P(X ≤ 1) =

P1

x=0

e−2.7(2.7)x

x! = 0.2487.

(c) λt = 13.5. So,

P(X > 10) = 1 − P(X ≤ 10) = 1 −

P10

x=0

e−13.5(13.5)x

x! = 1 − 0.2971 = 0.7129.

5.85 n = 15 and p = 0.05.

(a) P(X = 5) =

15

5

(0.05)5(1 − 0.05)10 = 0.000562.

(b) I would not believe the claim of 5% defective.

5.86 λ = 0.2, so λt = (0.2)(5) = 1.

(a) P(X ≤ 1) =

P1

x=0

e−1(1)x

x! = 2e−1 = 0.7358. Hence, P(X > 1) = 1−0.7358 = 0.2642.

(b) λ = 0.25, so λt = 1.25. P(X|le1) =

P1

x=0

e−1.25(1.25)x

x! = 0.6446.

5.87 (a) 1 − P(X ≤ 1) = 1 −

P1

x=0

100

x

(0.01)x(0.99)100−x = 1 − 0.7358 = 0.2642.

(b) P(X ≤ 1) =

P1

x=0

100

x

(0.05)x(0.95)100−x = 0.0371.

5.88 (a) 100 visits/60 minutes with λt = 5 visits/3 minutes. P(X = 0) = e−550

0! = 0.0067.

(b) P(X > 5) = 1 −

P5

x=0

e55x

x! = 1 − 0.6160 = 0.3840.

5.89 (a) P(X ≥ 1) = 1 − P(X = 0) = 1 −

4

0

1

6

0 5

6

4

= 1 − 0.4822 = 0.5177.

(b) P(X ≥ 1) = 1 − P(X = 0) = 1 −

24

0

1

36

0

35

36

24

= 1 − 0.5086 = 0.4914.

5.90 n = 5 and p = 0.4; P(X ≥ 3) = 1 − P(X ≤ 2) = 1 − 0.6826 = 0.3174.

5.91 (a) μ = bp = (200)(0.03) = 6.

(b) σ2 = npq = 5.82.

Solutions for Exercises in Chapter 5 69

(c) P(X = 0) = e6(6)0

0! = 0.0025 (using the Poisson approximation).

P(X = 0) = (0.97)200 = 0.0023 (using the binomial distribution).

5.92 (a) p10q0 = (0.99)10 = 0.9044.

(b) p10q12−10 = (0.99)10(0.01)2 = (0.9044)(0.0001) = 0.00009.

5.93 n = 75 with p = 0.999.

(a) X = the number of trials, and P(X = 75) = (0.999)75(0.001)0 = 0.9277.

(b) Y = the number of trials before the first failure (geometric distribution), and

P(Y = 20) = (0.001)(0.999)19 = 0.000981.

(c) 1 − P(no failures) = 1 − (0.001)0(0.999)10 = 0.01.

5.94 (a)

10

1

pq9 = (10)(0.25)(0.75)9 = 0.1877.

(b) Let X be the number of drills until the first success. X follows a geometric

distribution with p = 0.25. So, the probability of having the first 10 drills being

failure is q10 = (0.75)10 = 0.056. So, there is a small prospects for bankruptcy.

Also, the probability that the first success appears in the 11th drill is pq10 = 0.014

which is even smaller.

5.95 It is a negative binomial distribution.

x−1

k−1

pkqx−k =

6−1

2−1

(0.25)2(0.75)4 = 0.0989.

5.96 It is a negative binomial distribution.

x−1

k−1

pkqx−k =

4−1

2−1

(0.5)2(0.5)2 = 0.1875.

5.97 n = 1000 and p = 0.01, with μ = (1000)(0.01) = 10. P(X < 7) = P(X ≤ 6) = 0.1301.

5.98 n = 500;

(a) If p = 0.01,

P(X ≥ 15) = 1 − P(X ≤ 14) = 1 −

X14

x=0

500

x

(0.01)x(0.99)500−x = 0.00021.

This is a very rare probability and thus the original claim that p = 0.01 is questionable.

(b) P(X = 3) =

500

3

(0.01)3(0.99)497 = 0.1402.

(c) For (a), if p = 0.01, μ = (500)(0.01) = 5. So

P(X ≥ 15) = 1 − P(X ≤ 14) = 1 − 0.9998 = 0.0002.

For (b),

P(X = 3) = 0.2650 − 0.1247 = 0.1403.

5.99 N = 50 and n = 10.

70 Chapter 5 Some Discrete Probability Distributions

(a) k = 2; P(X ≥ 1) = 1 − P(X = 0) = 1 −

(2

0)(48

10)

(50

10)

= 1 − 0.6367 = 0.3633.

(b) Even though the lot contains 2 defectives, the probability of reject the lot is not

very high. Perhaps more items should be sampled.

(c) μ = (10)(2/50) = 0.4.

5.100 Suppose n items need to be sampled. P(X ≥ 1) = 1−

(2

0)(48

n )

(50

n )

= 1 − (50−n)(49−n)

(50)(49) ≥ 0.9.

The solution is n = 34.

5.101 Define X = number of screens will detect. Then X ∼ b(x; 3, 0.8).

(a) P(X = 0) = (1 − 0.8)3 = 0.008.

(b) P(X = 1) = (3)(0.2)2(0.8) = 0.096.

(c) P(X ≥ 2) = P(X = 2) + P(X = 3) = (3)(0.8)2(0.2) + (0.8)3 = 0.896.

5.102 (a) P(X = 0) = (1 − 0.8)n ≤ 0.0001 implies that n ≥ 6.

(b) (1 − p)3 ≤ 0.0001 implies p ≥ 0.9536.

5.103 n = 10 and p = 2

50 = 0.04.

P(X ≥ 1) = 1 − P(X = 0) ≈ 1 −

10

0

(0.04)0(1 − 0.04)10 = 1 − 0.6648 = 0.3351.

The approximation is not that good due to n

N = 0.2 is too large.

5.104 (a) P =

(2

2)(3

0)

(5

2)

= 0.1.

(b) P =

(2

1)(1

1)

(5

2)

= 0.2.

5.105 n = 200 with p = 0.00001.

(a) P(X ≥ 5) = 1 − P(X ≤ 4) = 1 −

P4

x=0

200

x

(0.00001)x(1 − 0.00001)200−x ≈ 0. This

is a rare event. Therefore, the claim does not seem right.

(b) μ = np = (200)(0.00001) = 0.02. Using Poisson approximation,

P(X ≥ 5) = 1 − P(X ≤ 4) ≈ 1 −

X4

x=0

e−0.02 (0.02)x

x!

= 0.

Chapter 6

Some Continuous Probability

Distributions

6.1 (a) Area=0.9236.

(b) Area=1 − 0.1867 = 0.8133.

(c) Area=0.2578 − 0.0154 = 0.2424.

(d) Area=0.0823.

(e) Area=1 − 0.9750 = 0.0250.

(f) Area=0.9591 − 0.3156 = 0.6435.

6.2 (a) The area to the left of z is 1−0.3622 = 0.6378 which is closer to the tabled value

0.6368 than to 0.6406. Therefore, we choose z = 0.35.

(b) From Table A.3, z = −1.21.

(c) The total area to the left of z is 0.5000+0.4838=0.9838. Therefore, from Table

A.3, z = 2.14.

(d) The distribution contains an area of 0.025 to the left of −z and therefore a total

area of 0.025+0.95=0.975 to the left of z. From Table A.3, z = 1.96.

6.3 (a) From Table A.3, k = −1.72.

(b) Since P(Z > k) = 0.2946, then P(Z < k) = 0.7054/ From Table A.3, we find

k = 0.54.

(c) The area to the left of z = −0.93 is found from Table A.3 to be 0.1762. Therefore,

the total area to the left of k is 0.1762+0.7235=0.8997, and hence k = 1.28.

6.4 (a) z = (17 − 30)/6 = −2.17. Area=1 − 0.0150 = 0.9850.

(b) z = (22 − 30)/6 = −1.33. Area=0.0918.

(c) z1 = (32−3)/6 = 0.33, z2 = (41−30)/6 = 1.83. Area = 0.9664−0.6293 = 0.3371.

(d) z = 0.84. Therefore, x = 30 + (6)(0.84) = 35.04.

71

72 Chapter 6 Some Continuous Probability Distributions

(e) z1 = −1.15, z2 = 1.15. Therefore, x1 = 30 + (6)(−1.15) = 23.1 and x2 =

30 + (6)(1.15) = 36.9.

6.5 (a) z = (15 − 18)/2.5 = −1.2; P(X < 15) = P(Z < −1.2) = 0.1151.

(b) z = −0.76, k = (2.5)(−0.76) + 18 = 16.1.

(c) z = 0.91, k = (2.5)(0.91) + 18 = 20.275.

(d) z1 = (17 − 18)/2.5 = −0.4, z2 = (21 − 18)/2.5 = 1.2;

P(17 < X < 21) = P(−0.4 < Z < 1.2) = 0.8849 − 0.3446 = 0.5403.

6.6 z1 = [(μ − 3σ) − μ]/σ = −3, z2 = [(μ + 3σ) − μ]/σ = 3;

P(μ − 3σ < Z < μ + 3σ) = P(−3 < Z < 3) = 0.9987 − 0.0013 = 0.9974.

6.7 (a) z = (32 − 40)/6.3 = −1.27; P(X > 32) = P(Z > −1.27) = 1 − 0.1020 = 0.8980.

(b) z = (28 − 40)/6.3 = −1.90, P(X < 28) = P(Z < −1.90) = 0.0287.

(c) z1 = (37 − 40)/6.3 = −0.48, z2 = (49 − 40)/6.3 = 1.43;

So, P(37 < X < 49) = P(−0.48 < Z < 1.43) = 0.9236 − 0.3156 = 0.6080.

6.8 (a) z = (31.7 − 30)/2 = 0.85; P(X > 31.7) = P(Z > 0.85) = 0.1977.

Therefore, 19.77% of the loaves are longer than 31.7 centimeters.

(b) z1 = (29.3 − 30)/2 = −0.35, z2 = (33.5 − 30)/2 = 1.75;

P(29.3 < X < 33.5) = P(−0.35 < Z < 1.75) = 0.9599 − 0.3632 = 0.5967.

Therefore, 59.67% of the loaves are between 29.3 and 33.5 centimeters in length.

(c) z = (25.5 − 30)/2 = −2.25; P(X < 25.5) = P(Z < −2.25) = 0.0122.

Therefore, 1.22% of the loaves are shorter than 25.5 centimeters in length.

6.9 (a) z = (224 − 200)/15 = 1.6. Fraction of the cups containing more than 224 millimeters

is P(Z > 1.6) = 0.0548.

(b) z1 = (191 − 200)/15 = −0.6, Z2 = (209 − 200)/15 = 0.6;

P(191 < X < 209) = P(−0.6 < Z < 0.6) = 0.7257 − 0.2743 = 0.4514.

(c) z = (230 − 200)/15 = 2.0; P(X > 230) = P(Z > 2.0) = 0.0228. Therefore,

(1000)(0.0228) = 22.8 or approximately 23 cups will overflow.

(d) z = −0.67, x = (15)(−0.67) + 200 = 189.95 millimeters.

6.10 (a) z = (10.075 − 10.000)/0.03 = 2.5; P(X > 10.075) = P(Z > 2.5) = 0.0062.

Therefore, 0.62% of the rings have inside diameters exceeding 10.075 cm.

(b) z1 = (9.97 − 10)/0.03 = −1.0, z2 = (10.03 − 10)/0.03 = 1.0;

P(9.97 < X < 10.03) = P(−1.0 < Z < 1.0) = 0.8413 − 0.1587 = 0.6826.

(c) z = −1.04, x = 10 + (0.03)(−1.04) = 9.969 cm.

6.11 (a) z = (30 − 24)/3.8 = 1.58; P(X > 30) = P(Z > 1.58) = 0.0571.

(b) z = (15 − 24)/3.8 = −2.37; P(X > 15) = P(Z > −2.37) = 0.9911. He is late

99.11% of the time.

Solutions for Exercises in Chapter 6 73

(c) z = (25 − 24)/3.8 = 0.26; P(X > 25) = P(Z > 0.26) = 0.3974.

(d) z = 1.04, x = (3.8)(1.04) + 24 = 27.952 minutes.

(e) Using the binomial distribution with p = 0.0571, we get

b(2; 3, 0.0571) =

3

2

(0.0571)2(0.9429) = 0.0092.

6.12 μ = 99.61 and σ = 0.08.

(a) P(99.5 < X < 99.7) = P(−1.375 < Z < 1.125) = 0.8697 − 0.08455 = 0.7852.

(b) P(Z > 1.645) = 0.05; x = (1.645)(0.08) + 99.61 = 99.74.

6.13 z = −1.88, x = (2)(−1.88) + 10 = 6.24 years.

6.14 (a) z = (159.75 − 174.5)/6.9 = −2.14; P(X < 159.75) = P(Z < −2.14) = 0.0162.

Therefore, (1000)(0.0162) = 16 students.

(b) z1 = (171.25 − 174.5)/6.9 = −0.47, z2 = (182.25 − 174.5)/6.9 = 1.12.

P(171.25 < X < 182.25) = P(−0.47 < Z < 1.12) = 0.8686 − 0.3192 = 0.5494.

Therefore, (1000)(0.5494) = 549 students.

(c) z1 = (174.75 − 174.5)/6.9 = 0.04, z2 = (175.25 − 174.5)/6.9 = 0.11.

P(174.75 < X < 175.25) = P(0.04 < Z < 0.11) = 0.5438 − 0.5160 = 0.0278.

Therefore, (1000)(0.0278)=28 students.

(d) z = (187.75 − 174.5)/6.9 = 1.92; P(X > 187.75) = P(Z > 1.92) = 0.0274.

Therefore, (1000)(0.0274) = 27 students.

6.15 μ = $15.90 and σ = $1.50.

(a) 51%, since P(13.75 < X < 16.22) = P

13.745−15.9

1.5 < Z < 16.225−15.9

1.5

= P(−1.437 < Z < 0.217) = 0.5871 − 0.0749 = 0.5122.

(b) $18.36, since P(Z > 1.645) = 0.05; x = (1.645)(1.50) + 15.90 + 0.005 = 18.37.

6.16 (a) z = (9.55 − 8)/0.9 = 1.72. Fraction of poodles weighing over 9.5 kilograms =

P(X > 9.55) = P(Z > 1.72) = 0.0427.

(b) z = (8.65 − 8)/0.9 = 0.72. Fraction of poodles weighing at most 8.6 kilograms =

P(X < 8.65) = P(Z < 0.72) = 0.7642.

(c) z1 = (7.25 − 8)/0.9 = −0.83 and z2 = (9.15 − 8)/0.9 = 1.28.

Fraction of poodles weighing between 7.3 and 9.1 kilograms inclusive

= P(7.25 < X < 9.15) = P(−0.83 < Z < 1.28) = 0.8997 − 0.2033 = 0.6964.

6.17 (a) z = (10, 175 − 10, 000)/100 = 1.75. Proportion of components exceeding 10.150

kilograms in tensile strength= P(X > 10, 175) = P(Z > 1.75) = 0.0401.

(b) z1 = (9, 775 − 10, 000)/100 = −2.25 and z2 = (10, 225 − 10, 000)/100 = 2.25.

Proportion of components scrapped= P(X < 9, 775) + P(X > 10, 225) = P(Z <

−2.25) + P(Z > 2.25) = 2P(Z < −2.25) = 0.0244.

74 Chapter 6 Some Continuous Probability Distributions

6.18 (a) x1 = μ + 1.3σ and x2 = μ − 1.3σ. Then z1 = 1.3 and z2 = −1.3. P(X >

μ+1.3σ)+P(X < 1.3σ) = P(Z > 1.3)+P(Z < −1.3) = 2P(Z < −1.3) = 0.1936.

Therefore, 19.36%.

(b) x1 = μ+0.52σ and x2 = μ−0.52σ. Then z1 = 0.52 and z2 = −0.52. P(μ−0.52σ <

X < μ + 0.52σ) = P(−0.52 < Z < 0.52) = 0.6985 − 0.3015 = 0.3970. Therefore,

39.70%.

6.19 z = (94.5 − 115)/12 = −1.71; P(X < 94.5) = P(Z < −1.71) = 0.0436. Therefore,

(0.0436)(600) = 26 students will be rejected.

6.20 f(x) = 1

B−A for A ≤ x ≤ B.

(a) μ =

R B

A

x

B−A dx = B2−A2

2(B−A) = A+B

2 .

(b) E(X2) =

R B

A

x2

B−A dx = B3−A3

3(B−A) .

So, σ2 = B3−A3

3(B−A) −

A+B

2

2

= 4(B2+AB+A2)−3(B2+2AB+A2)

12 = B2−2AB+A2

12 = (B−A)2

12 .

6.21 A = 7 and B = 10.

(a) P(X ≤ 8.8) = 8.8−7

3 = 0.60.

(b) P(7.4 < X < 9.5) = 9.5−7.4

3 = 0.70.

(c) P(X ≥ 8.5) = 10−8.5

3 = 0.50.

6.22 (a) P(X > 7) = 10−7

10 = 0.3.

(b) P(2 < X < 7) = 7−2

10 = 0.5.

6.23 (a) From Table A.1 with n = 15 and p = 0.2 we have

P(1 ≤ X ≤ 4) =

P4

x=0

b(x; 15, 0.2) − b(0; 15, 0.2) = 0.8358 − 0.0352 = 0.8006.

(b) By the normal-curve approximation we first find

μ = np = 3 and then σ2 = npq = (15)(0.2)(0.8) = 2.4. Then σ = 1.549.

Now, z1 = (0.5 − 3)/1.549 = −1.61 and z2 = (4.5 − 3)/1.549 = 0.97.

Therefore, P(1 ≤ X ≤ 4) = P(−1.61 ≤ Z ≤ 0.97) = 0.8340 − 0.0537 = 0.7803.

6.24 μ = np = (400)(1/2) = 200, σ = √npq =

p

(400)(1/2)(1/2) = 10.

(a) z1 = (184.5 − 200)/10 = −1.55 and z2 = (210.5 − 200)/10 = 1.05.

P(184.5 < X < 210.5) = P(−1.55 < Z < 1.05) = 0.8531 − 0.0606 = 0.7925.

(b) z1 = (204.5 − 200)/10 = 0.45 and z2 = (205.5 − 200)/10 = 0.55.

P(204.5 < X < 205.5) = P(0.45 < Z < 0.55) = 0.7088 − 0.6736 = 0.0352.

(c) z1 = (175.5 − 200)/10 = −2.45 and z2 = (227.5 − 200)/10 = 2.75.

P(X < 175.5) + P(X > 227.5) = P(Z < −2.45) + P(Z > 2.75)

= P(Z < −2.45) + 1 − P(Z < 2.75) = 0.0071 + 1 − 0.9970 = 0.0101.

Solutions for Exercises in Chapter 6 75

6.25 n = 100.

(a) p = 0.01 with μ = (100)(0.01) = 1 and σ =

p

(100)(0.01)(0.99) = 0.995.

So, z = (0.5 − 1)/0.995 = −0.503. P(X ≤ 0) ≈ P(Z ≤ −0.503) = 0.3085.

(b) p = 0.05 with μ = (100)(0.05) = 5 and σ =

p

(100)(0.05)(0.95) = 2.1794.

So, z = (0.5 − 5)/2.1794 = −2.06. P(X ≤ 0) ≈ P(X ≤ −2.06) = 0.0197.

6.26 μ = np = (100)(0.1) = 10 and σ =

p

(100)(0.1)(0.9) = 3.

(a) z = (13.5 − 10)/3 = 1.17; P(X > 13.5) = P(Z > 1.17) = 0.1210.

(b) z = (7.5 − 10)/3 = −0.83; P(X < 7.5) = P(Z < −0.83) = 0.2033.

6.27 μ = (100)(0.9) = 90 and σ =

p

(100)(0.9)(0.1) = 3.

(a) z1 = (83.5 − 90)/3 = −2.17 and z2 = (95.5 − 90)/3 = 1.83.

P(83.5 < X < 95.5) = P(−2.17 < Z < 1.83) = 0.9664 − 0.0150 = 0.9514.

(b) z = (85.5 − 90)/3 = −1.50; P(X < 85.5) = P(Z < −1.50) = 0.0668.

6.28 μ = (80)(3/4) = 60 and σ =

p

(80)(3/4)(1/4) = 3.873.

(a) z = (49.5 − 60)/3.873 = −2.71; P(X > 49.5) = P(Z > −2.71) = 1 − 0.0034 =

0.9966.

(b) z = (56.5 − 60)/3.873 = −0.90; P(X < 56.5) = P(Z < −0.90) = 0.1841.

6.29 μ = (1000)(0.2) = 200 and σ =

p

(1000)(0.2)(0.8) = 12.649.

(a) z1 = (169.5 − 200)/12.649 = −2.41 and z2 = (185.5 − 200)/12.649 = −1.15.

P(169.5 < X < 185.5) = P(−2.41 < Z < −1.15) = 0.1251 − 0.0080 = 0.1171.

(b) z1 = (209.5 − 200)/12.649 = 0.75 and z2 = (225.5 − 200)/12.649 = 2.02.

P(209.5 < X < 225.5) = P(0.75 < Z < 2.02) = 0.9783 − 0.7734 = 0.2049.

6.30 (a) μ = (100)(0.8) = 80 and σ =

p

(100)(0.8)(0.2) = 4 with z = (74.5 − 80)/4 =

−1.38.

P(Claim is rejected when p = 0.8) = P(Z < −1.38) = 0.0838.

(b) μ = (100)(0.7) = 70 and σ =

p

(100)(0.7)(0.3) = 4.583 with z = (74.5 −

70)/4.583 = 0.98.

P(Claim is accepted when p = 0.7) = P(Z > 0.98) = 1 − 0.8365 = 0.1635.

6.31 μ = (180)(1/6) = 30 and σ =

p

(180)(1/6)(5/6) = 5 with z = (35.5 − 30)/5 = 1.1.

P(X > 35.5) = P(Z > 1.1) = 1 − 0.8643 = 0.1357.

6.32 μ = (200)(0.05) = 10 and σ =

p

(200)(0.05)(0.95) = 3.082 with

z = (9.5 − 10)/3.082 = −0.16. P(X < 10) = P(Z < −0.16) = 0.4364.

6.33 μ = (400)(1/10) = 40 and σ =

p

(400)(1/10)(9/10) = 6.

76 Chapter 6 Some Continuous Probability Distributions

(a) z = (31.5 − 40)/6 = −1.42; P(X < 31.5) = P(Z < −1.42) = 0.0778.

(b) z = (49.5 − 40)/6 = 1.58; P(X > 49.5) = P(Z > 1.58) = 1 − 0.9429 = 0.0571.

(c) z1 = (34.5 − 40)/6 = −0.92 and z2 = (46.5 − 40)/6 = 1.08;

P(34.5 < X < 46.5) = P(−0.92 < Z < 1.08) = 0.8599 − 0.1788 = 0.6811.

6.34 μ = (180)(1/6) = 30 and σ =

p

(180)(1/6)(5/6) = 5.

(a) z = (24.5 − 30)/5 = −1.1; P(X > 24.5) = P(Z > −1.1) = 1 − 0.1357 = 0.8643.

(b) z1 = (32.5 − 30)/5 = 0.5 and z2 = (41.5 − 30)/5 = 2.3.

P(32.5 < X < 41.5) = P(0.5 < Z < 2.3) = 0.9893 − 0.6915 = 0.2978.

(c) z1 = (29.5 − 30)/5 = −0.1 and z2 = (30.5 − 30)/5 = 0.1.

P(29.5 < X < 30.5) = P(−0.1 < Z < 0.1) = 0.5398 − 0.4602 = 0.0796.

6.35 (a) p = 0.05, n = 100 with μ = 5 and σ =

p

(100)(0.05)(0.95) = 2.1794.

So, z = (2.5 − 5)/2.1794 = −1.147; P(X ≥ 2) ≈ P(Z ≥ −1.147) = 0.8749.

(b) z = (10.5 − 5)/2.1794 = 2.524; P(X ≥ 10) ≈ P(Z > 2.52) = 0.0059.

6.36 n = 200; X = The number of no shows with p = 0.02. z = 3−0.5−4 √(200)(0.02)(0.98)

= −0.76.

Therefore, P(airline overbooks the flight) = 1 − P(X ≥ 3) ≈ 1 − P(Z > −0.76) =

0.2236.

6.37 (a) P(X ≥ 230) = P

Z > 230−170

30

= 0.0228.

(b) Denote by Y the number of students whose serum cholesterol level exceed 230

among the 300. Then Y ∼ b(y; 300, 0.0228 with μ = (300)(0.0228) = 6.84 and

σ =

p

(300)(0.0228)(1 − 0.0228) = 2.5854. So, z = 8−0.5−6.84

2.5854 = 0.26 and

P(X ≥ 8) ≈ P(Z > 0.26) = 0.3974.

6.38 (a) Denote by X the number of failures among the 20. X ∼ b(x; 20, 0.01) and P(X >

1) = 1−b(0; 20, 0.01)−b(1; 20, 0.01) = 1−

20

0

(0.01)0(0.99)20−

20

1

(0.01)(0.99)19 =

0.01686.

(b) n = 500 and p = 0.01 with μ = (500)(0.01) = 5 and σ =

p

(500)(0.01)(0.99) =

2.2249. So, P(more than 8 failures) ≈ P(Z > (8.5−5)/2.2249) = P(Z > 1.57) =

1 − 0.9418 = 0.0582.

6.39 P(1.8 < X < 2.4) =

R 2.4

1.8 xe−x dx = [−xe−x − e−x]|2.4

1.8 = 2.8e−1.8 − 3.4e−2.4 = 0.1545.

6.40 P(X > 9) = 1

9

R

∞

9 x−x/3 dx =

−x

3 e−x/3 − e−x/3

∞9

= 4e−3 = 0.1992.

6.41 Setting α = 1/2 in the gamma distribution and integrating, we have

1

√β(1/2)

Z

∞

0

x−1/2e−x/ dx = 1.

Solutions for Exercises in Chapter 6 77

Substitute x = y2/2, dx = y dy, to give

(1/2) =

√2

√β

Z

∞

0

e−y2/2 dy = 2√π

1

√2π√β

Z

∞

0

e−y2/2 dy

= √π,

since the quantity in parentheses represents one-half of the area under the normal curve

n(y; 0,√β).

6.42 (a) P(X < 1) = 4

R 1

0 xe−2x dx = [−2xe−2x − e−2x]|1

0 = 1 − 3e−2 = 0.5940.

(b) P(X > 2) = 4

R

∞

0 xe−2x dx = [−2xe−2x − e−2x]|∞2 = 5e−4 = 0.0916.

6.43 (a) μ = αβ = (2)(3) = 6 million liters; σ2 = αβ2 = (2)(9) = 18.

(b) Water consumption on any given day has a probability of at least 3/4 of falling

in the interval μ ± 2σ = 6 ± 2√18 or from −2.485 to 14.485. That is from 0 to

14.485 million liters.

6.44 (a) μ = αβ = 6 and σ2 = αβ2 = 12. Substituting α = 6/β into the variance formula

we find 6β = 12 or β = 2 and then α = 3.

(b) P(X > 12) = 1

16

R

∞

12 x2e−x/2 dx. Integrating by parts twice gives

P(X > 12) =

1

16

−2x2e−x/2 − 8xe−x/2 − 16e−x/2 ∞

12 = 25e−6 = 0.0620.

6.45 P(X < 3) = 1

4

R 3

0 e−x/4 dx = −e−x/4

3

0 = 1 − e−3/4 = 0.5276.

Let Y be the number of days a person is served in less than 3 minutes. Then

P(Y ≥ 4) =

P6

x=4

b(y; 6, 1 − e−3/4) =

6

4

(0.5276)4(0.4724)2 +

6

5

(0.5276)5(0.4724)

+

6

6

(0.5276)6 = 0.3968.

6.46 P(X < 1) = 1

2

R 1

0 e−x/2 dx = −e−x/2

1

0 = 1 − e−1/2 = 0.3935. Let Y be the number

of switches that fail during the first year. Using the normal approximation we find

μ = (100)(0.3935) = 39.35, σ =

p

(100)(0.3935)(0.6065) = 4.885, and z = (30.5 − 39.35)/4.885 = −1.81. Therefore, P(Y ≤ 30) = P(Z < −1.81) = 0.0352.

6.47 (a) E(X) =

R

∞

0 x2e−x2/2 dx = −xe−x2/2

∞

0

+

R

∞

0 e−x2/2 dx

= 0 + √2π · 1 √2

R

∞

0 e−x2/2 dx =

√2

2 =

p

2 = 1.2533.

(b) P(X > 2) =

R

∞

2 xe−x2/2 dx = −e−x2/2

∞

2

= e−2 = 0.1353.

6.48 μ = E(T) = αβ

R

∞

0 t e− t dt. Let y = αt , then dy = αβt −1 dt and t = (y/α)1/ .

Then

μ =

Z

∞

0

(y/α)1/ e−y dy = α−1/

Z

∞

0

y(1+1/ )−1e−y dy = α−1/ (1 + 1/β).

78 Chapter 6 Some Continuous Probability Distributions

E(T2) = αβ

Z

∞

0

t +1e− t

dt =

Z

∞

0

(y/α)2/ e−y dy = α−2/

Z

∞

0

y(1+2/ )−1e−y dy

= α−2/ (1 + 2/β).

So, σ2 = E(T2) − μ2 = α−2/ {(1 + 2/β) − [(1 + 1/β)]2}.

6.49 R(t) = ce−R 1/√t dt = ce−2√t. However, R(0) = 1 and hence c = 1. Now

f(t) = Z(t)R(t) = e−2√t/√t, t > 0,

and

P(T > 4) =

Z

∞

4

e−2√t/√t dt = −e−2√t

∞

4

= e−4 = 0.0183.

6.50 f(x) = 12x2(1 − x), 0 < x < 1. Therefore,

P(X > 0.8) = 12

Z 1

0.8

x2(1 − x) dx = 0.1808.

6.51 α = 5; β = 10;

(a) αβ = 50.

(b) σ2 = αβ2 = 500; so σ = √500 = 22.36.

(c) P(X > 30) = 1

( )

R

∞

30 x −1e−x/ dx. Using the incomplete gamma with y =

x/β, then

1 − P(X ≤ 30) = 1 − P(Y ≤ 3) = 1 −

Z 3

0

y4e−y

(5)

dy = 1 − 0.185 = 0.815.

6.52 αβ = 10; σ =

p

αβ2√50 = 7.07.

(a) Using integration by parts,

P(X ≤ 50) =

1

β (α)

Z 50

0

x −1e−x/ dx =

1

25

Z 50

0

xe−x/5 dx = 0.9995.

(b) P(X < 10) = 1

( )

R 10

0 x −1e−x/ dx. Using the incomplete gamma with y =

x/β, we have

P(X < 10) = P(Y < 2) =

Z 2

0

ye−y dy = 0.5940.

6.53 μ = 3 seconds with f(x) = 1

3e−x/3 for x > 0.

Solutions for Exercises in Chapter 6 79

(a) P(X > 5) =

R

∞

5

1

3e−x/3 dx = 1

3

−3e−x/3

∞5

= e−5/3 = 0.1889.

(b) P(X > 10) = e−10/3 = 0.0357.

6.54 P(X > 270) = 1 −

ln 270−4

2

= 1 − (0.7992) = 0.2119.

6.55 μ = E(X) = e4+4/2 = e6; σ2 = e8+4(e4 − 1) = e12(e4 − 1).

6.56 β = 1/5 and α = 10.

(a) P(X > 10) = 1 − P(X ≤ 10) = 1 − 0.9863 = 0.0137.

(b) P(X > 2) before 10 cars arrive.

P(X ≤ 2) =

Z 2

0

1

β

x −1e−x/

(α)

dx.

Given y = x/β, then

P(X ≤ 2) = P(Y ≤ 10) =

Z 10

0

y −1e−y

(α)

dy =

Z 10

0

y10−1e−y

(10)

dy = 0.542,

with P(X > 2) = 1 − P(X ≤ 2) = 1 − 0.542 = 0.458.

6.57 (a) P(X > 1) = 1 − P(X ≤ 1) = 1 − 10

R 1

0 e−10x dx = e−10 = 0.000045.

(b) μ = β = 1/10 = 0.1.

6.58 Assume that Z(t) = αβt −1, for t > 0. Then we can write f(t) = Z(t)R(t), where

R(t) = ce−R Z(t) dt = ce−R t −1dt = ce− t . From the condition that R(0) = 1, we find

that c = 1. Hence R(t) = e t and f(t) = αβt −1e− t , for t > 0. Since

Z(t) =

f(t)

R(t)

,

where

R(t) = 1 − F(t) = 1 −

Z t

0

αβx −1e− x

dx = 1 +

Z t

0

de− x

= e− t

,

then

Z(t) =

αβt −1e− t

e− t = αβt −1, t > 0.

6.59 μ = np = (1000)(0.49) = 490, σ = √npq =

p

(1000)(0.49)(0.51) = 15.808.

z1 =

481.5 − 490

15.808

= −0.54, z2 =

510.5 − 490

15.808

= 1.3.

P(481.5 < X < 510.5) = P(−0.54 < Z < 1.3) = 0.9032 − 0.2946 = 0.6086.

80 Chapter 6 Some Continuous Probability Distributions

6.60 P(X > 1/4) =

R

∞

1/4 6e−6x dx = −e−6x|∞ 1/4 = e−1.5 = 0.223.

6.61 P(X < 1/2) = 108

R 1/2

0 x2e−6x dx. Letting y = 6x and using Table A.24 we have

P(X < 1/2) = P(Y < 3) =

Z 3

0

y2e−y dy = 0.577.

6.62 Manufacturer A:

P(X ≥ 10000) = P

Z ≥

100000 − 14000

2000

= P(Z ≥ −2) = 0.9772.

Manufacturer B:

P(X ≥ 10000) = P

Z ≥

10000 − 13000

1000

= P(Z ≥ −3) = 0.9987.

Manufacturer B will produce the fewest number of defective rivets.

6.63 Using the normal approximation to the binomial with μ = np = 650 and σ = √npq =

15.0831. So,

P(590 ≤ X ≤ 625) = P(−10.64 < Z < −8.92) ≈ 0.

6.64 (a) μ = β = 100 hours.

(b) P(X ≥ 200) = 0.01

R

∞

200 e−0.01x dx = e−2 = 0.1353.

6.65 (a) μ = 85 and σ = 4. So, P(X < 80) = P(Z < −1.25) = 0.1056.

(b) μ = 79 and σ = 4. So, P(X ≥ 80) = P(Z > 0.25) = 0.4013.

6.66 1/β = 1/5 hours with α = 2 failures and β = 5 hours.

(a) αβ = (2)(5) = 10.

(b) P(X ≥ 12) =

R

∞

12

1

52(2)xe−x/5 dx = 1

25

R

∞

12 xe−x/5 dx =

−x

5 e−x/5 − e−x/5

∞ 1

2

= 0.3084.

6.67 Denote by X the elongation. We have μ = 0.05 and σ = 0.01.

(a) P(X ≥ 0.1) = P

Z ≥ 0.1−0.05

0.01

= P(Z ≥ 5) ≈ 0.

(b) P(X ≤ 0.04) = P

Z ≤ 0.04−0.05

0.01

= P(Z ≤ −1) = 0.1587.

(c) P(0.025 ≤ X ≤ 0.065) = P(−2.5 ≤ Z ≤ 1.5) = 0.9332 − 0.0062 = 0.9270.

6.68 Let X be the error. X ∼ n(x; 0, 4). So,

P(fails) = 1 − P(−10 < X < 10) = 1 − P(−2.25 < Z < 2.25) = 2(0.0122) = 0.0244.

Solutions for Exercises in Chapter 6 81

6.69 Let X be the time to bombing with μ = 3 and σ = 0.5. Then

P(1 ≤ X ≤ 4) = P

1 − 3

0.5 ≤ Z ≤

4 − 3

0.5

= P(−4 ≤ Z ≤ 2) = 0.9772.

P(of an undesirable product) is 1 − 0.9772 = 0.0228. Hence a product is undesirable

is 2.28% of the time.

6.70 α = 2 and β = 100. P(X ≤ 200) = 1

2

R 200

0 xe−x/ dx. Using the incomplete gamma

table and let y = x/β,

R 2

0 ye−y dy = 0.594.

6.71 μ = αβ = 200 hours and σ2 = αβ2 = 20, 000 hours.

6.72 X follows a lognormal distribution.

P(X ≥ 50, 000) = 1 −

ln 50, 000 − 5

2

= 1 − (2.9099) = 1 − 0.9982 = 0.0018.

6.73 The mean of X, which follows a lognormal distribution is μ = E(X) = eμ+ 2/2 = e7.

6.74 μ = 10 and σ = √50.

(a) P(X ≤ 50) = P(Z ≤ 5.66) ≈ 1.

(b) P(X ≤ 10) = 0.5.

(c) The results are very similar.

6.75 (a) Since f(y) ≥ 0 and

R 1

0 10(1 − y)9 dy = − (1 − y)10|1

0 = 1, it is a density function.

(b) P(Y > 0.6) = − (1 − y)10|1

0.6 = (0.4)10 = 0.0001.

(c) α = 1 and β = 10.

(d) μ =

+ = 1

11 = 0.0909.

(e) σ2 =

( + )2( + +1) = (1)(10)

(1+10)2(1+10+1) = 0.006887.

6.76 (a) μ = 1

10

R

∞

0 ze−z/10 dz = − ze−z/10

∞0

+

R

∞

0 e−z/10 dz = 10.

(b) Using integral by parts twice, we get

E(Z2) =

1

10

Z

∞

0

z2e−z/10 dz = 200.

So, σ2 = E(Z2) − μ2 = 200 − (10)2 = 100.

(c) P(Z > 10) = − ez/10

∞ 1

0

=

e−1

=

0.3679.

6.77 This is an exponential distribution with β = 10.

(a) μ = β = 10.

82 Chapter 6 Some Continuous Probability Distributions

(b) σ2 = β2 = 100.

6.78 μ = 0.5 seconds and σ = 0.4 seconds.

(a) P(X > 0.3) = P

Z > 0.3−0.5

0.4

= P(Z > −0.5) = 0.6915.

(b) P(Z > −1.645) = 0.95. So, −1.645 = x−0.5

0.4 yields x = −0.158 seconds. The

negative number in reaction time is not reasonable. So, it means that the normal

model may not be accurate enough.

6.79 (a) For an exponential distribution with parameter β,

P(X > a + b | X > a) =

P(X > a + b)

P(X > a)

=

e−a−b

e−a = e−b = P(X > b).

So, P(it will breakdown in the next 21 days | it just broke down) = P(X > 21) =

e−21/15 = e−1.4 = 0.2466.

(b) P(X > 30) = e−30/15 = e−2 = 0.1353.

6.80 α = 2 and β = 50. So,

P(X ≤ 10) = 100

Z 10

0

x49e−2x50

dx.

Let y = 2x50 with x = (y/2)1/50 and dx = 1

21/50(50)y−49/50 dy.

P(X ≤ 10) =

100

21/50(50)

Z (2)1050

0

y

2

49/50

y−49/50e−y dy =

Z (2)1050

0

e−y dy ≈ 1.

6.81 The density function of a Weibull distribution is

f(y) = αβy −1e− y

, y > 0.

So, for any y ≥ 0,

F(y) =

Z y

0

f(t) dt = αβ

Z y

0

t −1e− t

dt.

Let z = t which yields t = z1/ and dt = 1

z1/ −1 dz. Hence,

F(y) = αβ

Z y

0

z1−1/ 1

β

z1/ −1e− z dz = α

Z y

0

e− z dz = 1 − e− y

.

On the other hand, since de− y = −αβy −1e− y , the above result follows immediately.

Solutions for Exercises in Chapter 6 83

6.82 One of the basic assumptions for the exponential distribution centers around the “lackof-

memory” property for the associated Poisson distribution. Thus the drill bit of

problem 6.80 is assumed to have no punishment through wear if the exponential distribution

applies. A drill bit is a mechanical part that certainly will have significant

wear over time. Hence the exponential distribution would not apply.

6.83 The chi-squared distribution is a special case of the gamma distribution when α = v/2

and β = 2, where v is the degrees of the freedom of the chi-squared distribution.

So, the mean of the chi-squared distribution, using the property from the gamma

distribution, is μ = αβ = (v/2)(2) = v, and the variance of the chi-squared distribution

is σ2 = αβ2 = (v/2)(2)2 = 2v.

6.84 Let X be the length of time in seconds. Then Y = ln(X) follows a normal distribution

with μ = 1.8 and σ = 2.

(a) P(X > 20) = P(Y > ln 20) = P(Z > (ln 20 − 1.8)/2) = P(Z > 0.60) = 0.2743.

P(X > 60) = P(Y > ln 60) = P(Z > (ln 60 − 1.8)/2) = P(Z > 1.15) = 0.1251.

(b) The mean of the underlying normal distribution is e1.8+4/2 = 44.70 seconds. So,

P(X < 44.70) = P(Z < (ln 44.70 − 1.8)/2) = P(Z < 1) = 0.8413.

Chapter 7

Functions of Random Variables

7.1 From y = 2x − 1 we obtain x = (y + 1)/2, and given x = 1, 2, and 3, then

g(y) = f[(y + 1)/2] = 1/3, for y = 1, 3, 5.

7.2 From y = x2, x = 0, 1, 2, 3, we obtain x = √y,

g(y) = f(√y) =

3

√y

2

5

√y

3

5

3−√y

, fory = 0, 1, 4, 9.

7.3 The inverse functions of y1 = x1 + x2 and y2 = x1 − x2 are x1 = (y1 + y2)/2 and

x2 = (y1 − y2)/2. Therefore,

g(y1, y2) =

2

y1+y2

2 , y1−y2

2 , 2 − y1

1

4

(y1+y2)/2

1

3

(y1−y2)/2

5

12

2−y1

,

where y1 = 0, 1, 2, y2 = −2,−1, 0, 1, 2, y2 ≤ y1 and y1 + y2 = 0, 2, 4.

7.4 Let W = X2. The inverse functions of y = x1x2 and w = x2 are x1 = y/w and x2 = w,

where y/w = 1, 2. Then

g(y,w) = (y/w)(w/18) = y/18, y = 1, 2, 3, 4, 6; w = 1, 2, 3, y/w = 1, 2.

In tabular form the joint distribution g(y,w) and marginal h(y) are given by

y

g(y,w) 1 2 3 4 6

1 1/18 2/18

w 2 2/18 4/18

3 3/18 6/18

h(y) 1/18 2/9 1/6 2/9 1/3

85

86 Chapter 7 Functions of Random Variables

The alternate solutions are:

P(Y = 1) = f(1, 1) = 1/18,

P(Y = 2) = f(1, 2) + f(2, 1) = 2/18 + 2/18 = 2/9,

P(Y = 3) = f(1, 3) = 3/18 = 1/6,

P(Y = 4) = f(2, 2) = 4/18 = 2/9,

P(Y = 6) = f(2, 3) = 6/18 = 1/3.

7.5 The inverse function of y = −2 ln x is given by x = e−y/2 from which we obtain

|J| = | − e−y/2/2| = e−y/2/2. Now,

g(y) = f(ey/2)|J| = e−y/2/2, y > 0,

which is a chi-squared distribution with 2 degrees of freedom.

7.6 The inverse function of y = 8x3 is x = y1/3/2, for 0 < y < 8 from which we obtain

|J| = y−2/3/6. Therefore,

g(y) = f(y1/3/2)|J| = 2(y1/3/2)(y−2/3/6) =

1

6

y−1/3, 0 < y < 8.

7.7 To find k we solve the equation k

R

∞

0 v2e−bv2 dv = 1. Let x = bv2, then dx = 2bv dv

and dv = x−1/2

2√b

dx. Then the equation becomes

k

2b3/2

Z

∞

0

x3/2−1e−x dx = 1, or

k(3/2)

2b3/2 = 1.

Hence k = 4b3/2

(1/2) .

Now the inverse function of w = mv2/2 is v =

p

2w/m, for w > 0, from which we

obtain |J| = 1/√2mw. It follows that

g(w) = f(

p

2w/m)|J| =

4b3/2

(1/2)

(2w/m)e−2bw/m =

1

(m/2b)3/2(3/2)

w3/2−1e−(2b/m)w,

for w > 0, which is a gamma distribution with α = 3/2 and β = m/2b.

7.8 (a) The inverse of y = x2 is x = √y, for 0 < y < 1, from which we obtain |J| = 1/2√y.

Therefore,

g(y) = f(√y)|J| = 2(1 − √y)/2√y = y−1/2 − 1, 0 < y < 1.

(b) P(Y < 1) =

R 1

0 (y−1/2 − 1) dy = (2y1/2 − y)

1

0 = 0.5324.

7.9 (a) The inverse of y = x + 4 is x = y − 4, for y > 4, from which we obtain |J| = 1.

Therefore,

g(y) = f(y − 4)|J| = 32/y3, y > 4.

Solutions for Exercises in Chapter 7 87

(b) P(Y > 8) = 32

R

∞

8 y−3 dy = − 16y−2|∞8 = 1

4 .

7.10 (a) Let W = X. The inverse functions of z = x + y and w = x are x = w and

y = z − w, 0 < w < z, 0 < z < 1, from which we obtain

J =

@x

@w

@x

@z

@y

@w

@y

@z

=

1 0

−1 1

= 1.

Then g(w, z) = f(w, z − w)|J| = 24w(z − w), for 0 < w < z and 0 < z < 1. The

marginal distribution of Z is

f1(z) =

Z 1

0

24(z − w)w dw = 4z3, 0 < z < 1.

(b) P(1/2 < Z < 3/4) = 4

R 3/4

1/2 z3 dz = 65/256.

7.11 The amount of kerosene left at the end of the day is Z = Y − X. Let W = Y . The

inverse functions of z = y − x and w = y are x = w − z and y = w, for 0 < z < w and

0 < w < 1, from which we obtain

J =

@x

@w

@x

@z

@y

@w

@y

@z

=

1 −1

1 0

= 1.

Now,

g(w, z) = g(w − z,w) = 2, 0 < z < w, 0 < w < 1,

and the marginal distribution of Z is

h(z) = 2

Z 1

z

dw = 2(1 − z), 0 < z < 1.

7.12 Since X1 and X2 are independent, the joint probability distribution is

f(x1, x2) = f(x1)f(x2) = e−(x1+x2), x1 > 0, x2 > 0.

The inverse functions of y1 = x1 + x2 and y2 = x1/(x1 + x2) are x1 = y1y2 and

x2 = y1(1 − y2), for y1 > 0 and 0 < y2 < 1, so that

J =

∂x1/∂y1 ∂x1/∂y2

∂x2/∂y1 ∂x2/∂y2

=

y2 y1

1 − y2 −y1

= −y1.

Then, g(y1, y2) = f(y1y2, y1(1−y2))|J| = y1e−y1 , for y1 > 0 and 0 < y2 < 1. Therefore,

g(y1) =

Z 1

0

y1e−y1 dy2 = y1e−y1 , y1 > 0,

and

g(y2) =

Z

∞

0

y1e−y1 dy1 = (2) = 1, 0 < y2 < 1.

Since g(y1, y2) = g(y1)g(y2), the random variables Y1 and Y2 are independent.

88 Chapter 7 Functions of Random Variables

7.13 Since I and R are independent, the joint probability distribution is

f(i, r) = 12ri(1 − i), 0 < i < 1, 0 < r < 1.

Let V = R. The inverse functions of w = i2r and v = r are i =

p

w/v and r = v, for

w < v < 1 and 0 < w < 1, from which we obtain

J =

∂i/∂w ∂i/∂v

∂r/∂w ∂r/∂v

=

1

2√vw

.

Then,

g(w, v) = f(

p

w/v, v)|J| = 12v

p

w/v(1 −

p

w/v)

1

2√vw

= 6(1 −

p

w/v),

for w < v < 1 and 0 < w < 1, and the marginal distribution of W is

h(w) = 6

Z 1

w

(1 −

p

w/v) dv = 6 (v − 2√wv)

v

=

1

v=w = 6 + 6w − 12√w, 0 < w < 1.

7.14 The inverse functions of y = x2 are given by x1 = √y and x2 = −√y from which we

obtain J1 = 1/2√y and J2 = 1/2√y. Therefore,

g(y) = f(√y)|J1| + f(−√y)|J2| =

1 + √y

2 ·

1

2√y

+

1 − √y

2 ·

1

2√y

= 1/2√y,

for 0 < y < 1.

7.15 The inverse functions of y = x2 are x1 = √y, x2 = −√y for 0 < y < 1 and x1 = √y

for 0 < y < 4. Now |J1| = |J2| = |J3| = 1/2√y, from which we get

g(y) = f(√y)|J1| + f(−√y)|J2| =

2(√y + 1)

9 ·

1

2√y

+

2(−√y + 1)

9 ·

1

2√y

=

2

9√y

,

for 0 < y < 1 and

g(y) = f(√y)|J3| =

2(√y + 1)

9 ·

1

2√y

=

√y + 1

9√y

, for 1 < y < 4.

7.16 Using the formula we obtain

μ

′

r = E(Xr) =

Z

∞

0

xr ·

x −1e−x/

β (α)

dx =

β +r(α + r)

β (α)

Z

∞

0

x +r−1e−x/

β +r(α + r)

dx

=

βr(α + r)

(α)

,

since the second integrand is a gamma density with parametersα + r and β.

Solutions for Exercises in Chapter 7 89

7.17 The moment-generating function of X is

MX(t) = E(etX) =

1

k

Xk

x=1

etx =

et(1 − ekt)

k(1 − et)

,

by summing the geometric series of k terms.

7.18 The moment-generating function of X is

MX(t) = E(etX) = p

∞X

x=1

etxqx−1 =

p

q

∞X

x=1

(etq)x =

pet

1 − qet ,

by summing an infinite geometric series. To find out the moments, we use

μ = M

′

X(0) =

(1 − qet)pet + pqe2t

(1 − qet)2

t=0

=

(1 − q)p + pq

(1 − q)2 =

1

p

,

and

μ

′

2 = M

′′

X(0) =

(1 − qet)2pet + 2pqe2t(1 − qet)

(1 − qet)4

t=0

=

2 − p

p2 .

So, σ2 = μ′

2 − μ2 = q

p2 .

7.19 The moment-generating function of a Poisson random variable is

MX(t) = E(etX) =

∞X

x=0

etxe−μμx

x!

= e−μ

∞X

x=0

(μet)x

x!

= e−μeμet

= eμ(et−1).

So,

μ = M

′

X(0) = μ eμ(et−1)+t

t=0

= μ,

μ

′

2 = M

′′

X(0) = μeμ(et−1)+t(μet + 1)

t=0

= μ(μ + 1),

and

σ2 = μ

′

2 − μ2 = μ(μ + 1) − μ2 = μ.

7.20 From MX(t) = e4(et−1) we obtain μ = 6, σ2 = 4, and σ = 2. Therefore,

P(μ − 2σ < X < μ + 2σ) = P(0 < X < 8) =

X7

x=1

p(x; 4) = 0.9489 − 0.0183 = 0.9306.

90 Chapter 7 Functions of Random Variables

7.21 Using the moment-generating function of the chi-squared distribution, we obtain

μ = M

′

X(0) = v(1 − 2t)−v/2−1

t=0 = v,

μ

′

2 = M

′′

X(0) = v(v + 2) (1 − 2t)−v/2−2

t=0 = v(v + 2).

So, σ2 = μ′

2 − μ2 = v(v + 2) − v2 = 2v.

7.22

MX(t) =

Z

∞

−∞

etxf(x) dx =

Z

∞

−∞

1 + tx +

t2x2

2!

+ · · · +

trxr

r!

+ · · ·

f(x) dx

=

Z

∞

−∞

f(x) dx + t

Z

∞

−∞

xf(x) dx +

t2

2

Z

∞

−∞

x2f(x) dx

+ · · · +

tr

r!

Z

∞

−∞

xrf(x) dx + · · · = 1 + μt + μ

′

1

t2

2!

+ · · · + μ

′

r

tr

r!

+ · · · .

7.23 The joint distribution of X and Y is fX,Y (x, y) = e−x−y for x > 0 and y > 0. The

inverse functions of u = x + y and v = x/(x + y) are x = uv and y = u(1 − v) with

J =

v u

1 − v −u

= u for u > 0 and 0 < v < 1. So, the joint distribution of U and V is

fU,V (u, v) = ue−uv · e−u(1−v) = ue−u,

for u > 0 and 0 < v < 1.

(a) fU (u) =

R 1

0 ue−u dv = ue−u for u > 0, which is a gamma distribution with

parameters 2 and 1.

(b) fV (v) =

R

∞

0 ue−u du = 1 for 0 < v < 1. This is a uniform (0,1) distribution.

Chapter 8

Fundamental Sampling Distributions

and Data Descriptions

8.1 (a) Responses of all people in Richmond who have telephones.

(b) Outcomes for a large or infinite number of tosses of a coin.

(c) Length of life of such tennis shoes when worn on the professional tour.

(d) All possible time intervals for this lawyer to drive from her home to her office.

8.2 (a) Number of tickets issued by all state troopers in Montgomery County during the

Memorial holiday weekend.

(b) Number of tickets issued by all state troopers in South Carolina during the Memorial

holiday weekend.

8.3 (a) ¯x = 2.4.

(b) ¯x = 2.

(c) m = 3.

8.4 (a) ¯x = 8.6 minutes.

(b) ¯x = 9.5 minutes.

(c) Mode are 5 and 10 minutes.

8.5 (a) ¯x = 3.2 seconds.

(b) ¯x = 3.1 seconds.

8.6 (a) ¯x = 35.7 grams.

(b) ¯x = 32.5 grams.

(c) Mode=29 grams.

8.7 (a) ¯x = 53.75.

91

92 Chapter 8 Fundamental Sampling Distributions and Data Descriptions

(b) Modes are 75 and 100.

8.8 ¯x = 22.2 days, ˜x = 14 days and m = 8 days. ˜x is the best measure of the center of the

data. The mean should not be used on account of the extreme value 95, and the mode

is not desirable because the sample size is too small.

8.9 (a) Range = 15 − 5 = 10.

(b) s2 =

n

n P

i

=

1

x2

i −(

n P

i

=

1

xi)2

n(n−1) = (10)(838)−862

(10)(9) = 10.933. Taking the square root, we have

s = 3.307.

8.10 (a) Range = 4.3 − 2.3 = 2.0.

(b) s2 =

n

n P

i

=

1

x2

i −(

n P

i

=

1

xi)2

n(n−1) = (9)(96.14)−28.82

(9)(8) = 0.498.

8.11 (a) s2 = 1

n−1

Pn

x=1

(xi − ¯x)2 = 1

14 [(2 − 2.4)2 + (1 − 2.4)2 + · · · + (2 − 2.4)2] = 2.971.

(b) s2 =

n

n P

i

=

1

x2

i −(

n P

i

=

1

xi)2

n(n−1) = (15)(128)−362

(15)(14) = 2.971.

8.12 (a) ¯x = 11.69 milligrams.

(b) s2 =

n

n P

i

=

1

x2

i −(

n P

i

=

1

xi)2

n(n−1) = (8)(1168.21)−93.52

(8)(7) = 10.776.

8.13 s2 =

n

n P

i

=

1

x2

i −(

n P

i

=

1

xi)2

n(n−1) = (2)(148.55)−53.32

(20)(19) = 0.342 and hence s = 0.585.

8.14 (a) Replace Xi in S2 by Xi + c for i = 1, 2, . . . , n. Then ¯X becomes ¯X + c and

S2 =

1

n − 1

Xn

i=1

[(Xi + c) − ( ¯X + c)]2 =

1

n − 1

Xn

i=1

(Xi − ¯X )2.

(b) Replace Xi by cXi in S2 for i = 1, 2, . . . , n. Then ¯X becomes c ¯X and

S2 =

1

n − 1

Xn

i=1

(cXi − c ¯X )2 =

c2

n − 1

Xn

i=1

(Xi − ¯X )2.

8.15 s2 =

n

n P

i

=

1

x2

i −(

n P

i

=

1

xi)2

n(n−1) = (6)(207)−332

(6)(5) = 5.1.

(a) Multiplying each observation by 3 gives s2 = (9)(5.1) = 45.9.

(b) Adding 5 to each observation does not change the variance. Hence s2 = 5.1.

8.16 Denote by D the difference in scores.

Solutions for Exercises in Chapter 8 93

(a) ¯D = 25.15.

(b) ˜D = 31.00.

8.17 z1 = −1.9, z2 = −0.4. Hence,

P(μ ¯X − 1.9σ ¯X < ¯X < μ ¯X − 0.4σ ¯X ) = P(−1.9 < Z < −0.4) = 0.3446 − 0.0287 = 0.3159.

8.18 n = 54, μ ¯X = 4, σ2

¯X

= σ2/n = (8/3)/54 = 4/81 with σ ¯X = 2/9. So,

z1 = (4.15 − 4)/(2/9) = 0.68, and z2 = (4.35 − 4)/(2/9) = 1.58,

and

P(4.15 < ¯X < 4.35) = P(0.68 < Z < 1.58) = 0.9429 − 0.7517 = 0.1912.

8.19 (a) For n = 64, σ ¯X = 5.6/8 = 0.7, whereas for n = 196, σ ¯X = 5.6/14 = 0.4.

Therefore, the variance of the sample mean is reduced from 0.49 to 0.16 when the

sample size is increased from 64 to 196.

(b) For n = 784, σ ¯X = 5.6/28 = 0.2, whereas for n = 49, σ ¯X = 5.6/7 = 0.8.

Therefore, the variance of the sample mean is increased from 0.04 to 0.64 when

the sample size is decreased from 784 to 49.

8.20 n = 36, σ ¯X = 2. Hence σ = √nσ ¯X = (6)(2) = 12. If σ ¯X = 1.2, then 1.2 = 12/√n and

n = 100.

8.21 μ ¯X = μ = 240, σ ¯X = 15/√40 = 2.372. Therefore, μ ¯X ±2σ ¯X = 240±(2)(2.372) or from

235.257 to 244.743, which indicates that a value of x = 236 milliliters is reasonable

and hence the machine needs not be adjusted.

8.22 (a) μ ¯X = μ = 174.5, σ ¯X = σ/√n = 6.9/5 = 1.38.

(b) z1 = (172.45 − 174.5)/1.38 = −1.49, z2 = (175.85 − 174.5)/1.38 = 0.98. So,

P(172.45 < ¯X < 175.85) = P(−1.49 < Z < 0.98) = 0.8365 − 0.0681 = 0.7684.

Therefore, the number of sample means between 172.5 and 175.8 inclusive is

(200)(0.7684) = 154.

(c) z = (171.95 − 174.5)/1.38 = −1.85. So,

P( ¯X < 171.95) = P(Z < −1.85) = 0.0322.

Therefore, about (200)(0.0322) = 6 sample means fall below 172.0 centimeters.

8.23 (a) μ =

P

xf(x) = (4)(0.2) + (5)(0.4) + (6)(0.3) + (7)(0.1) = 5.3, and

σ2 =

P

(x − μ)2f(x) = (4 − 5.3)2(0.2) + (5 − 5.3)2(0.4) + (6 − 5.3)2(0.3) + (7 −

5.3)2(0.1) = 0.81.

94 Chapter 8 Fundamental Sampling Distributions and Data Descriptions

(b) With n = 36, μ ¯X = μ = 5.3 and σ ¯X = σ2/n = 0.81/36 = 0.0225.

(c) n = 36, μ ¯X = 5.3, σ ¯X = 0.9/6 = 0.15, and z = (5.5 − 5.3)/0.15 = 1.33. So,

P( ¯X < 5.5) = P(Z < 1.33) = 0.9082.

8.24 n = 36, μ ¯X = 40, σ ¯X = 2/6 = 1/3 and z = (40.5 − 40)/(1/3) = 1.5. So,

P

X36

i=1

Xi > 1458

!

= P( ¯X > 40.5) = P(Z > 1.5) = 1 − 0.9332 = 0.0668.

8.25 (a) P(6.4 < ¯X < 7.2) = P(−1.8 < Z < 0.6) = 0.6898.

(b) z = 1.04, ¯x = z(σ/√n) + μ = (1.04)(1/3) + 7 = 7.35.

8.26 n = 64, μ ¯X = 3.2, σ ¯X = σ/√n = 1.6/8 = 0.2.

(a) z = (2.7 − 3.2)/0.2 = −2.5, P( ¯X < 2.7) = P(Z < −2.5) = 0.0062.

(b) z = (3.5 − 3.2)/0.2 = 1.5, P( ¯X > 3.5) = P(Z > 1.5) = 1 − 0.9332 = 0.0668.

(c) z1 = (3.2 − 3.2)/0.2 = 0, z2 = (3.4 − 3.2)/0.2 = 1.0,

P(3.2 < ¯X < 3.4) = P(0 < Z < 1.0) = 0.9413 − 0.5000 = 0.3413.

8.27 n = 50, ¯x = 0.23 and σ = 0.1. Now, z = (0.23 − 0.2)/(0.1/√50) = 2.12; so

P( ¯X ≥ 0.23) = P(Z ≥ 2.12) = 0.0170.

Hence the probability of having such observations, given the mean μ = 0.20, is small.

Therefore, the mean amount to be 0.20 is not likely to be true.

8.28 μ1−μ2 = 80−75 = 5, σ ¯X1− ¯X2 =

p

25/25 + 9/36 = 1.118, z1 = (3.35−5)/1.118 = −1.48

and z2 = (5.85 − 5)/1.118 = 0.76. So,

P(3.35 < ¯X1 − ¯X2 < 5.85) = P(−1.48 < Z < 0.76) = 0.7764 − 0.0694 = 0.7070.

8.29 μ ¯X1− ¯X2 = 72 − 28 = 44, σ ¯X1− ¯X2 =

p

100/64 + 25/100 = 1.346 and z = (44.2 − 44)/1.346 = 0.15. So, P( ¯X1 − ¯X2 < 44.2) = P(Z < 0.15) = 0.5596.

8.30 μ1 − μ2 = 0, σ ¯X1− ¯X2 = 50

p

1/32 + 1/50 = 11.319.

(a) z1 = −20/11.319 = −1.77, z2 = 20/11.319 = 1.77, so

P(| ¯X1 − ¯X2| > 20) = 2P(Z < −1.77) = (2)(0.0384) = 0.0768.

(b) z1 = 5/11.319 = 0.44 and z2 = 10/11.319 = 0.88. So,

P(−10 < ¯X1 − ¯X2 < −5) + P(5 < ¯X1 − ¯X2 < 10) = 2P(5 < ¯X1 − ¯X2 < 10) =

2P(0.44 < Z < 0.88) = 2(0.8106 − 0.6700) = 0.2812.

8.31 The normal quantile-quantile plot is shown as

Solutions for Exercises in Chapter 8 95

−2 −1 0 1 2

700 800 900 1000 1100 1200 1300

Normal Q−Q Plot

Theoretical Quantiles

Sample Quantiles

8.32 (a) If the two population mean drying times are truly equal, the probability that the

difference of the two sample means is 1.0 is 0.0013, which is very small. This means

that the assumption of the equality of the population means are not reasonable.

(b) If the experiment was run 10,000 times, there would be (10000)(0.0013) = 13

experiments where ¯XA − ¯XB would be at least 1.0.

8.33 (a) n1 = n2 = 36 and z = 0.2/

p

1/36 + 1/36 = 0.85. So,

P( ¯XB − ¯XA ≥ 0.2) = P(Z ≥ 0.85) = 0.1977.

(b) Since the probability in (a) is not negligible, the conjecture is not true.

8.34 The normal quantile-quantile plot is shown as

−2 −1 0 1 2

6.65 6.70 6.75 6.80

Normal Q−Q Plot

Theoretical Quantiles

Sample Quantiles

96 Chapter 8 Fundamental Sampling Distributions and Data Descriptions

8.35 (a) When the population equals the limit, the probability of a sample mean exceeding

the limit would be 1/2 due the symmetry of the approximated normal distribution.

(b) P( ¯X ≥ 7960 | μ = 7950) = P(Z ≥ (7960 − 7950)/(100/√25)) = P(Z ≥ 0.5) =

0.3085. No, this is not very strong evidence that the population mean of the

process exceeds the government limit.

8.36 (a) σ ¯XA− ¯XB =

q

52

30 + 52

30 = 1.29 and z = 4−0

1.29 = 3.10. So,

P( ¯XA − ¯XB > 4 | μA = μB) = P(Z > 3.10) = 0.0010.

Such a small probability means that the difference of 4 is not likely if the two

population means are equal.

(b) Yes, the data strongly support alloy A.

8.37 Since the probability that ¯X ≤ 775 is 0.0062, given that μ = 800 is true, it suggests

that this event is very rare and it is very likely that the claim of μ = 800 is not true.

On the other hand, if μ is truly, say, 760, the probability

P( ¯X ≤ 775 | μ = 760) = P(Z ≤ (775 − 760)/(40/√16)) = P(Z ≤ 1.5) = 0.9332,

which is very high.

8.38 Define Wi = lnXi for i = 1, 2, . . . . Using the central limit theorem, Z = ( ¯W − μW1)/(σW1/√n) ∼ n(z; 0, 1). Hence ¯W follows, approximately, a normal distribution

when n is large. Since

¯W

=

1

n

Xn

i=1

ln(Xi) =

1

n

ln

Yn

i=1

Xi

!

=

1

n

ln(Y ),

then it is easily seen that Y follows, approximately, a lognormal distribution.

8.39 (a) 27.488.

(b) 18.475.

(c) 36.415.

8.40 (a) 16.750.

(b) 30.144.

(c) 26.217.

8.41 (a) χ2

= χ2

0.99 = 0.297.

(b) χ2

= χ2

0.025 = 32.852.

(c) χ2

0.05 = 37.652. Therefore, α = 0.05−0.045 = 0.005. Hence, χ2

= χ2

0.005 = 46.928.

8.42 (a) χ2

= χ2

0.01 = 38.932.

Solutions for Exercises in Chapter 8 97

(b) χ2

= χ2

0.05 = 12.592.

(c) χ2

0.01 = 23.209 and χ2

0.025 = 20.483 with α = 0.01 + 0.015 = 0.025.

8.43 (a) P(S2 > 9.1) = P

(n−1)S2

2 > (24)(9.1)

6

= P(χ2 > 36.4) = 0.05.

(b) P(3.462 < S2 < 10.745) = P

(24)(3.462)

6 < (n−1)S2

2 < (24)(10.745)

6

= P(13.848 < χ2 < 42.980) = 0.95 − 0.01 = 0.94.

8.44 χ2 = (19)(20)

8 = 47.5 while χ2

0.01 = 36.191. Conclusion values are not valid.

8.45 Since (n−1)S2

2 is a chi-square statistic, it follows that

σ2

(n−1)S2/ 2 =

(n − 1)2

σ4 σ2

S2 = 2(n − 1).

Hence, σ2

S2 = 2 4

n−1 , which decreases as n increases.

8.46 (a) 2.145.

(b) −1.372.

(c) −3.499.

8.47 (a) P(T < 2.365) = 1 − 0.025 = 0.975.

(b) P(T > 1.318) = 0.10.

(c) P(T < 2.179) = 1 − 0.025 = 0.975, P(T < −1.356) = P(T > 1.356) = 0.10.

Therefore, P(−1.356 < T < 2.179) = 0.975 − 0.010 = 0.875.

(d) P(T > −2.567) = 1 − P(T > 2.567) = 1 − 0.01 = 0.99.

8.48 (a) Since t0.01 leaves an area of 0.01 to the right, and −t0.005 an area of 0.005 to the

left, we find the total area to be 1 − 0.01 − 0.005 = 0.985 between −t0.005 and

t0.01. Hence, P(−t0.005 < T < t0.01) = 0.985.

(b) Since −t0.025 leaves an area of 0.025 to the left, the desired area is 1−0.025 = 0.975.

That is, P(T > −t0.025) = 0.975.

8.49 (a) From Table A.4 we note that 2.069 corresponds to t0.025 when v = 23. Therefore,

−t0.025 = −2.069 which means that the total area under the curve to the left of

t = k is 0.025 + 0.965 = 0.990. Hence, k = t0.01 = 2.500.

(b) From Table A.4 we note that 2.807 corresponds to t0.005 when v = 23. Therefore

the total area under the curve to the right of t = k is 0.095+0.005 = 0.10. Hence,

k = t0.10 = 1.319.

(c) t0.05 = 1.714 for 23 degrees of freedom.

98 Chapter 8 Fundamental Sampling Distributions and Data Descriptions

8.50 From Table A.4 we find t0.025 = 2.131 for v = 15 degrees of freedom. Since the value

t =

27.5 − 30

5/4

= −2.00

falls between −2.131 and 2.131, the claim is valid.

8.51 t = (24 − 20)/(4.1/3) = 2.927, t0.01 = 2.896 with 8 degrees of freedom. Conclusion:

no, μ > 20.

8.52 ¯x = 0.475, s2 = 0.0336 and t = (0.475 − 0.5)/0.0648 = −0.39. Hence

P( ¯X < 0.475) = P(T < −0.39) ≈ 0.35.

So, the result is inconclusive.

8.53 (a) 2.71.

(b) 3.51.

(c) 2.92.

(d) 1/2.11 = 0.47.

(e) 1/2.90 = 0.34.

8.54 s2

1 = 10.441 and s2

2 = 1.846 which gives f = 5.66. Since, from Table A.6, f0.05(9, 7) =

3.68 and f0.01(9, 7) = 6.72, the probability of P(F > 5.66) should be between 0.01 and

0.05, which is quite small. Hence the variances may not be equal. Furthermore, if a

computer software can be used, the exact probability of F > 5.66 can be found 0.0162,

or if two sides are considered, P(F < 1/5.66) + P(F > 5.66) = 0.026.

8.55 s2

1 = 15750 and s2

2 = 10920 which gives f = 1.44. Since, from Table A.6, f0.05(4, 5) =

5.19, the probability of F > 1.44 is much bigger than 0.05, which means that the two

variances may be considered equal. The actual probability of F > 1.44 is 0.3436 and

P(F < 1/1.44) + P(F > 1.44) = 0.7158.

8.56 The box-and-whisker plot is shown below.

5 10 15 20

Box−and−Whisker Plot

Solutions for Exercises in Chapter 8 99

The sample mean = 12.32 and the sample standard deviation = 6.08.

8.57 The moment-generating function for the gamma distribution is given by

MX(t) = E(etX) =

1

β (α)

Z

∞

0

etxx −1e−x/ dx

=

1

β (1/β − t)

1

(1/β − t)− (α)

Z

∞

0

x −1e−x( 1

−t) dx

=

1

(1 − βt)

Z

∞

0

x −1e−x/(1/ −t)−1

[(1/β − t)−1] (α)

dx =

1

(1 − βt) ,

for t < 1/β, since the last integral is one due to the integrand being a gamma density

function. Therefore, the moment-generating function of an exponential distribution,

by substituting α to 1, is given by MX(t) = (1−θt)−1. Hence, the moment-generating

function of Y can be expressed as

MY (t) = MX1(t)MX2 (t) · · ·MXn(t) =

Yn

i=1

(1 − θt)−1 = (1 − θt)−n,

which is seen to be the moment-generating function of a gamma distribution with

α = n and β = θ.

8.58 The variance of the carbon monoxide contents is the same as the variance of the coded

measurements. That is, s2 = (15)(199.94)−392

(15)(14) = 7.039, which results in s = 2.653.

8.59 P

S2

1

S2

2

< 4.89

= P

S2

1/ 2

S2

2/ 2 < 4.89

= P(F < 4.89) = 0.99, where F has 7 and 11

degrees of freedom.

8.60 s2 = 114, 700, 000.

8.61 Let X1 and X2 be Poisson variables with parameters λ1 = 6 and λ2 = 6 representing the

number of hurricanes during the first and second years, respectively. Then Y = X1+X2

has a Poisson distribution with parameter λ = λ1 + λ2 = 12.

(a) P(Y = 15) = e−121215

15! = 0.0724.

(b) P(Y ≤ 9) =

P9

y=0

e−1212y

y! = 0.2424.

8.62 Dividing each observation by 1000 and then subtracting 55 yields the following data:

−7, −2, −10, 6, 4, 1, 8, −6, −2, and −1. The variance of this coded data is

(10)(311) − (−9)2

(10)(9)

= 33.656.

Hence, with c = 1000, we have

s2 = (1000)2(33.656) = 33.656 × 106,

and then s = 5801 kilometers.

100 Chapter 8 Fundamental Sampling Distributions and Data Descriptions

8.63 The box-and-whisker plot is shown below.

5 10 15 20 25

Box−and_Whisker Plot

The sample mean is 2.7967 and the sample standard deviation is 2.2273.

8.64 P

S2

1

S2

2

> 1.26

= P

S2

1/ 2

1

S2

2/ 2

2

> (15)(1.26)

10

= P(F > 1.89) ≈ 0.05, where F has 24 and 30

degrees of freedom.

8.65 No outliers.

8.66 The value 32 is a possible outlier.

8.67 μ = 5,000 psi, σ = 400 psi, and n = 36.

(a) Using approximate normal distribution (by CLT),

P(4800 < ¯X < 5200) = P

4800 − 5000

400/√36

< Z <

5200 − 5000

400/√36

= P(−3 < Z < 3) = 0.9974.

(b) To find a z such that P(−z < Z < z) = 0.99, we have P(Z < z) = 0.995, which

results in z = 2.575. Hence, by solving 2.575 = 5100−5000

400/√n we have n ≥ 107. Note

that the value n can be affected by the z values picked (2.57 or 2.58).

8.68 ¯x = 54,100 and s = 5801.34. Hence

t =

54100 − 53000

5801.34/√10

= 0.60.

So, P( ¯X ≥ 54, 100) = P(T ≥ 0.60) is a value between 0.20 and 0.30, which is not a

rare event.

8.69 nA = nB = 20, ¯xA = 20.50, ¯xB = 24.50, and σA = σB = 5.

(a) P( ¯XA − ¯XB ≥ 4.0 | μA = μB) = P(Z > (24.5 − 20.5)/

p

52/20 + 52/20)

= P(Z > 4.5/(5/√10)) = P(Z > 2.85) = 0.0022.

Solutions for Exercises in Chapter 8 101

(b) It is extremely unlikely that μA = μB.

8.70 (a) nA = 30, ¯xA = 64.5% and σA = 5%. Hence,

P( ¯XA ≤ 64.5 | μA = 65) = P(Z < (64.5 − 65)/(5/√30)) = P(Z < −0.55)

= 0.2912.

There is no evidence that the μA is less than 65%.

(b) nB = 30, ¯xB = 70% and σB = 5%. It turns out σ ¯XB− ¯XA =

q

52

30 + 52

30 = 1.29%.

Hence,

P( ¯XB − ¯XA ≥ 5.5 | μA = μB) = P

Z ≥

5.5

1.29

= P(Z ≥ 4.26) ≈ 0.

It does strongly support that μB is greater than μA.

(c) i) Since σ ¯XB = 5 √30

= 0.9129, ¯XB ∼ n(x; 65%, 0.9129%).

ii) ¯XA − ¯XB ∼ n(x; 0, 1.29%).

iii)

¯X

A− ¯XB

√2/30 ∼ n(z; 0, 1).

8.71 P( ¯XB ≥ 70) = P

Z ≥ 70−65

0.9129

= P(Z ≥ 5.48) ≈ 0.

8.72 It is known, from Table A.3, that P(−1.96 < Z < 1.96) = 0.95. Given μ = 20 and

σ = √9 = 3, we equate 1.96 = 20.1−20

3/√n to obtain n =

(3)(1.96)

0.1

2

= 3457.44 ≈ 3458.

8.73 It is known that P(−2.575 < Z < 2.575) = 0.99. Hence, by equating 2.575 = 0.05

1/√n, we

obtain n =

2.575

0.05

2 = 2652.25 ≈ 2653.

8.74 μ = 9 and σ = 1. Then

P(9 − 1.5 < X < 9 + 1.5) = P(7.5 < X < 10.5) = P(−1.5 < Z < 1.5)

= 0.9322 − 0.0668 = 0.8654.

Thus the proportion of defective is 1 − 0.8654 = 0.1346. To meet the specifications

99% of the time, we need to equate 2.575 = 1.5

, since P(−2.575 < Z < 2.575) = 0.99.

Therefore, σ = 1.5

2.575 = 0.5825.

8.75 With the 39 degrees of freedom,

P(S2 ≤ 0.188 | σ2 = 1.0) = P(χ2 ≤ (39)(0.188)) = P(χ2 ≤ 7.332) ≈ 0,

which means that it is impossible to observe s2 = 0.188 with n = 40 for σ2 = 1.

Note that Table A.5 does not provide any values for the degrees of freedom to be larger

than 30. However, one can deduce the conclusion based on the values in the last line

of the table. Also, computer software gives the value of 0.

Chapter 9

One- and Two-Sample Estimation

Problems

9.1 From Example 9.1 on page 271, we know that E(S2) = σ2. Therefore,

E(S′2) = E

n − 1

n

S2

=

n − 1

n

E(S2) =

n − 1

n

σ2.

9.2 (a) E(X) = np; E( ˆ P) = E(X/n) = E(X)/n = np/n = p.

(b) E(P′) = E(X)+√n/2

n+√n = np+√n/2

n+√n 6= p.

9.3 lim

n→∞

np+√n/2

n+√n = lim

n→∞

p+1/2√n

1+1/√n = p.

9.4 n = 30, ¯x = 780, and σ = 40. Also, z0.02 = 2.054. So, a 96% confidence interval for the

population mean can be calculated as

780 − (2.054)(40/√30) < μ < 780 + (2.054)(40/√30),

or 765 < μ < 795.

9.5 n = 75, ¯x = 0.310, σ = 0.0015, and z0.025 = 1.96. A 95% confidence interval for the

population mean is

0.310 − (1.96)(0.0015/√75) < μ < 0.310 + (1.96)(0.0015/√75),

or 0.3097 < μ < 0.3103.

9.6 n = 50, ¯x = 174.5, σ = 6.9, and z0.01 = 2.33.

(a) A 98% confidence interval for the population mean is

174.5 − (2.33)(6.9/√50) < μ < 174.5 + (2.33)(6.9/√50), or 172.23 < μ < 176.77.

(b) e < (2.33)(6.9)/√50 = 2.27.

103

104 Chapter 9 One- and Two-Sample Estimation Problems

9.7 n = 100, ¯x = 23, 500, σ = 3900, and z0.005 = 2.575.

(a) A 99% confidence interval for the population mean is

23, 500 − (2.575)(3900/10) < μ < 23, 500 + (2.575)(3900/10), or

22, 496 < μ < 24, 504.

(b) e < (2.575)(3900/10) = 1004.

9.8 n = [(2.05)(40)/10]2 = 68 when rounded up.

9.9 n = [(1.96)(0.0015)/0.0005]2 = 35 when rounded up.

9.10 n = [(1.96)(40)/15]2 = 28 when rounded up.

9.11 n = [(2.575)(5.8)/2]2 = 56 when rounded up.

9.12 n = 20, ¯x = 11.3, s = 2.45, and t0.025 = 2.093 with 19 degrees of freedom. A 95%

confidence interval for the population mean is

11.3 − (2.093)(2.45/√20) < μ < 11.3 + (2.093)(2.45/√20),

or 10.15 < μ < 12.45.

9.13 n = 9, ¯x = 1.0056, s = 0.0245, and t0.005 = 3.355 with 8 degrees of freedom. A 99%

confidence interval for the population mean is

1.0056 − (3.355)(0.0245/3) < μ < 1.0056 + (3.355)(0.0245/3),

or 0.978 < μ < 1.033.

9.14 n = 10, ¯x = 230, s = 15, and t0.005 = 3.25 with 9 degrees of freedom. A 99% confidence

interval for the population mean is

230 − (3.25)(15/√10) < μ < 230 + (3.25)(15/√10),

or 214.58 < μ < 245.42.

9.15 n = 12, ¯x = 48.50, s = 1.5, and t0.05 = 1.796 with 11 degrees of freedom. A 90%

confidence interval for the population mean is

48.50 − (1.796)(1.5/√12) < μ < 48.50 + (1.796)(1.5/√12),

or 47.722 < μ < 49.278.

9.16 n = 12, ¯x = 79.3, s = 7.8, and t0.025 = 2.201 with 11 degrees of freedom. A 95%

confidence interval for the population mean is

79.3 − (2.201)(7.8/√12) < μ < 79.3 + (2.201)(7.8/√12),

or 74.34 < μ < 84.26.

Solutions for Exercises in Chapter 9 105

9.17 n = 25, ¯x = 325.05, s = 0.5, γ = 5%, and 1 − α = 90%, with k = 2.208. So, 325.05 ±

(2.208)(0.5) yields (323.946, 326.151). Thus, we are 95% confident that this tolerance

interval will contain 90% of the aspirin contents for this brand of buffered aspirin.

9.18 n = 15, ¯x = 3.7867, s = 0.9709, γ = 1%, and 1 − α = 95%, with k = 3.507. So, by

calculating 3.7867 ± (3.507)(0.9709) we obtain (0.382, 7.192) which is a 99% tolerance

interval that will contain 95% of the drying times.

9.19 n = 100, ¯x = 23,500, s = 3, 900, 1 − α = 0.99, and γ = 0.01, with k = 3.096. The

tolerance interval is 23,500 ± (3.096)(3,900) which yields 11,425 < μ < 35,574.

9.20 n = 12, ¯x = 48.50, s = 1.5, 1 − α = 0.90, and γ = 0.05, with k = 2.655. The tolerance

interval is 48.50 ± (2.655)(1.5) which yields (44.52, 52.48).

9.21 By definition, MSE = E(ˆ − θ)2 which can be expressed as

MSE = E[ˆ − E(ˆ ) + E(ˆ ) − θ]2

= E[ˆ − E(ˆ )]2 + E[E(ˆ ) − θ]2 + 2E[ˆ − E(ˆ )]E[E(ˆ ) − θ].

The third term on the right hand side is zero since E[ˆ − E(ˆ )] = E[ˆ ] − E(ˆ ) = 0.

Hence the claim is valid.

9.22 (a) The bias is E(S′2) − σ2 = n−1

n σ2 − σ2 = 2

n .

(b) lim

n→∞

Bias = lim

n→∞

2

n = 0.

9.23 Using Theorem 8.4, we know that X2 = (n−1)S2

2 follows a chi-squared distribution with

n − 1 degrees of freedom, whose variance is 2(n − 1). So, V ar(S2) = V ar

2

n−1X2

=

2

n−1σ4, and V ar(S′2) = V ar

n−1

n S2

=

n−1

n

2

V ar(S2) = 2(n−1) 4

n2 . Therefore, the

variance of S′2 is smaller.

9.24 Using Exercises 9.21 and 9.23,

MSE(S2)

MSE(S′2)

=

V ar(S2) + [Bias(S2)]2

V ar(S′2) + [Bias(S′2)]2 =

2σ4/(n − 1)

2(n − 1)σ4/n2 + σ4/n2

= 1 +

3n − 1

2n2 − 3n + 1

,

which is always larger than 1 when n is larger than 1. Hence the MSE of S′2 is usually

smaller.

9.25 n = 20, ¯x = 11.3, s = 2.45, and t0.025 = 2.093 with 19 degrees of freedom. A 95%

prediction interval for a future observation is

11.3 ± (2.093)(2.45)

p

1 + 1/20 = 11.3 ± 5.25,

which yields (6.05, 16.55).

106 Chapter 9 One- and Two-Sample Estimation Problems

9.26 n = 12, ¯x = 79.3, s = 7.8, and t0.025 = 2.201 with 11 degrees of freedom. A 95%

prediction interval for a future observation is

79.3 ± (2.201)(7.8)

p

1 + 1/12 = 79.3 ± 17.87,

which yields (61.43, 97.17).

9.27 n = 15, ¯x = 3.7867, s = 0.9709, and t0.025 = 2.145 with 14 degrees of freedom. A 95%

prediction interval for a new observation is

3.7867 ± (2.145)(0.9709)

p

1 + 1/15 = 3.7867 ± 2.1509,

which yields (1.6358, 5.9376).

9.28 n = 9, ¯x = 1.0056, s = 0.0245, 1 − α = 0.95, and γ = 0.05, with k = 3.532. The

tolerance interval is 1.0056 ± (3.532)(0.0245) which yields (0.919, 1.092).

9.29 n = 15, ¯x = 3.84, and s = 3.07. To calculate an upper 95% prediction limit, we

obtain t0.05 = 1.761 with 14 degrees of freedom. So, the upper limit is 3.84 +

(1.761)(3.07)

p

1 + 1/15 = 3.84 + 5.58 = 9.42. This means that a new observation

will have a chance of 95% to fall into the interval (−∞, 9.42). To obtain an upper

95% tolerance limit, using 1 − α = 0.95 and γ = 0.05, with k = 2.566, we get

3.84 + (2.566)(3.07) = 11.72. Hence, we are 95% confident that (−∞, 11.72) will

contain 95% of the orthophosphorous measurements in the river.

9.30 n = 50, ¯x = 78.3, and s = 5.6. Since t0.05 = 1.677 with 49 degrees of freedom,

the bound of a lower 95% prediction interval for a single new observation is 78.3 − (1.677)(5.6)

p

1 + 1/50 = 68.91. So, the interval is (68.91,∞). On the other hand,

with 1 − α = 95% and γ = 0.01, the k value for a one-sided tolerance limit is 2.269

and the bound is 78.3 − (2.269)(5.6) = 65.59. So, the tolerance interval is (65.59,∞).

9.31 Since the manufacturer would be more interested in the mean tensile strength for future

products, it is conceivable that prediction interval and tolerance interval may be more

interesting than just a confidence interval.

9.32 This time 1 − α = 0.99 and γ = 0.05 with k = 3.126. So, the tolerance limit is

78.3−(3.126)(5.6) = 60.79. Since 62 exceeds the lower bound of the interval, yes, this

is a cause of concern.

9.33 In Exercise 9.27, a 95% prediction interval for a new observation is calculated as

(1.6358, 5.9377). Since 6.9 is in the outside range of the prediction interval, this new

observation is likely to be an outlier.

9.34 n = 12, ¯x = 48.50, s = 1.5, 1 − α = 0.95, and γ = 0.05, with k = 2.815. The lower

bound of the one-sided tolerance interval is 48.50−(2.815)(1.5) = 44.275. Their claim

is not necessarily correct.

Solutions for Exercises in Chapter 9 107

9.35 n1 = 25, n2 = 36, ¯x1 = 80, ¯x2 = 75, σ1 = 5, σ2 = 3, and z0.03 = 1.88. So, a 94%

confidence interval for μ1 − μ2 is

(80 − 75) − (1.88)

p

25/25 + 9/36 < μ1 − μ2 < (80 − 75) + (1.88)

p

25/25 + 9/36,

which yields 2.9 < μ1 − μ2 < 7.1.

9.36 nA = 50, nB = 50, ¯xA = 78.3, ¯xB = 87.2, σA = 5.6, and σB = 6.3. It is known that

z0.025 = 1.96. So, a 95% confidence interval for the difference of the population means

is

(87.2 − 78.3) ± 1.96

p

5.62/50 + 6.32/50 = 8.9 ± 2.34,

or 6.56 < μA − μB < 11.24.

9.37 n1 = 100, n2 = 200, ¯x1 = 12.2, ¯x2 = 9.1, s1 = 1.1, and s2 = 0.9. It is known that

z0.01 = 2.327. So

(12.2 − 9.1) ± 2.327

p

1.12/100 + 0.92/200 = 3.1 ± 0.30,

or 2.80 < μ1 −μ2 < 3.40. The treatment appears to reduce the mean amount of metal

removed.

9.38 n1 = 12, n2 = 10, ¯x1 = 85, ¯x2 = 81, s1 = 4, s2 = 5, and sp = 4.478 with t0.05 = 1.725

with 20 degrees of freedom. So

(85 − 81) ± (1.725)(4.478)

p

1/12 + 1/10 = 4 ± 3.31,

which yields 0.69 < μ1 − μ2 < 7.31.

9.39 n1 = 12, n2 = 18, ¯x1 = 84, ¯x2 = 77, s1 = 4, s2 = 6, and sp = 5.305 with t0.005 = 2.763

with 28 degrees of freedom. So,

(84 − 77) ± (2.763)(5.305)

p

1/12 + 1/18 = 7 ± 5.46,

which yields 1.54 < μ1 − μ2 < 12.46.

9.40 n1 = 10, n2 = 10, ¯x1 = 0.399, ¯x2 = 0.565, s1 = 0.07279, s2 = 0.18674, and sp = 0.14172

with t0.025 = 2.101 with 18 degrees of freedom. So,

(0.565 − 0.399) ± (2.101)(0.14172)

p

1/10 + 1/10 = 0.166 ± 0.133,

which yields 0.033 < μ1 − μ2 < 0.299.

9.41 n1 = 14, n2 = 16, ¯x1 = 17, ¯x2 = 19, s2

1 = 1.5, s2

2 = 1.8, and sp = 1.289 with t0.005 =

2.763 with 28 degrees of freedom. So,

(19 − 17) ± (2.763)(1.289)

p

1/16 + 1/14 = 2 ± 1.30,

which yields 0.70 < μ1 − μ2 < 3.30.

108 Chapter 9 One- and Two-Sample Estimation Problems

9.42 n1 = 12, n2 = 10, ¯x1 = 16, ¯x2 = 11, s1 = 1.0, s2 = 0.8, and sp = 0.915 with t0.05 = 1.725

with 20 degrees of freedom. So,

(16 − 11) ± (1.725)(0.915)

p

1/12 + 1/10 = 5 ± 0.68,

which yields 4.3 < μ1 − μ2 < 5.7.

9.43 nA = nB = 12, ¯xA = 36, 300, ¯xB = 38, 100, sA = 5, 000, sB = 6, 100, and

v =

50002/12 + 61002/12

(50002/12)2

11 + (61002/12)2

11

= 21,

with t0.025 = 2.080 with 21 degrees of freedom. So,

(36, 300 − 38, 100) ± (2.080)

r

50002

12

+

61002

12

= −1, 800 ± 4, 736,

which yields −6, 536 < μA − μB < 2, 936.

9.44 n = 8, ¯ d = −1112.5, sd = 1454, with t0.005 = 3.499 with 7 degrees of freedom. So,

−1112.5 ± (3.499)

1454

√8

= −1112.5 ± 1798.7,

which yields −2911.2 < μD < 686.2.

9.45 n = 9, ¯ d = 2.778, sd = 4.5765, with t0.025 = 2.306 with 8 degrees of freedom. So,

2.778 ± (2.306)

4.5765

√9

= 2.778 ± 3.518,

which yields −0.74 < μD < 6.30.

9.46 nI = 5, nII = 7, ¯xI = 98.4, ¯xII = 110.7, sI = 8.375, and sII = 32.185, with

v =

(8.7352/5 + 32.1852/7)2

(8.7352/5)2

4 + (32.1852/7)2

6

= 7

So, t0.05 = 1.895 with 7 degrees of freedom.

(110.7 − 98.4) ± 1.895

p

8.7352/5 + 32.1852/7 = 12.3 ± 24.2,

which yields −11.9 < μII − μI < 36.5.

9.47 n = 10, ¯ d = 14.89%, and sd = 30.4868, with t0.025 = 2.262 with 9 degrees of freedom.

So,

14.89 ± (2.262)

30.4868

√10

= 14.89 ± 21.81,

which yields −6.92 < μD < 36.70.

Solutions for Exercises in Chapter 9 109

9.48 nA = nB = 20, ¯xA = 32.91, ¯xB = 30.47, sA = 1.57, sB = 1.74, and Sp = 1.657.

(a) t0.025 ≈ 2.042 with 38 degrees of freedom. So,

(32.91 − 30.47) ± (2.042)(1.657)

p

1/20 + 1/20 = 2.44 ± 1.07,

which yields 1.37 < μA − μB < 3.51.

(b) Since it is apparent that type A battery has longer life, it should be adopted.

9.49 nA = nB = 15, ¯xA = 3.82, ¯xB = 4.94, sA = 0.7794, sB = 0.7538, and sp = 0.7667 with

t0.025 = 2.048 with 28 degrees of freedom. So,

(4.94 − 3.82) ± (2.048)(0.7667)

p

1/15 + 1/15 = 1.12 ± 0.57,

which yields 0.55 < μB − μA < 1.69.

9.50 n1 = 8, n2 = 13, ¯x1 = 1.98, ¯x2 = 1.30, s1 = 0.51, s2 = 0.35, and sp = 0.416. t0.025 =

2.093 with 19 degrees of freedom. So,

(1.98 − 1.30) ± (2.093)(0.416)

p

1/8 + 1/13 = 0.68 ± 0.39,

which yields 0.29 < μ1 − μ2 < 1.07.

9.51 (a) n = 200, ˆp = 0.57, ˆq = 0.43, and z0.02 = 2.05. So,

0.57 ± (2.05)

r

(0.57)(0.43)

200

= 0.57 ± 0.072,

which yields 0.498 < p < 0.642.

(b) Error ≤ (2.05)

q

(0.57)(0.43)

200 = 0.072.

9.52 n = 500.ˆp = 485

500 = 0.97, ˆq = 0.03, and z0.05 = 1.645. So,

0.97 ± (1.645)

r

(0.97)(0.03)

500

= 0.97 ± 0.013,

which yields 0.957 < p < 0.983.

9.53 n = 1000, ˆp = 228

1000 = 0.228, ˆq = 0.772, and z0.005 = 2.575. So,

0.228 ± (2.575)

r

(0.228)(0.772)

1000

= 0.228 ± 0.034,

which yields 0.194 < p < 0.262.

9.54 n = 100, ˆp = 8

100 = 0.08, ˆq = 0.92, and z0.01 = 2.33. So,

0.08 ± (2.33)

r

(0.08)(0.92)

100

= 0.08 ± 0.063,

which yields 0.017 < p < 0.143.

110 Chapter 9 One- and Two-Sample Estimation Problems

9.55 (a) n = 40, ˆp = 34

40 = 0.85, ˆq = 0.15, and z0.025 = 1.96. So,

0.85 ± (1.96)

r

(0.85)(0.15)

40

= 0.85 ± 0.111,

which yields 0.739 < p < 0.961.

(b) Since p = 0.8 falls in the confidence interval, we can not conclude that the new

system is better.

9.56 n = 100, ˆp = 24

100 = 0.24, ˆq = 0.76, and z0.005 = 2.575.

(a) 0.24 ± (2.575)

q

(0.24)(0.76)

100 = 0.24 ± 0.110, which yields 0.130 < p < 0.350.

(b) Error ≤ (2.575)

q

(0.24)(0.76)

100 = 0.110.

9.57 n = 1600, ˆp = 2

3 , ˆq = 1

3 , and z0.025 = 1.96.

(a) 2

3 ± (1.96)

q

(2/3)(1/3)

1600 = 2

3 ± 0.023, which yields 0.644 < p < 0.690.

(b) Error ≤ (1.96)

q

(2/3)(1/3)

1600 = 0.023.

9.58 n = (1.96)2(0.32)(0.68)

(0.02)2 = 2090 when round up.

9.59 n = (2.05)2(0.57)(0.43)

(0.02)2 = 2576 when round up.

9.60 n = (2.575)2(0.228)(0.772)

(0.05)2 = 467 when round up.

9.61 n = (2.33)2(0.08)(0.92)

(0.05)2 = 160 when round up.

9.62 n = (1.96)2

(4)(0.01)2 = 9604 when round up.

9.63 n = (2.575)2

(4)(0.01)2 = 16577 when round up.

9.64 n = (1.96)2

(4)(0.04)2 = 601 when round up.

9.65 nM = nF = 1000, ˆpM = 0.250, ˆqM = 0.750, ˆpF = 0.275, ˆqF = 0.725, and z0.025 = 1.96.

So

(0.275 − 0.250) ± (1.96)

r

(0.250)(0.750)

1000

+

(0.275)(0.725)

1000

= 0.025 ± 0.039,

which yields −0.0136 < pF − pM < 0.0636.

Solutions for Exercises in Chapter 9 111

9.66 n1 = 250, n2 = 175, ˆp1 = 80

250 = 0.32, ˆp2 = 40

175 = 0.2286, and z0.05 = 1.645. So,

(0.32 − 0.2286) ± (1.645)

r

(0.32)(0.68)

250

+

(0.2286)(0.7714)

175

= 0.0914 ± 0.0713,

which yields 0.0201 < p1 − p2 < 0.1627. From this study we conclude that there is

a significantly higher proportion of women in electrical engineering than there is in

chemical engineering.

9.67 n1 = n2 = 500, ˆp1 = 120

500 = 0.24, ˆp2 = 98

500 = 0.196, and z0.05 = 1.645. So,

(0.24 − 0.196) ± (1.645)

r

(0.24)(0.76)

500

+

(0.196)(0.804)

500

= 0.044 ± 0.0429,

which yields 0.0011 < p1 − p2 < 0.0869. Since 0 is not in this confidence interval, we

conclude, at the level of 90% confidence, that inoculation has an effect on the incidence

of the disease.

9.68 n5◦C = n15◦C = 20, ˆp5◦C = 0.50, ˆp15◦C = 0.75, and z0.025 = 1.96. So,

(0.5 − 0.75) ± (1.96)

r

(0.50)(0.50)

20

+

(0.75)(0.25)

20

= −0.25 ± 0.2899,

which yields −0.5399 < p5◦C − p15◦C < 0.0399. Since this interval includes 0, the

significance of the difference cannot be shown at the confidence level of 95%.

9.69 nnow = 1000, ˆpnow = 0.2740, n91 = 760, ˆp91 = 0.3158, and z0.025 = 1.96. So,

(0.2740 − 0.3158) ± (1.96)

r

(0.2740)(0.7260)

1000

+

(0.3158)(0.6842)

760

= −0.0418 ± 0.0431,

which yields −0.0849 < pnow − p91 < 0.0013. Hence, at the confidence level of 95%,

the significance cannot be shown.

9.70 n90 = n94 = 20, ˆp90 = 0.337, and ˆ094 = 0.362

(a) n90 ˆp90 = (20)(0.337) ≈ 7 and n94 ˆp94 = (20)(0.362) ≈ 7.

(b) Since z0.025 = 1.96, (0.337−0.362)±(1.96)

q

(0.337)(0.663)

20 + (0.362)(0.638)

20 = −0.025±

0.295, which yields −0.320 < p90 − p94 < 0.270. Hence there is no evidence, at

the confidence level of 95%, that there is a change in the proportions.

9.71 s2 = 0.815 with v = 4 degrees of freedom. Also, χ2

0.025 = 11.143 and χ2

0.975 = 0.484.

So,

(4)(0.815)

11.143

< σ2 <

(4)(0.815)

0.484

, which yields 0.293 < σ2 < 6.736.

Since this interval contains 1, the claim that σ2 seems valid.

112 Chapter 9 One- and Two-Sample Estimation Problems

9.72 s2 = 16 with v = 19 degrees of freedom. It is known χ2

0.01 = 36.191 and χ2

0.99 = 7.633.

Hence

(19)(16)

36.191

< σ2 <

(19)(16)

7.633

, or 8.400 < σ2 < 39.827.

9.73 s2 = 6.0025 with v = 19 degrees of freedom. Also, χ2

0.025 = 32.852 and χ2

0.975 = 8.907.

Hence,

(19)(6.0025)

32.852

< σ2 <

(19)(6.0025)

8.907

, or 3.472 < σ2 < 12.804,

9.74 s2 = 0.0006 with v = 8 degrees of freedom. Also, χ2

0.005 = 21.955 and χ2

0.995 = 1.344.

Hence,

(8)(0.0006)

21.955

< σ2 <

(8)(0.0006)

1.344

, or 0.00022 < σ2 < 0.00357.

9.75 s2 = 225 with v = 9 degrees of freedom. Also, χ2

0.005 = 23.589 and χ2

0.995 = 1.735.

Hence,

(9)(225)

23.589

< σ2 <

(9)(225)

1.735

, or 85.845 < σ2 < 1167.147,

which yields 9.27 < σ < 34.16.

9.76 s2 = 2.25 with v = 11 degrees of freedom. Also, χ2

0.05 = 19.675 and χ2

0.95 = 4.575.

Hence,

(11)(2.25)

19.675

< σ2 <

(11)(2.25)

4.575

, or 1.258 < σ2 < 5.410.

9.77 s2

1 = 1.00, s2

2 = 0.64, f0.01(11, 9) = 5.19, and f0.01(9, 11) = 4.63. So,

1.00/0.64

5.19

<

σ2

1

σ2

2

< (1.00/0.64)(4.63), or 0.301 <

σ2

1

σ2

2

< 7.234,

which yields 0.549 < 1

2

< 2.690.

9.78 s2

1 = 50002, s2

2 = 61002, and f0.05(11, 11) = 2.82. (Note: this value can be found by

using “=finv(0.05,11,11)” in Microsoft Excel.) So,

5000

6100

2 1

2.82

<

σ2

1

σ2

2

<

5000

6100

2

(2.82), or 0.238 <

σ2

1

σ2

2

< 1.895.

Since the interval contains 1, it is reasonable to assume that σ2

1 = σ2

2.

9.79 s2

I = 76.3, s2

II = 1035.905, f0.05(4, 6) = 4.53, and f0.05(6, 4) = 6.16. So,

76.3

1035.905

1

4.53

<

σ2

I

σ2

II

<

76.3

1035.905

(6.16), or 0.016 <

σ2

I

σ2

II

< 0.454.

Hence, we may assume that σ2

I 6= σ2

II .

Solutions for Exercises in Chapter 9 113

9.80 sA = 0.7794, sB = 0.7538, and f0.025(14, 14) = 2.98 (Note: this value can be found by

using “=finv(0.025,14,14)” in Microsoft Excel.) So,

0.7794

0.7538

2

1

2.98

<

σ2

A

σ2

B

<

0.7794

0.7538

2

(2.98), or 0.623 <

σ2

A

σ2

B

< 3.186.

Hence, it is reasonable to assume the equality of the variances.

9.81 The likelihood function is

L(x1, . . . , xn) =

Yn

i=1

f(xi; p) =

Yn

i=1

pxi(1 − p)1−xi = pn¯x(1 − p)n(1−¯x).

Hence, ln L = n[¯x ln(p) + (1 − ¯x) ln(1 − p)]. Taking derivative with respect to p

and setting the derivative to zero, we obtain @ ln(L)

@p = n

¯x

p − 1−¯x

1−p

= 0, which yields

¯x

p − 1−¯x

1−p = 0. Therefore, ˆp = ¯x.

9.82 (a) The likelihood function is

L(x1, . . . , xn) =

Yn

i=1

f(xi; α, β) = (αβ)n

Yn

i=1

x −1

i e− x

i

= (αβ)ne−

n P

i

=

1

x

i

Yn

i=1

xi

! −1

.

(b) So, the log-likelihood can be expressed as

ln L = n[ln(α) + ln(β)] − α

Xn

i=1

x

i + (β − 1)

Xn

i=1

ln(xi).

To solve for the maximum likelihood estimate, we need to solve the following two

equations

∂ lnL

∂α

= 0, and

∂ ln L

∂β

= 0.

9.83 (a) The likelihood function is

L(x1, . . . , xn) =

Yn

i=1

f(xi; μ, σ) =

Yn

i=1

1

√2πσxi

e−[ln(xi)−μ]2

2 2

=

1

(2π)n/2σn

Qn

i=1

xi

exp

(

−

1

2σ2

Xn

i=1

[ln(xi) − μ]2

)

.

114 Chapter 9 One- and Two-Sample Estimation Problems

(b) It is easy to obtain

ln L = −

n

2

ln(2π) −

n

2

ln(σ2) −

Xn

i=1

ln(xi) −

1

2σ2

Xn

i=1

[ln(xi) − μ]2.

So, setting 0 = @ lnL

@μ = 1

2

Pn

i=1

[ln(xi) − μ], we obtain ˆμ = 1

n

Pn

i=1

ln(xi), and setting

0 = @ lnL

@ 2 = − n

2 2 + 1

2 4

Pn

i=1

[ln(xi) − μ]2, we get ˆσ2 = 1

n

Pn

i=1

[ln(xi) − ˆμ]2.

9.84 (a) The likelihood function is

L(x1, . . . , xn) =

1

βn (α)n

Yn

i=1

x −1

i e−xi/

=

1

βn (α)n

Yn

i=1

xi

! −1

e−

n P

i

=

1

(xi/ )

.

(b) Hence

ln L = −nα ln(β) − n ln((α)) + (α − 1)

Xn

i=1

ln(xi) −

1

β

Xn

i=1

xi.

Taking derivatives of ln L with respect to α and β, respectively and setting both

as zeros. Then solve them to obtain the maximum likelihood estimates.

9.85 L(x) = px(1 − p)1−x, and ln L = x ln(p) + (1 − x) ln(1 − p), with @ lnL

@p = x

p − 1−x

1−p = 0,

we obtain ˆp = x = 1.

9.86 From the density function b∗(x; p) =

x−1

k−1

pk(1 − p)x−k, we obtain

ln L = ln

x − 1

k − 1

+ k ln p + (n − k) ln(1 − p).

Setting @ lnL

@p = k

p − n−k

1−p = 0, we obtain ˆp = k

n.

9.87 For the estimator S2,

V ar(S2) =

1

(n − 1)2 V ar

"

Xn

i=1

(xi − ¯x)2

#

=

1

(n − 1)2V ar(σ2χ2

n−1)

=

1

(n − 1)2 σ4[2(n − 1)] =

2σ4

n − 1

.

For the estimator b σ2, we have

V ar( b σ2) =

2σ4(n − 1)

n2 .

Solutions for Exercises in Chapter 9 115

9.88 n = 7, ¯ d = 3.557, sd = 2.776, and t0.025 = 2.447 with 6 degrees of freedom. So,

3.557 ± (2.447)

2.776

√7

= 3.557 ± 2.567,

which yields 0.99 < μD < 6.12. Since 0 is not in the interval, the claim appears valid.

9.89 n = 75, x = 28, hence ˆp = 28

75 = 0.3733. Since z0.025 = 1.96, a 95% confidence interval

for p can be calculate as

0.3733 ± (1.96)

r

(0.3733)(0.6267)

75

= 0.3733 ± 0.1095,

which yields 0.2638 < p < 0.4828. Since the interval contains 0.421, the claim made

by the Roanoke Times seems reasonable.

9.90 n = 12, ¯ d = 40.58, sd = 15.791, and t0.025 = 2.201 with 11 degrees of freedom. So,

40.58 ± (2.201)

15.791

√12

= 40.58 ± 10.03,

which yields 30.55 < μD < 50.61.

9.91 n = 6, ¯ d = 1.5, sd = 1.543, and t0.025 = 2.571 with 5 degrees of freedom. So,

1.5 ± (2.571)

1.543

√6

= 1.5 ± 1.62,

which yields −0.12 < μD < 3.12.

9.92 n = 12, ¯ d = 417.5, sd = 1186.643, and t0.05 = 1.796 with 11 degrees of freedom. So,

417.5 ± (1.796)

1186.643

√12

= 417.5 ± 615.23,

which yields −197.73 < μD < 1032.73.

9.93 np = nu = 8, ¯xp = 86, 250.000, ¯xu = 79, 837.500, σp = σu = 4, 000, and z0.025 = 1.96.

So,

(86250 − 79837.5) ± (1.96)(4000)

p

1/8 + 1/8 = 6412.5 ± 3920,

which yields 2, 492.5 < μp −μu < 10, 332.5. Hence, polishing does increase the average

endurance limit.

9.94 nA = 100, nB = 120, ˆpA = 24

100 = 0.24, ˆpB = 36

120 = 0.30, and z0.025 = 1.96. So,

(0.30 − 0.24) ± (1.96)

r

(0.24)(0.76)

100

+

(0.30)(0.70)

120

= 0.06 ± 0.117,

which yields −0.057 < pB − pA < 0.177.

116 Chapter 9 One- and Two-Sample Estimation Problems

9.95 nN = nO = 23, s2

N = 105.9271, s2

O = 77.4138, and f0.025(22, 22) = 2.358. So,

105.9271

77.4138

1

2.358

<

σ2

N

σ2

O

<

105.9271

77.4138

(2.358), or 0.58 <

σ2

N

σ2

O

< 3.23.

For the ratio of the standard deviations, the 95% confidence interval is approximately

0.76 <

σN

σO

< 1.80.

Since the intervals contain 1 we will assume that the variability did not change with

the local supplier.

9.96 nA = nB = 6, ¯xA = 0.1407, ¯xB = 0.1385, sA = 0.002805, sB = 0.002665, and sp =

002736. Using a 90% confidence interval for the difference in the population means,

t0.05 = 1.812 with 10 degrees of freedom, we obtain

(0.1407 − 0.1385) ± (1.812)(0.002736)

p

1/6 + 1/6 = 0.0022 ± 0.0029,

which yields −0.0007 < μA−μB < 0.0051. Since the 90% confidence interval contains 0,

we conclude that wire A was not shown to be better than wire B, with 90% confidence.

9.97 To calculate the maximum likelihood estimator, we need to use

ln L = ln

e−nμμ

n P

i

=

1

xi

Qn

i=1

xi!

= −nμ + ln(μ)

Xn

i=1

xi − ln(

Yn

i=1

xi!).

Taking derivative with respect to μ and setting it to zero, we obtain ˆμ = 1

n

Pn

i=1

xi = ¯x.

On the other hand, using the method of moments, we also get ˆμ = ¯x.

9.98 ˆμ = ¯x and ˆσ2 = 1

n−1

Pn

i=1

(xi − ¯x)2.

9.99 Equating ¯x = eμ+ 2/2 and s2 = (e2μ+ 2 )(e 2−1), we get ln(¯x) = μ+ 2

2 , or ˆμ = ln(¯x)−ˆ 2

2 .

On the other hand, ln s2 = 2μ + σ2 + ln(e 2 − 1). Plug in the form of ˆμ, we obtain

ˆσ2 = ln

1 + s2

¯x2

.

9.100 Setting ¯x = αβ and s2 = αβ2, we get ˆα = ¯x2

s2 , and ˆ β = s2

¯x .

9.101 n1 = n2 = 300, ¯x1 = 102300, ¯x2 = 98500, s1 = 5700, and s2 = 3800.

(a) z0.005 = 2.575. Hence,

(102300 − 98500) ± (2.575)

r

57002

300

+

38002

300

= 3800 ± 1018.46,

which yields 2781.54 < μ1 − μ2 < 4818.46. There is a significant difference in

salaries between the two regions.

Solutions for Exercises in Chapter 9 117

(b) Since the sample sizes are large enough, it is not necessary to assume the normality

due to the Central Limit Theorem.

(c) We assumed that the two variances are not equal. Here we are going to obtain

a 95% confidence interval for the ratio of the two variances. It is known that

f0.025(299, 299) = 1.255. So,

5700

3800

2 1

1.255

<

σ2

1

σ2

2

<

5700

3800

2

(1.255), or 1.793 <

σ2

1

σ2

2

< 2.824.

Since the confidence interval does not contain 1, the difference between the variances

is significant.

9.102 The error in estimation, with 95% confidence, is (1.96)(4000)

q

2

n. Equating this quantity

to 1000, we obtain

n = 2

(1.96)(4000)

1000

2

= 123,

when round up. Hence, the sample sizes in Review Exercise 9.101 is sufficient to

produce a 95% confidence interval on μ1 − μ2 having a width of $1,000.

9.103 n = 300, ¯x = 6.5 and s = 2.5. Also, 1 − α = 0.99 and 1 − γ = 0.95. Using Table A.7,

k = 2.522. So, the limit of the one-sided tolerance interval is 6.5+(2.522)(2.5) = 12.805.

Since this interval contains 10, the claim by the union leaders appears valid.

9.104 n = 30, x = 8, and z0.025 = 1.96. So,

4

15 ± (1.96)

r

(4/15)(11/15)

30

=

4

15 ± 0.158,

which yields 0.108 < p < 0.425.

9.105 n = (1.96)2(4/15)(11/15)

0.052 = 301, when round up.

9.106 n1 = n2 = 100, ˆp1 = 0.1, and ˆp2 = 0.06.

(a) z0.025 = 1.96. So,

(0.1 − 0.06) ± (1.96)

r

(0.1)(0.9)

100

+

(0.06)(0.94)

100

= 0.04 ± 0.075,

which yields −0.035 < p1 − p2 < 0.115.

(b) Since the confidence interval contains 0, it does not show sufficient evidence that

p1 > p2.

118 Chapter 9 One- and Two-Sample Estimation Problems

9.107 n = 20 and s2 = 0.045. It is known that χ2

0.025 = 32.825 and χ2

0.975 = 8.907 with 19

degrees of freedom. Hence the 95% confidence interval for σ2 can be expressed as

(19)(0.045)

32.825

< σ2 <

(19)(0.045)

8.907

, or 0.012 < σ2 < 0.045.

Therefore, the 95% confidence interval for σ can be approximated as

0.110 < σ < 0.212.

Since 0.3 falls outside of the confidence interval, there is strong evidence that the

process has been improved in variability.

9.108 nA = nB = 15, ¯yA = 87, sA = 5.99, ¯yB = 75, sB = 4.85, sp = 5.450, and t0.025 = 2.048

with 28 degrees of freedom. So,

(87 − 75) ± (2.048)(5.450)

r

1

15

+

1

15

= 12 ± 4.076,

which yields 7.924 < μA − μB < 16.076. Apparently, the mean operating costs of type

A engines are higher than those of type B engines.

9.109 Since the unbiased estimators of σ2

1 and σ2

2 are S2

1 and S2

2 , respectively,

E(S2) =

1

n1 + n2 − 2

[(n1 − 1)E(S2

1 ) + (n2 − 1)E(S2

2 )] =

(n1 − 1)σ2

1 + (n2 − 1)σ2

2

n1 + n2 − 2

.

If we assume σ2

1 = σ2

2 = σ2, the right hand side of the above is σ2, which means that

S2 is unbiased for σ2.

9.110 n = 15, ¯x = 3.2, and s = 0.6.

(a) t0.01 = 2.624 with 14 degrees of freedom. So, a 99% left-sided confidence interval

has an upper bound of 3.2 + (2.624) 0.6 √15

= 3.607 seconds. We assumed normality

in the calculation.

(b) 3.2 + (2.624)(0.6)

q

1 + 1

15 = 4.826. Still, we need to assume normality in the

distribution.

(c) 1 − α = 0.99 and 1 − γ = 0.95. So, k = 3.520 with n = 15. So, the upper bound

is 3.2 + (3.520)(0.6) = 5.312. Hence, we are 99% confident to claim that 95% of

the pilot will have reaction time less than 5.312 seconds.

9.111 n = 400, x = 17, so ˆp = 17

400 = 0.0425.

(a) z0.025 = 1.96. So,

0.0425 ± (1.96)

r

(0.0425)(0.9575)

400

= 0.0425 ± 0.0198,

which yields 0.0227 < p < 0.0623.

Solutions for Exercises in Chapter 9 119

(b) z0.05 = 1.645. So, the upper bound of a left-sided 95% confidence interval is

0.0425 + (1.645)

q

(0.0425)(0.9575)

400 = 0.0591.

(c) Using both intervals, we do not have evidence to dispute suppliers’ claim.

Chapter 10

One- and Two-Sample Tests of

Hypotheses

10.1 (a) Conclude that fewer than 30% of the public are allergic to some cheese products

when, in fact, 30% or more are allergic.

(b) Conclude that at least 30% of the public are allergic to some cheese products

when, in fact, fewer than 30% are allergic.

10.2 (a) The training course is effective.

(b) The training course is effective.

10.3 (a) The firm is not guilty.

(b) The firm is guilty.

10.4 (a) α = P(X ≤ 5 | p = 0.6)+P(X ≥ 13 | p = 0.6) = 0.0338+(1−0.9729) = 0.0609.

(b) β = P(6 ≤ X ≤ 12 | p = 0.5) = 0.9963 − 0.1509 = 0.8454.

β = P(6 ≤ X ≤ 12 | p = 0.7) = 0.8732 − 0.0037 = 0.8695.

(c) This test procedure is not good for detecting differences of 0.1 in p.

10.5 (a) α = P(X < 110 | p = 0.6) + P(X > 130 | p = 0.6) = P(Z < −1.52) + P(Z >

1.52) = 2(0.0643) = 0.1286.

(b) β = P(110 < X < 130 | p = 0.5) = P(1.34 < Z < 4.31) = 0.0901.

β = P(110 < X < 130 | p = 0.7) = P(−4.71 < Z < −1.47) = 0.0708.

(c) The probability of a Type I error is somewhat high for this procedure, although

Type II errors are reduced dramatically.

10.6 (a) α = P(X ≤ 3 | p = 0.6) = 0.0548.

(b) β = P(X > 3 | p = 0.3) = 1 − 0.6496 = 0.3504.

β = P(X > 3 | p = 0.4) = 1 − 0.3823 = 0.6177.

β = P(X > 3 | p = 0.5) = 1 − 0.1719 = 0.8281.

121

122 Chapter 10 One- and Two-Sample Tests of Hypotheses

10.7 (a) α = P(X ≤ 24 | p = 0.6) = P(Z < −1.59) = 0.0559.

(b) β = P(X > 24 | p = 0.3) = P(Z > 2.93) = 1 − 0.9983 = 0.0017.

β = P(X > 24 | p = 0.4) = P(Z > 1.30) = 1 − 0.9032 = 0.0968.

β = P(X > 24 | p = 0.5) = P(Z > −0.14) = 1 − 0.4443 = 0.5557.

10.8 (a) n = 12, p = 0.7, and α = P(X > 11) = 0.0712 + 0.0138 = 0.0850.

(b) n = 12, p = 0.9, and β = P(X ≤ 10) = 0.3410.

10.9 (a) n = 100, p = 0.7, μ = np = 70, and σ = √npq =

p

(100)(0.7)(0.3) = 4.583.

Hence z = 82.5−70

4.583 = 0.3410. Therefore,

α = P(X > 82) = P(Z > 2.73) = 1 − 0.9968 = 0.0032.

(b) n = 100, p = 0.9, μ = np = 90, and σ = √npq =

p

(100)(0.9)(0.1) = 3. Hence

z = 82.5−90

3 = −2.5. So,

β = P(X ≤ 82) = P(X < −2.5) = 0.0062.

10.10 (a) n = 7, p = 0.4, α = P(X ≤ 2) = 0.4199.

(b) n = 7, p = 0.3, β = P(X ≥ 3) = 1 − P(X ≤ 2) = 1 − 0.6471 = 0.3529.

10.11 (a) n = 70, p = 0.4, μ = np = 28, and σ = √npq = 4.099, with z = 23.5−28

4.099 = −1.10.

Then α = P(X < 24) = P(Z < −1.10) = 0.1357.

(b) n = 70, p = 0.3, μ = np = 21, and σ = √npq = 3.834, with z = 23.5−21

3.834 = 0.65

Then β = P(X ≥ 24) = P(Z > 0.65) = 0.2578.

10.12 (a) n = 400, p = 0.6, μ = np = 240, and σ = √npq = 9.798, with

z1 =

259.5 − 240

9.978

= 1.990, and z2 =

220.5 − 240

9.978

= −1.990.

Hence,

α = 2P(Z < −1.990) = (2)(0.0233) = 0.0466.

(b) When p = 0.48, then μ = 192 and σ = 9.992, with

z1 =

220.5 − 192

9.992

= 2.852, and z2 =

259.5 − 192

9.992

= 6.755.

Therefore,

β = P(2.852 < Z < 6.755) = 1 − 0.9978 = 0.0022.

10.13 From Exercise 10.12(a) we have μ = 240 and σ = 9.798. We then obtain

z1 =

214.5 − 240

9.978

= −2.60, and z2 =

265.5 − 240

9.978

= 2.60.

Solutions for Exercises in Chapter 10 123

So

α = 2P(Z < −2.60) = (2)(0.0047) = 0.0094.

Also, from Exercise 10.12(b) we have μ = 192 and σ = 9.992, with

z1 =

214.5 − 192

9.992

= 2.25, and z2 =

265.5 − 192

9.992

= 7.36.

Therefore,

β = P(2.25 < Z < 7.36) = 1 − 0.9878 = 0.0122.

10.14 (a) n = 50, μ = 15, σ = 0.5, and σ ¯X = 0.5 √50

= 0.071, with z = 14.9−15

0.071 = −1.41.

Hence, α = P(Z < −1.41) = 0.0793.

(b) If μ = 14.8, z = 14.9−14.8

0.071 = 1.41. So, β = P(Z > 1.41) = 0.0793.

If μ = 14.9, then z = 0 and β = P(Z > 0) = 0.5.

10.15 (a) μ = 200, n = 9, σ = 15 and σ ¯X = 15

3 = 5. So,

z1 =

191 − 200

5

= −1.8, and z2 =

209 − 200

5

= 1.8,

with α = 2P(Z < −1.8) = (2)(0.0359) = 0.0718.

(b) If μ = 215, then z − 1 = 191−215

5 = −4.8 and z2 = 209−215

5 = −1.2, with

β = P(−4.8 < Z < −1.2) = 0.1151 − 0 = 0.1151.

10.16 (a) When n = 15, then σ ¯X = 15

5 = 3, with μ = 200 and n = 25. Hence

z1 =

191 − 200

3

= −3, and z2 =

209 − 200

3

= 3,

with α = 2P(Z < −3) = (2)(0.0013) = 0.0026.

(b) When μ = 215, then z − 1 = 191−215

3 = −8 and z2 = 209−215

3 = −2, with

β = P(−8 < Z < −2) = 0.0228 − 0 = 0.0228.

10.17 (a) n = 50, μ = 5000, σ = 120, and σ ¯X = 120 √50

= 16.971, with z = 4970−5000

16.971 = −1.77

and α = P(Z < −1.77) = 0.0384.

(b) If μ = 4970, then z = 0 and hence β = P(Z > 0) = 0.5.

If μ = 4960, then z = 4970−4960

16.971 = 0.59 and β = P(Z > 0.59) = 0.2776.

10.18 The OC curve is shown next.

124 Chapter 10 One- and Two-Sample Tests of Hypotheses

180 190 200 210 220

0.0 0.2 0.4 0.6 0.8

OC curve

m

Probability of accepting the null hypothesis

10.19 The hypotheses are

H0 : μ = 800,

H1 : μ 6= 800.

Now, z = 788−800

40/√30

= −1.64, and P-value= 2P(Z < −1.64) = (2)(0.0505) = 0.1010.

Hence, the mean is not significantly different from 800 for α < 0.101.

10.20 The hypotheses are

H0 : μ = 5.5,

H1 : μ < 5.5.

Now, z = 5.23−5.5

0.24/√64

= −9.0, and P-value= P(Z < −9.0) ≈ 0. The White Cheddar

Popcorn, on average, weighs less than 5.5oz.

10.21 The hypotheses are

H0 : μ = 40 months,

H1 : μ < 40 months.

Now, z = 38−40

5.8/√64

= −2.76, and P-value= P(Z < −2.76) = 0.0029. Decision: reject

H0.

10.22 The hypotheses are

H0 : μ = 162.5 centimeters,

H1 : μ 6= 162.5 centimeters.

Now, z = 165.2−162.5

6.9/√50

= 2.77, and P-value= 2P(Z > 2.77) = (2)(0.0028) = 0.0056.

Decision: reject H0 and conclude that μ 6= 162.5.

Solutions for Exercises in Chapter 10 125

10.23 The hypotheses are

H0 : μ = 20, 000 kilometers,

H1 : μ > 20, 000 kilometers.

Now, z = 23,500−20,000

3900/√100

= 8.97, and P-value= P(Z > 8.97) ≈ 0. Decision: reject H0 and

conclude that μ 6= 20, 000 kilometers.

10.24 The hypotheses are

H0 : μ = 8,

H1 : μ > 8.

Now, z = 8.5−8

2.25/√225

= 3.33, and P-value= P(Z > 3.33) = 0.0004. Decision: Reject H0

and conclude that men who use TM, on average, mediate more than 8 hours per week.

10.25 The hypotheses are

H0 : μ = 10,

H1 : μ 6= 10.

α = 0.01 and df = 9.

Critical region: t < −3.25 or t > 3.25.

Computation: t = 10.06−10

0.246/√10

= 0.77.

Decision: Fail to reject H0.

10.26 The hypotheses are

H0 : μ = 220 milligrams,

H1 : μ > 220 milligrams.

α = 0.01 and df = 9.

Critical region: t > 1.729.

Computation: t = 224−220

24.5/√20

= 4.38.

Decision: Reject H0 and claim μ > 220 milligrams.

10.27 The hypotheses are

H0 : μ1 = μ2,

H1 : μ1 > μ2.

Since sp =

q

(29)(10.5)2+(29)(10.2)2

58 = 10.35, then

P

"

T >

34.0

10.35

p

1/30 + 1/30

#

= P(Z > 12.72) ≈ 0.

Hence, the conclusion is that running increases the mean RMR in older women.

126 Chapter 10 One- and Two-Sample Tests of Hypotheses

10.28 The hypotheses are

H0 : μC = μA,

H1 : μC > μA,

with sp =

q

(24)(1.5)2+(24)(1.25)2

48 = 1.3807. We obtain t = 20.0−12.0

1.3807√2/25

= 20.48. Since

P(T > 20.48) ≈ 0, we conclude that the mean percent absorbency for the cotton fiber

is significantly higher than the mean percent absorbency for acetate.

10.29 The hypotheses are

H0 : μ = 35 minutes,

H1 : μ < 35 minutes.

α = 0.05 and df = 19.

Critical region: t < −1.729.

Computation: t = 33.1−35

4.3/√20

= −1.98.

Decision: Reject H0 and conclude that it takes less than 35 minutes, on the average,

to take the test.

10.30 The hypotheses are

H0 : μ1 = μ2,

H1 : μ1 6= μ2.

Since the variances are known, we obtain z = 81−76 √5.22/25+3.52/36

= 4.22. So, P-value≈ 0

and we conclude that μ1 > μ2.

10.31 The hypotheses are

H0 : μA − μB = 12 kilograms,

H1 : μA − μB > 12 kilograms.

α = 0.05.

Critical region: z > 1.645.

Computation: z = (86.7−77.8)−12 √(6.28)2/50+(5.61)2/50

= −2.60. So, fail to reject H0 and conclude that

the average tensile strength of thread A does not exceed the average tensile strength

of thread B by 12 kilograms.

10.32 The hypotheses are

H0 : μ1 − μ2 = $2, 000,

H1 : μ1 − μ2 > $2, 000.

Solutions for Exercises in Chapter 10 127

α = 0.01.

Critical region: z > 2.33.

Computation: z = (70750−65200)−2000 √(6000)2/200+(5000)2/200

= 6.43, with a P-value= P(Z > 6.43) ≈

0. Reject H0 and conclude that the mean salary for associate professors in research

institutions is $2000 higher than for those in other institutions.

10.33 The hypotheses are

H0 : μ1 − μ2 = 0.5 micromoles per 30 minutes,

H1 : μ1 − μ2 > 0.5 micromoles per 30 minutes.

α = 0.01.

Critical region: t > 2.485 with 25 degrees of freedom.

Computation: s2

p = (14)(1.5)2+(11)(1.2)2

25 = 1.8936, and t = (8.8−7.5)−0.5

√1.8936√1/15+1/12

= 1.50. Do

not reject H0.

10.34 The hypotheses are

H0 : μ1 − μ2 = 8,

H1 : μ1 − μ2 < 8.

Computation: s2

p = (10)(4.7)2+(16)(6.1)2

26 = 31.395, and t = (85−79)−8

√31.395√1/11+1/17

= −0.92.

Using 28 degrees of freedom and Table A.4, we obtain that 0.15 < P-value < 0.20.

Decision: Do not reject H0.

10.35 The hypotheses are

H0 : μ1 − μ2 = 0,

H1 : μ1 − μ2 < 0.

α = 0.05

Critical region: t < −1.895 with 7 degrees of freedom.

Computation: sp =

q

(3)(1.363)+(4)(3.883)

7 = 1.674, and t = 2.075−2.860

1.674√1/4+1/5

= −0.70.

Decision: Do not reject H0.

10.36 The hypotheses are

H0 : μ1 = μ2,

H1 : μ1 6= μ2.

Computation: sp =

q

51002+59002

2 = 5515, and t = 37,900−39,800

5515√1/12+1/12

= −0.84.

Using 22 degrees of freedom and since 0.20 < P(T < −0.84) < 0.3, we obtain 0.4 <

P-value < 0.6. Decision: Do not reject H0.

128 Chapter 10 One- and Two-Sample Tests of Hypotheses

10.37 The hypotheses are

H0 : μ1 − μ2 = 4 kilometers,

H1 : μ1 − μ2 6= 4 kilometers.

α = 0.10 and the critical regions are t < −1.725 or t > 1.725 with 20 degrees of

freedom.

Computation: t = 5−4

(0.915)√1/12+1/10

= 2.55.

Decision: Reject H0.

10.38 The hypotheses are

H0 : μ1 − μ2 = 8,

H1 : μ1 − μ2 < 8.

α = 0.05 and the critical region is t < −1.714 with 23 degrees of freedom.

Computation: sp =

q

(9)(3.2)2+(14)(2.8)2

23 = 2.963, and t = 5.5−8

2.963√1/10+1/15

= −2.07.

Decision: Reject H0 and conclude that μ1 − μ2 < 8 months.

10.39 The hypotheses are

H0 : μII − μI = 10,

H1 : μII − μI > 10.

α = 0.1.

Degrees of freedom is calculated as

v =

(78.8/5 + 913.333/7)2

(78.8/5)2/4 + (913/333/7)2/6

= 7.38,

hence we use 7 degrees of freedom with the critical region t > 2.998.

Computation: t = (110−97.4)−10 √78.800/5+913.333/7

= 0.22.

Decision: Fail to reject H0.

10.40 The hypotheses are

H0 : μS = μN,

H1 : μS 6= μN.

Degrees of freedom is calculated as

v =

(0.3914782/8 + 0.2144142/24)2

(0.3914782/8)2/7 + (0.2144142/24)2/23

= 8.

Computation: t = 0.97625−0.91583 √0.3914782/8+0.2144142/24

= −0.42. Since 0.3 < P(T < −0.42) < 0.4,

we obtain 0.6 < P-value < 0.8.

Decision: Fail to reject H0.

Solutions for Exercises in Chapter 10 129

10.41 The hypotheses are

H0 : μ1 = μ2,

H1 : μ1 6= μ2.

α = 0.05.

Degrees of freedom is calculated as

v =

(7874.3292/16 + 2479/5032/12)2

(7874.3292/16)2/15 + (2479.5032/12)2/11

= 19 degrees of freedom.

Critical regions t < −2.093 or t > 2.093.

Computation: t = 9897.500−4120.833 √7874.3292/16+2479.5032/12

= 2.76.

Decision: Reject H0 and conclude that μ1 > μ2.

10.42 The hypotheses are

H0 : μ1 = μ2,

H1 : μ1 6= μ2.

α = 0.05.

Critical regions t < −2.776 or t > 2.776, with 4 degrees of freedom.

Computation: ¯ d = −0.1, sd = 0.1414, so t = −0.1

0.1414/√5

= −1.58.

Decision: Do not reject H0 and conclude that the two methods are not significantly

different.

10.43 The hypotheses are

H0 : μ1 = μ2,

H1 : μ1 > μ2.

Computation: ¯ d = 0.1417, sd = 0.198, t = 0.1417

0.198/√12

= 2.48 and 0.015 < P-value < 0.02

with 11 degrees of freedom.

Decision: Reject H0 when a significance level is above 0.02.

10.44 The hypotheses are

H0 : μ1 − μ2 = 4.5 kilograms,

H1 : μ1 − μ2 < 4.5 kilograms.

Computation: ¯ d = 3.557, sd = 2.776, t = 3.557−4.5

2.778/√7

= −0.896, and 0.2 < P-value < 0.3

with 6 degrees of freedom.

Decision: Do not reject H0.

130 Chapter 10 One- and Two-Sample Tests of Hypotheses

10.45 The hypotheses are

H0 : μ1 = μ2,

H1 : μ1 < μ2.

Computation: ¯ d = −54.13, sd = 83.002, t = −54.13

83.002/√15

= −2.53, and 0.01 < P-value <

0.015 with 14 degrees of freedom.

Decision: Reject H0.

10.46 The hypotheses are

H0 : μ1 = μ2,

H1 : μ1 6= μ2.

α = 0.05.

Critical regions are t < −2.365 or t > 2.365 with 7 degrees of freedom.

Computation: ¯ d = 198.625, sd = 210.165, t = 198.625

210.165/√8

= 2.67.

Decision: Reject H0; length of storage influences sorbic acid residual concentrations.

10.47 n = (1.645+1.282)2(0.24)2

0.32 = 5.48. The sample size needed is 6.

10.48 β = 0.1, σ = 5.8, δ = 35.9 − 40 = −4.1. Assume α = 0.05 then z0.05 = 1.645,

z0.10 = 1.28. Therefore,

n =

(1.645 + 1.28)2(5.8)2

(−4.1)2 = 17.12 ≈ 18 due to round up.

10.49 1 − β = 0.95 so β = 0.05, δ = 3.1 and z0.01 = 2.33. Therefore,

n =

(1.645 + 2.33)2(6.9)2

3.12 = 78.28 ≈ 79 due to round up.

10.50 β = 0.05, δ = 8, α = 0.05, z0.05 = 1.645, σ1 = 6.28 and σ2 = 5.61. Therefore,

n =

(1.645 + 1.645)2(6.282 + 5.612)

82 = 11.99 ≈ 12 due to round up.

10.51 n = 1.645+0.842)2(2.25)2

[(1.2)(2.25)]2 = 4.29. The sample size would be 5.

10.52 σ = 1.25, α = 0.05, β = 0.1, δ = 0.5, so = 0.5

1.25 = 0.4. Using Table A.8 we find

n = 68.

10.53 (a) The hypotheses are

H0 : Mhot −Mcold = 0,

H1 : Mhot −Mcold 6= 0.

Solutions for Exercises in Chapter 10 131

(b) Use paired T-test and find out t = 0.99 with 0.3 < P-value < 0.4. Hence, fail to

reject H0.

10.54 Using paired T-test, we find out t = 2.4 with 8 degrees of freedom. So, 0.02 <

P-value < 0.025. Reject H0; breathing frequency significantly higher in the presence

of CO.

10.55 The hypotheses are

H0 : p = 0.40,

H1 : p > 0.40.

Denote by X for those who choose lasagna.

P-value = P(X ≥ 9 | p = 0.40) = 0.4044.

The claim that p = 0.40 is not refuted.

10.56 The hypotheses are

H0 : p = 0.40,

H1 : p > 0.40.

α = 0.05.

Test statistic: binomial variable X with p = 0.4 and n = 15.

Computation: x = 8 and np0 = (15)(0.4) = 6. Therefore, from Table A.1,

P-value = P(X ≥ 8 | p = 0.4) = 1 − P(X ≤ 7 | p = 0.4) = 0.2131,

which is larger than 0.05.

Decision: Do not reject H0.

10.57 The hypotheses are

H0 : p = 0.5,

H1 : p < 0.5.

P-value = P(X ≤ 5 | p = 0.05) = 0.0207.

Decision: Reject H0.

10.58 The hypotheses are

H0 : p = 0.6,

H1 : p < 0.6.

So

P-value ≈ P

Z <

110 − (200)(0.6) p

(200)(0.6)(0.4)

!

= P(Z < −1.44) = 0.0749.

Decision: Fail to reject H0.

132 Chapter 10 One- and Two-Sample Tests of Hypotheses

10.59 The hypotheses are

H0 : p = 0.2,

H1 : p < 0.2.

Then

P-value ≈ P

Z <

136 − (1000)(0.2) p

(1000)(0.2)(0.8)

!

= P(Z < −5.06) ≈ 0.

Decision: Reject H0; less than 1/5 of the homes in the city are heated by oil.

10.60 The hypotheses are

H0 : p = 0.25,

H1 : p > 0.25.

α = 0.05.

Computation:

P-value ≈ P

Z >

28 − (90)(0.25) p

(90)(0.25)(0.75)

!

= P(Z > 1, 34) = 0.091.

Decision: Fail to reject H0; No sufficient evidence to conclude that p > 0.25.

10.61 The hypotheses are

H0 : p = 0.8,

H1 : p > 0.8.

α = 0.04.

Critical region: z > 1.75.

Computation: z = 250−(300)(0.8) √(300)(0.8)(0.2)

= 1.44.

Decision: Fail to reject H0; it cannot conclude that the new missile system is more

accurate.

10.62 The hypotheses are

H0 : p = 0.25,

H1 : p > 0.25.

α = 0.05.

Critical region: z > 1.645.

Computation: z = 16−(48)(0.25) √(48)(0.25)(0.75)

= 1.333.

Decision: Fail to reject H0. On the other hand, we can calculate

P-value = P(Z > 1.33) = 0.0918.

Solutions for Exercises in Chapter 10 133

10.63 The hypotheses are

H0 : p1 = p2,

H1 : p1 6= p2.

Computation: ˆp = 63+59

100+125 = 0.5422, z = (63/100)−(59/125) √(0.5422)(0.4578)(1/100+1/125)

= 2.36, with

P-value = 2P(Z > 2.36) = 0.0182.

Decision: Reject H0 at level 0.0182. The proportion of urban residents who favor the

nuclear plant is larger than the proportion of suburban residents who favor the nuclear

plant.

10.64 The hypotheses are

H0 : p1 = p2,

H1 : p1 > p2.

Computation: ˆp = 240+288

300+400 = 0.7543, z = (240/300)−(288/400) √(0.7543)(0.2457)(1/300+1/400)

= 2.44, with

P-value = P(Z > 2.44) = 0.0073.

Decision: Reject H0. The proportion of couples married less than 2 years and planning

to have children is significantly higher than that of couples married 5 years and planning

to have children.

10.65 The hypotheses are

H0 : pU = pR,

H1 : pU > pR.

Computation: ˆp = 20+10

200+150 = 0.085714, z = (20/200)−(10/150) √(0.085714)(0.914286)(1/200+1/150)

= 1.10, with

P-value = P(Z > 1.10) = 0.1357.

Decision: Fail to reject H0. It cannot be shown that breast cancer is more prevalent

in the urban community.

10.66 The hypotheses are

H0 : p1 = p2,

H1 : p1 > p2.

Computation: ˆp = 29+56

120+280 = 0.2125, z = (29/120)−(56/280)) √(0.2125)(0.7875)(1/120+1/280)

= 0.93, with

P-value = P(Z > 0.93) = 0.1762.

Decision: Fail to reject H0. There is no significant evidence to conclude that the new

medicine is more effective.

134 Chapter 10 One- and Two-Sample Tests of Hypotheses

10.67 The hypotheses are

H0 : σ2 = 0.03,

H1 : σ2 6= 0.03.

Computation: χ2 = (9)(0.24585)2

0.03 = 18.13. Since 0.025 < P(χ2 > 18.13) < 0.05 with 9

degrees of freedom, 0.05 < P-value = 2P(χ2 > 18.13) < 0.10.

Decision: Fail to reject H0; the sample of 10 containers is not sufficient to show that

σ2 is not equal to 0.03.

10.68 The hypotheses are

H0 : σ = 6,

H1 : σ < 6.

Computation: χ2 = (19)(4.51)2

36 = 10.74. Using the table, 1−0.95 < P(χ2 < 10.74) < 0.1

with 19 degrees of freedom, we obtain 0.05 < P-value < 0.1.

Decision: Fail to reject H0; there was not sufficient evidence to conclude that the

standard deviation is less then 6 at level α = 0.05 level of significance.

10.69 The hypotheses are

H0 : σ2 = 4.2 ppm,

H1 : σ2 6= 4.2 ppm.

Computation: χ2 = (63)(4.25)2

4.2 = 63.75. Since 0.3 < P(χ2 > 63.75) < 0.5 with 63

degrees of freedom, P-value = 2P(χ2 > 18.13) > 0.6 (In Microsoft Excel, if you type

“=2*chidist(63.75,63)”, you will get the P-value as 0.8898.

Decision: Fail to reject H0; the variance of aflotoxins is not significantly different from

4.2 ppm.

10.70 The hypotheses are

H0 : σ = 1.40,

H1 : σ > 1.40.

Computation: χ2 = (11)(1.75)2

1.4 = 17.19. Using the table, 0.1 < P(χ2 > 17.19) < 0.2

with 11 degrees of freedom, we obtain 0.1 < P-value < 0.2.

Decision: Fail to reject H0; the standard deviation of the contributions from the sanitation

department is not significantly greater than $1.40 at the α = 0.01 level of

significance.

10.71 The hypotheses are

H0 : σ2 = 1.15,

H1 : σ2 > 1.15.

Solutions for Exercises in Chapter 10 135

Computation: χ2 = (24)(2.03)2

1.15 = 42.37. Since 0.01 < P(χ2 > 42.37) < 0.02 with 24

degrees of freedom, 0.01 < P-value < 0.02.

Decision: Reject H0; there is sufficient evidence to conclude, at level α = 0.05, that

the soft drink machine is out of control.

10.72 (a) The hypotheses are

H0 : σ = 10.0,

H1 : σ 6= 10.0.

Computation: z = 11.9−10.0

10.0/√200

= 2.69. So P-value = P(Z < −2.69)+P(Z > 2.69) =

0.0072. There is sufficient evidence to conclude that the standard deviation is

different from 10.0.

(b) The hypotheses are

H0 : σ2 = 6.25,

H1 : σ2 < 6.25.

Computation: z = 2.1−2.5

2.5/√144

= −1.92. P-value = P(Z < −1.92) = 0.0274.

Decision: Reject H0; the variance of the distance achieved by the diesel model is

less than the variance of the distance achieved by the gasoline model.

10.73 The hypotheses are

H0 : σ2

1 = σ2

2,

H1 : σ2

1 > σ2

2.

Computation: f = (6.1)2

(5.3)2 = 1.33. Since f0.05(10, 13) = 2.67 > 1.33, we fail to reject

H0 at level α = 0.05. So, the variability of the time to assemble the product is not

significantly greater for men. On the other hand, if you use “=fdist(1.33,10,13)”, you

will obtain the P-value = 0.3095.

10.74 The hypotheses are

H0 : σ2

1 = σ2

2,

H1 : σ2

1 6= σ2

2.

Computation: f = (7874.329)2

(2479.503)2 = 10.09. Since f0.01(15, 11) = 4.25, the P-value >

(2)(0.01) = 0.02. Hence we reject H0 at level α = 0.02 and claim that the variances

for the two locations are significantly different. The P-value = 0.0004.

10.75 The hypotheses are

H0 : σ2

1 = σ2

2,

H1 : σ2

1 6= σ2

2.

136 Chapter 10 One- and Two-Sample Tests of Hypotheses

Computation: f = 78.800

913.333 = 0.086. Since P-value = 2P(f < 0.086) = (2)(0.0164) =

0.0328 for 4 and 6 degrees of freedom, the variability of running time for company 1 is

significantly less than, at level 0.0328, the variability of running time for company 2.

10.76 The hypotheses are

H0 : σA = σB,

H1 : σA 6= σB.

Computation: f = (0.0125)

0.0108 = 1.15. Since P-value = 2P(f > 1.15) = (2)(0.424) = 0.848

for 8 and 8 degrees of freedom, the two instruments appear to have similar variability.

10.77 The hypotheses are

H0 : σ1 = σ2,

H1 : σ1 6= σ2.

Computation: f = (0.0553)2

(0.0125)2 = 19.67. Since P-value = 2P(f > 19.67) = (2)(0.0004) =

0.0008 for 7 and 7 degrees of freedom, production line 1 is not producing as consistently

as production 2.

10.78 The hypotheses are

H0 : σ1 = σ2,

H1 : σ1 6= σ2.

Computation: s1 = 291.0667 and s2 = 119.3946, f = (291.0667)2

(119.3946)2 = 5.54. Since

P-value = 2P(f > 5.54) = (2)(0.0002) = 0.0004 for 19 and 19 degrees of freedom,

hydrocarbon emissions are more consistent in the 1990 model cars.

10.79 The hypotheses are

H0 : die is balanced,

H1 : die is unbalanced.

α = 0.01.

Critical region: χ2 > 15.086 with 5 degrees of freedom.

Computation: Since ei = 30, for i = 1, 2, . . . , 6, then

χ2 =

(28 − 30)2

30

+

(36 − 30)2

30

+ · · · +

(23 − 30)2

30

= 4.47.

Decision: Fail to reject H0; the die is balanced.

Solutions for Exercises in Chapter 10 137

10.80 The hypotheses are

H0 : coin is balanced,

H1 : coin is not balanced.

α = 0.05.

Critical region: χ2 > 3.841 with 1 degrees of freedom.

Computation: Since ei = 30, for i = 1, 2, . . . , 6, then

χ2 =

(63 − 50)2

50

+

(37 − 50)2

50

= 6.76.

Decision: Reject H0; the coin is not balanced.

10.81 The hypotheses are

H0 : nuts are mixed in the ratio 5:2:2:1,

H1 : nuts are not mixed in the ratio 5:2:2:1.

α = 0.05.

Critical region: χ2 > 7.815 with 3 degrees of freedom.

Computation:

Observed 269 112 74 45

Expected 250 100 100 50

χ2 =

(269 − 250)2

250

+

(112 − 100)2

100

+

(74 − 100)2

100

+

(45 − 50)2

50

= 10.14.

Decision: Reject H0; the nuts are not mixed in the ratio 5:2:2:1.

10.82 The hypotheses are

H0 : Distribution of grades is uniform,

H1 : Distribution of grades is not uniform.

α = 0.05.

Critical region: χ2 > 9.488 with 4 degrees of freedom.

Computation: Since ei = 20, for i = 1, 2, . . . , 5, then

χ2 =

(14 − 20)2

20

+

(18 − 20)2

20

+ · · ·

(16 − 20)2

20

= 10.0.

Decision: Reject H0; the distribution of grades is not uniform.

138 Chapter 10 One- and Two-Sample Tests of Hypotheses

10.83 The hypotheses are

H0 : Data follows the binomial distribution b(y; 3, 1/4),

H1 : Data does not follows the binomial distribution.

α = 0.01.

Computation: b(0; 3, 1/4) = 27/64, b(1; 3, 1/4) = 27/64, b(2; 3, 1/4) = 9/64, and

b(3; 3, 1/4) = 1/64. Hence e1 = 27, e2 = 27, e3 = 9 and e4 = 1. Combining the

last two classes together, we obtain

χ2 =

(21 − 27)2

27

+

(31 − 27)2

27

+

(12 − 10)2

10

= 2.33.

Critical region: χ2 > 9.210 with 2 degrees of freedom.

Decision: Fail to reject H0; the data is from a distribution not significantly different

from b(y; 3, 1/4).

10.84 The hypotheses are

H0 : Data follows the hypergeometric distribution h(x; 8, 3, 5),

H1 : Data does not follows the hypergeometric distribution.

α = 0.05.

Computation: h(0; 8, 3, 5) = 1/56, b(1; 8, 3, 5) = 15/56, b(2; 8, 3, 5) = 30/56, and

b(3; 8, 3, 5) = 10/56. Hence e1 = 2, e2 = 30, e3 = 60 and e4 = 20. Combining

the first two classes together, we obtain

χ2 =

(32 − 32)2

32

+

(55 − 60)2

60

+

(25 − 20)2

20

= 1.67.

Critical region: χ2 > 5.991 with 2 degrees of freedom.

Decision: Fail to reject H0; the data is from a distribution not significantly different

from h(y; 8, 3, 5).

10.85 The hypotheses are

H0 : f(x) = g(x; 1/2) for x = 1, 2, . . . ,

H1 : f(x) 6= g(x; 1/2).

α = 0.05.

Computation: g(x; 1/2) = 1

2x , for x = 1, 2, . . . , 7 and P(X ≥ 8) = 1

27 . Hence e1 = 128,

e2 = 64, e3 = 32, e4 = 16, e5 = 8, e6 = 4, e7 = 2 and e8 = 2. Combining the last three

classes together, we obtain

χ2 =

(136 − 128)2

128

+

(60 − 64)2

64

+

(34 − 32)2

32

+

(12 − 16)2

16

+

(9 − 8)2

8

+

(5 − 8)2

8

= 3.125

Critical region: χ2 > 11.070 with 5 degrees of freedom.

Decision: Fail to reject H0; f(x) = g(x; 1/2), for x = 1, 2, . . .

Solutions for Exercises in Chapter 10 139

10.88 The hypotheses are

H0 : Distribution of grades is normal n(x; 65, 21),

H1 : Distribution of grades is not normal.

α = 0.05.

Computation:

z values P(Z < z) P(zi−1 < Z < zi) ei oi

z1 = 19.5−65

21 = −2.17

z2 = 29.5−65

21 = −1.69

z3 = 39.5−65

21 = −1.21

0.0150

0.0454

0.1131

0.0150

0.0305

0.0676

0.9

1.8

4.1

6.8

3

2

3

8

z4 = 49.5−65

21 = −0.74

z5 = 59.5−65

21 = −0.26

z6 = 69.5−65

21 = 0.21

z7 = 79.5−65

21 = 0.69

z8 = 89.5−65

21 = 1.17

z9 = ∞

0.2296

0.3974

0.5832

0.7549

0.8790

1.0000

0.1165

0.1678

0.1858

0.1717

0.1241

0.1210

7.0

10.1

11.1

10.3

7.4

7.3

4

5

11

14

14

4

A goodness-of-fit test with 6 degrees of freedom is based on the following data:

oi 8 4 5 11 14 14 4

ei 6.8 7.0 10.1 11.1 10.3 7.4 7.3

Critical region: χ2 > 12.592.

χ2 =

(8 − 6.8)2

6.8

+

(4 − 7.0)2

7.0

+ · · · +

(4 − 7.3)2

7.3

= 12.78.

Decision: Reject H0; distribution of grades is not normal.

10.89 From the data we have

z values P(Z < z) P(zi−1 < Z < zi) ei oi

z1 = 0.795−1.8

0.4 = −2.51

z2 = 0.995−1.8

0.4 = −2.01

z3 = 1.195−1.8

0.4 = −1.51

z4 = 1.395−1.8

0.4 = −1.01

0.0060

0.0222

0.0655

0.1562

0.0060

0.0162

0.0433

0.0907

0.2

0.6

1.7

3.6

6.1

1

1

1

2

5

z5 = 1.595−1.8

0.4 = −0.51

z6 = 1.795−1.8

0.4 = −0.01

z7 = 1.995−1.8

0.4 = 0.49

z8 = 2.195−1.8

0.4 = 0.99

0.3050

0.4960

0.6879

0.8389

0.1488

0.1910

0.1919

0.1510

6.0

7.6

7.7

6.0

4

13

8

5

z9 = 2.395−1.8

0.4 = 1.49

z10 = ∞

0.9319

1.0000

0.0930

0.0681

3.7

2.7

6.4

3

2

5

140 Chapter 10 One- and Two-Sample Tests of Hypotheses

The hypotheses are

H0 : Distribution of nicotine contents is normal n(x; 1.8, 0.4),

H1 : Distribution of nicotine contents is not normal.

α = 0.01.

Computation: A goodness-of-fit test with 5 degrees of freedom is based on the following

data:

oi 5 4 13 8 5 5

ei 6.1 6.0 7.6 7.7 6.0 6.4

Critical region: χ2 > 15.086.

χ2 =

(5 − 6.1)2

6.1

+

(4 − 6.0)2

6.0

+ · · · +

(5 − 6.4)2

6.4

= 5.19.

Decision: Fail to reject H0; distribution of nicotine contents is not significantly different

from n(x; 1.8, 0.4).

10.90 The hypotheses are

H0 : Presence or absence of hypertension is independent of smoking habits,

H1 : Presence or absence of hypertension is not independent of smoking habits.

α = 0.05.

Critical region: χ2 > 5.991 with 2 degrees of freedom.

Computation:

Observed and expected frequencies

Nonsmokers Moderate Smokers Heavy Smokers Total

Hypertension 21 (33.4) 36 (30.0) 30 (23.6) 87

No Hypertension 48 (35.6) 26 (32.0) 19 (25.4) 93

Total 69 62 49 180

χ2 =

(21 − 33.4)2

33.4

+ · · · +

(19 − 25.4)2

25.4

= 14.60.

Decision: Reject H0; presence or absence of hypertension and smoking habits are not

independent.

10.91 The hypotheses are

H0 : A person’s gender and time spent watching television are independent,

H1 : A person’s gender and time spent watching television are not independent.

α = 0.01.

Critical region: χ2 > 6.635 with 1 degrees of freedom.

Computation:

Solutions for Exercises in Chapter 10 141

Observed and expected frequencies

Male Female Total

Over 25 hours 15 (20.5) 29 (23.5) 44

Under 25 hours 27 (21.5) 19 (24.5) 46

Total 42 48 90

χ2 =

(15 − 20.5)2

20.5

+

(29 − 23.5)2

23.5

+

(27 − 21.5)2

21.5

+

(19 − 24.5)2

24.5

= 5.47.

Decision: Fail to reject H0; a person’s gender and time spent watching television are

independent.

10.92 The hypotheses are

H0 : Size of family is independent of level of education of father,

H1 : Size of family and the education level of father are not independent.

α = 0.05.

Critical region: χ2 > 9.488 with 4 degrees of freedom.

Computation:

Observed and expected frequencies

Number of Children

Education 0–1 2–3 Over 3 Total

Elementary 14 (18.7) 37 (39.8) 32 (24.5) 83

Secondary 19 (17.6) 42 (37.4) 17 (23.0) 78

College 12 (8.7) 17 (18.8) 10 (11.5) 39

Total 45 96 59 200

χ2 =

(14 − 18.7)2

18.7

+

(37 − 39.8)2

39.8

+ · · · +

(10 − 11.5)2

11.5

= 7.54.

Decision: Fail to reject H0; size of family is independent of level of education of father.

10.93 The hypotheses are

H0 : Occurrence of types of crime is independent of city district,

H1 : Occurrence of types of crime is dependent upon city district.

α = 0.01.

Critical region: χ2 > 21.666 with 9 degrees of freedom.

Computation:

Observed and expected frequencies

District Assault Burglary Larceny Homicide Total

1 162 (186.4) 118 (125.8) 451 (423.5) 18 (13.3) 749

2 310 (380.0) 196 (256.6) 996 (863.4) 25 (27.1) 1527

3 258 (228.7) 193 (154.4) 458 (519.6) 10 (16.3) 919

4 280 (214.9) 175 (145.2) 390 (488.5) 19 (15.3) 864

Total 1010 682 2295 72 4059

142 Chapter 10 One- and Two-Sample Tests of Hypotheses

χ2 =

(162 − 186.4)2

186.4

+

(118 − 125.8)2

125.8

+ · · · +

(19 − 15.3)2

15.3

= 124.59.

Decision: Reject H0; occurrence of types of crime is dependent upon city district.

10.94 The hypotheses are

H0 : The three cough remedies are equally effective,

H1 : The three cough remedies are not equally effective.

α = 0.05.

Critical region: χ2 > 9.488 with 4 degrees of freedom.

Computation:

Observed and expected frequencies

NyQuil Robitussin Triaminic Total

No Relief

Some Relief

Total Relief

11 (11)

32 (29)

7 (10)

13 (11)

28 (29)

9 (10)

9 (11)

27 (29)

14 (10)

33

87

30

Total 50 50 50 150

χ2 =

(11 − 11)2

11

+

(13 − 11)2

11

+ · · · +

(14 − 10)2

10

= 3.81.

Decision: Fail to reject H0; the three cough remedies are equally effective.

10.95 The hypotheses are

H0 : The attitudes among the four counties are homogeneous,

H1 : The attitudes among the four counties are not homogeneous.

Computation:

Observed and expected frequencies

County

Attitude Craig Giles Franklin Montgomery Total

Favor 65 (74.5) 66 (55.9) 40 (37.3) 34 (37.3) 205

Oppose 42 (53.5) 30 (40.1) 33 (26.7) 42 (26.7) 147

No Opinion 93 (72.0) 54 (54.0) 27 (36.0) 24 (36.0) 198

Total 200 150 100 100 550

χ2 =

(65 − 74.5)2

74.5

+

(66 − 55.9)2

55.9

+ · · · +

(24 − 36.0)2

36.0

= 31.17.

Since P-value = P(χ2 > 31.17) < 0.001 with 6 degrees of freedom, we reject H0 and

conclude that the attitudes among the four counties are not homogeneous.

Solutions for Exercises in Chapter 10 143

10.96 The hypotheses are

H0 : The proportions of widows and widowers are equal with respect to the different

time period,

H1 : The proportions of widows and widowers are not equal with respect to the

different time period.

α = 0.05.

Critical region: χ2 > 5.991 with 2 degrees of freedom.

Computation:

Observed and expected frequencies

Years Lived Widow Widower Total

Less than 5 25 (32) 39 (32) 64

5 to 10 42 (41) 40 (41) 82

More than 10 33 (26) 21 (26) 54

Total 100 100 200

χ2 =

(25 − 32)2

32

+

(39 − 32)2

32

+ · · · +

(21 − 26)2

26

= 5.78.

Decision: Fail to reject H0; the proportions of widows and widowers are equal with

respect to the different time period.

10.97 The hypotheses are

H0 : Proportions of household within each standard of living category are equal,

H1 : Proportions of household within each standard of living category are not equal.

α = 0.05.

Critical region: χ2 > 12.592 with 6 degrees of freedom.

Computation:

Observed and expected frequencies

Period Somewhat Better Same Not as Good Total

1980: Jan.

May.

Sept.

1981: Jan.

72 (66.6)

63 (66.6)

47 (44.4)

40 (44.4)

144 (145.2)

135 (145.2)

100 (96.8)

105 (96.8)

84 (88.2)

102 (88.2)

53 (58.8)

55 (58.8)

300

300

200

200

Total 222 484 294 1000

χ2 =

(72 − 66.6)2

66.6

+

(144 − 145.2)2

145.2

+ · · · +

(55 − 58.8)2

58.8

= 5.92.

Decision: Fail to reject H0; proportions of household within each standard of living

category are equal.

144 Chapter 10 One- and Two-Sample Tests of Hypotheses

10.98 The hypotheses are

H0 : Proportions of voters within each attitude category are the same for each of the

three states,

H1 : Proportions of voters within each attitude category are not the same for each of

the three states.

α = 0.05.

Critical region: χ2 > 9.488 with 4 degrees of freedom.

Computation:

Observed and expected frequencies

Support Do not Support Undecided Total

Indiana

Kentucky

Ohio

82 (94)

107 (94)

93 (94)

97 (79)

66 (79)

74 (79)

21 (27)

27 (27)

33 (27)

200

200

200

Total 282 237 81 600

χ2 =

(82 − 94)2

94

+

(97 − 79)2

79

+ · · · +

(33 − 27)2

27

= 12.56.

Decision: Reject H0; the proportions of voters within each attitude category are not

the same for each of the three states.

10.99 The hypotheses are

H0 : Proportions of voters favoring candidate A, candidate B, or undecided are the

same for each city,

H1 : Proportions of voters favoring candidate A, candidate B, or undecided are not

the same for each city.

α = 0.05.

Critical region: χ2 > 5.991 with 2 degrees of freedom.

Computation:

Observed and expected frequencies

Richmond Norfolk Total

Favor A

Favor B

Undecided

204 (214.5)

211 (204.5)

85 (81)

225 (214.5)

198 (204.5)

77 (81)

429

409

162

Total 500 500 1000

Solutions for Exercises in Chapter 10 145

χ2 =

(204 − 214.5)2

214.5

+

(225 − 214.5)2

214.5

+ · · · +

(77 − 81)2

81

= 1.84.

Decision: Fail to reject H0; the proportions of voters favoring candidate A, candidate

B, or undecided are not the same for each city.

10.100 The hypotheses are

H0 : p1 = p2 = p3,

H1 : p1, p2, and p3 are not all equal.

α = 0.05.

Critical region: χ2 > 5.991 with 2 degrees of freedom.

Computation:

Observed and expected frequencies

Denver Phoenix Rochester Total

Watch Soap Operas

Do not Watch

52 (48)

148 (152)

31 (36)

119 (114)

37 (36)

113 (114)

120

380

Total 200 150 150 500

χ2 =

(52 − 48)2

48

+

(31 − 36)2

36

+ · · · +

(113 − 114)2

114

= 1.39.

Decision: Fail to reject H0; no difference among the proportions.

10.101 The hypotheses are

H0 : p1 = p2,

H1 : p1 > p2.

α = 0.01.

Critical region: z > 2.33.

Computation: ˆp1 = 0.31, ˆp2 = 0.24, ˆp = 0.275, and

z =

0.31 − 0.24 p

(0.275)(0.725)(1/100 + 1/100)

= 1.11.

Decision: Fail to reject H0; proportions are the same.

10.102 Using paired t-test, we observe that t = 1.55 with P-value > 0.05. Hence, the data

was not sufficient to show that the oxygen consumptions was higher when there was

little or not CO.

10.103 (a) H0 : μ = 21.8, H1 : μ 6= 21.8; critical region in both tails.

(b) H0 : p = 0.2, H1 : p > 0.2; critical region in right tail.

146 Chapter 10 One- and Two-Sample Tests of Hypotheses

(c) H0 : μ = 6.2, H1 : μ > 6.2; critical region in right tail.

(d) H0 : p = 0.7, H1 : p < 0.7; critical region in left tail.

(e) H0 : p = 0.58, H1 : p 6= 0.58; critical region in both tails.

(f) H0 : μ = 340, H1 : μ < 340; critical region in left tail.

10.104 The hypotheses are

H0 : p1 = p2,

H1 : p1 > p2.

α = 0.05.

Critical region: z > 1.645.

Computation: ˆp1 = 0.24, ˆp2 = 0.175, ˆp = 0.203, and

z =

0.24 − 0.175 p

(0.203)(0.797)(1/300 + 1/400)

= 2.12.

Decision: Reject H0; there is statistical evidence to conclude that more Italians prefer

white champagne at weddings.

10.105 n1 = n2 = 5, ¯x1 = 165.0, s1 = 6.442, ¯x2 = 139.8, s2 = 12.617, and sp = 10.02. Hence

t =

165 − 139.8

(10.02)

p

1/5 + 1/5

= 3.98.

This is a one-sided test. Therefore, 0.0025 < P-value < 0.005 with 8 degrees of

freedom. Reject H0; the speed is increased by using the facilitation tools.

10.106 (a) H0 : p = 0.2, H1 : p > 0.2; critical region in right tail.

(b) H0 : μ = 3, H1 : μ 6= 3; critical region in both tails.

(c) H0 : p = 0.15, H1 : p < 0.15; critical region in left tail.

(d) H0 : μ = $10, H1 : μ > $10; critical region in right tail.

(e) H0 : μ = 9, H1 : μ 6= 9; critical region in both tails.

10.107 The hypotheses are

H0 : p1 = p2 = p3,

H1 : p1, p2, and p3 are not all equal.

α = 0.01.

Critical region: χ2 > 9.210 with 2 degrees of freedom.

Computation:

Solutions for Exercises in Chapter 10 147

Observed and expected frequencies

Distributor

Nuts 1 2 3 Total

Peanuts

Other

345 (339)

155 (161)

313 (339)

187 (161)

359 (339)

141 (161)

1017

483

Total 500 500 500 1500

χ2 =

(345 − 339)2

339

+

(313 − 339)2

339

+ · · · +

(141 − 161)2

161

= 10.19.

Decision: Reject H0; the proportions of peanuts for the three distributors are not equal.

10.108 The hypotheses are

H0 : p1 − p2 = 0.03,

H1 : p1 − p2 > 0.03.

Computation: ˆp1 = 0.60 and ˆp2 = 0.48.

z =

(0.60 − 0.48) − 0.03 p

(0.60)(0.40)/200 + (0.48)(0.52)/500

= 2.18.

P-value = P(Z > 2.18) = 0.0146.

Decision: Reject H0 at level higher than 0.0146; the difference in votes favoring the

proposal exceeds 3%.

10.109 The hypotheses are

H0 : p1 = p2 = p3 = p4,

H1 : p1, p2, p3, and p4 are not all equal.

α = 0.01.

Critical region: χ2 > 11.345 with 3 degrees of freedom.

Computation:

Observed and expected frequencies

Preference Maryland Virginia Georgia Alabama Total

Yes

No

65 (74)

35 (26)

71 (74)

29 (26)

78 (74)

22 (26)

82 (74)

18 (26)

296

104

Total 100 100 100 100 400

χ2 =

(65 − 74)2

74

+

(71 − 74)2

74

+ · · · +

(18 − 26)2

26

= 8.84.

Decision: Fail to reject H0; the proportions of parents favoring Bibles in elementary

schools are the same across states.

148 Chapter 10 One- and Two-Sample Tests of Hypotheses

10.110 ¯ d = −2.905, sd = 3.3557, and t = ¯ d

sd/√n = −2.12. Since 0.025 < P(T > 2.12) < 0.05

with 5 degrees of freedom, we have 0.05 < P-value < 0.10. There is no significant

change in WBC leukograms.

10.111 n1 = 15, ¯x1 = 156.33, s1 = 33.09, n2 = 18, ¯x2 = 170.00 and s2 = 30.79. First we do the

f-test to test equality of the variances. Since f = s21

s22

= 1.16 and f0.05(15, 18) = 2.27,

we conclude that the two variances are equal.

To test the difference of the means, we first calculate sp = 31.85. Therefore, t =

156.33−170.00

(31.85)√1/15+1/18

= −1.23 with a P-value > 0.10.

Decision: H0 cannot be rejected at 0.05 level of significance.

10.112 n1 = n2 = 10, ¯x1 = 7.95, s1 = 1.10, ¯x2 = 10.26 and s2 = 0.57. First we do the f-test to

test equality of the variances. Since f = s21

s22

= 3.72 and f0.05(9, 9) = 3.18, we conclude

that the two variances are not equal at level 0.10.

To test the difference of the means, we first find the degrees of freedom v = 13 when

round up. Also, t = 7.95−10.26 √1.102/10+0.572/10

= −5.90 with a P-value < 0.0005.

Decision: Reject H0; there is a significant difference in the steel rods.

10.113 n1 = n2 = 10, ¯x1 = 21.5, s1 = 5.3177, ¯x2 = 28.3 and s2 = 5.8699. Since f = s21

s22

=

0.8207 and f0.05(9, 9) = 3.18, we conclude that the two variances are equal.

sp = 5.6001 and hence t = 21.5−28.3

(5.6001)√1/10+1/10

= −2.71 with 0.005 < P-value < 0.0075.

Decision: Reject H0; the high income neighborhood produces significantly more wastewater

to be treated.

10.114 n1 = n2 = 16, ¯x1 = 48.1875, s1 = 4.9962, ¯x2 = 43.7500 and s2 = 4.6833. Since

f = s21

s22

= 1.1381 and f0.05(15, 15) = 2.40, we conclude that the two variances are equal.

sp = 4.8423 and hence t = 48.1875−43.7500

(4.8423)√1/16+1/16

= 2.59. This is a two-sided test. Since

0.005 < P(T > 2.59) < 0.0075, we have 0.01 < P-value < 0.015.

Decision: Reject H0; there is a significant difference in the number of defects.

10.115 The hypotheses are:

H0 : μ = 24 × 10−4 gm,

H1 : μ < 24 × 10−4 gm.

t = 22.8−24

4.8/√50

= −1.77 with 0.025 < P-value < 0.05. Hence, at significance level of

α = 0.05, the mean concentration of PCB in malignant breast tissue is less than

24 × 10−4 gm.

Chapter 11

Simple Linear Regression and

Correlation

11.1 (a)

P

i

xi = 778.7,

P

i

yi = 2050.0,

P

i

x2

i = 26, 591.63,

P

i

xiyi = 65, 164.04, n = 25.

Therefore,

b =

(25)(65, 164.04)− (778.7)(2050.0)

(25)(26, 591.63) − (778.7)2 = 0.5609,

a =

2050 − (0.5609)(778.7)

25

= 64.53.

(b) Using the equation ˆy = 64.53 + 0.5609x with x = 30, we find ˆy = 64.53 +

(0.5609)(30) = 81.40.

(c) Residuals appear to be random as desired.

10 20 30 40 50 60

−30 −20 −10 0 10 20 30

Arm Strength

Residual

11.2 (a)

P

i

xi = 707,

P

i

yi = 658,

P

i

x2

i = 57, 557,

P

i

xiyi = 53, 258, n = 9.

b =

(9)(53, 258) − (707)(658)

(9)(57, 557) − (707)2 = 0.7771,

a =

658 − (0.7771)(707)

9

= 12.0623.

149

150 Chapter 11 Simple Linear Regression and Correlation

Hence ˆy = 12.0623 + 0.7771x.

(b) For x = 85, ˆy = 12.0623 + (0.7771)(85) = 78.

11.3 (a)

P

i

xi = 16.5,

P

i

yi = 100.4,

P

i

x2

i = 25.85,

P

i

xiyi = 152.59, n = 11. Therefore,

b =

(11)(152.59) − (16.5)(100.4)

(11)(25.85) − (16.5)2 = 1.8091,

a =

100.4 − (1.8091)(16.5)

11

= 6.4136.

Hence ˆy = 6.4136 + 1.8091x

(b) For x = 1.75, ˆy = 6.4136 + (1.8091)(1.75) = 9.580.

(c) Residuals appear to be random as desired.

1.0 1.2 1.4 1.6 1.8 2.0

−1.0 −0.5 0.0 0.5 1.0

Temperature

Residual

11.4 (a)

P

i

xi = 311.6,

P

i

yi = 297.2,

P

i

x2

i = 8134.26,

P

i

xiyi = 7687.76, n = 12.

b =

(12)(7687.26) − (311.6)(297.2)2

= − 0.6861,

a =

297.2 − (−0.6861)(311.6)

12

= 42.582.

Hence ˆy = 42.582 − 0.6861x.

(b) At x = 24.5, ˆy = 42.582 − (0.6861)(24.5) = 25.772.

11.5 (a)

P

i

xi = 675,

P

i

yi = 488,

P

i

x2

i = 37, 125,

P

i

xiyi = 25, 005, n = 18. Therefore,

b =

(18)(25, 005) − (675)(488)

(18)(37, 125) − (675)2 = 0.5676,

a =

488 − (0.5676)(675)

18

= 5.8254.

Hence ˆy = 5.8254 + 0.5676x

Solutions for Exercises in Chapter 11 151

(b) The scatter plot and the regression line are shown below.

0 20 40 60

10 20 30 40 50

Temperature

Grams

y^ = 5.8254 + 0.5676x

(c) For x = 50, ˆy = 5.8254 + (0.5676)(50) = 34.205 grams.

11.6 (a) The scatter plot and the regression line are shown below.

40 50 60 70 80 90

20 40 60 80

Placement Test

Course Grade

y^ = 32.5059 + 0.4711x

(b)

P

i

xi = 1110,

P

i

yi = 1173,

P

i

x2

i = 67, 100,

P

i

xiyi = 67, 690, n = 20. Therefore,

b =

(20)(67, 690) − (1110)(1173)

(20)(67, 100) − (1110)2 = 0.4711,

a =

1173 − (0.4711)(1110)

20

= 32.5059.

Hence ˆy = 32.5059 + 0.4711x

(c) See part (a).

(d) For ˆy = 60, we solve 60 = 32.5059 + 0.4711x to obtain x = 58.466. Therefore,

students scoring below 59 should be denied admission.

11.7 (a) The scatter plot and the regression line are shown here.

152 Chapter 11 Simple Linear Regression and Correlation

20 25 30 35 40 45 50

400 450 500 550

Advertising Costs

Sales

y^ = 343.706 + 3.221x

(b)

P

i

xi = 410,

P

i

yi = 5445,

P

i

x2

i = 15, 650,

P

i

xiyi = 191, 325, n = 12. Therefore,

b =

(12)(191, 325)− (410)(5445)

(12)(15, 650) − (410)2 = 3.2208,

a =

5445 − (3.2208)(410)

12

= 343.7056.

Hence ˆy = 343.7056 + 3.2208x

(c) When x = $35, ˆy = 343.7056 + (3.2208)(35) = $456.43.

(d) Residuals appear to be random as desired.

20 25 30 35 40 45 50

−100 −50 0 50

Advertising Costs

Residual

11.8 (a) ˆy = −1.70 + 1.81x.

(b) ˆx = (54 + 1.71)/1.81 = 30.78.

11.9 (a)

P

i

xi = 45,

P

i

yi = 1094,

P

i

x2

i = 244.26,

P

i

xiyi = 5348.2, n = 9.

b =

(9)(5348.2) − (45)(1094)

(9)(244.26) − (45)2 = −6.3240,

a =

1094 − (−6.3240)(45)

9

= 153.1755.

Hence ˆy = 153.1755 − 6.3240x.

Solutions for Exercises in Chapter 11 153

(b) For x = 4.8, ˆy = 153.1755 − (6.3240)(4.8) = 123.

11.10 (a) ˆz = cdw, ln ˆz = ln c + (ln d)w; setting ˆy = ln z, a = ln c, b = ln d, and ˆy = a + bx,

we have

x = w 1 2 2 3 5 5

y = ln z 8.7562 8.6473 8.6570 8.5932 8.5142 8.4960

P

i

xi = 18,

P

i

yi = 51.6639,

P

i

x2

i = 68,

P

i

xiyi = 154.1954, n = 6.

b = ln d =

(6)(154.1954) − (18)(51.6639)

(6)(68) − (18)2 = −0.0569,

a = ln c =

51.6639 − (−0.0569)(18)

6

= 8.7813.

Now c = e8.7813 = 6511.3364, d = e−0.0569 = 0.9447, and ˆz = 6511.3364×0.9447w.

(b) For w = 4, ˆz = 6511.3364 × 0.94474 = $5186.16.

11.11 (a) The scatter plot and the regression line are shown here.

1300 1400 1500 1600 1700 1800

3000 3500 4000 4500 5000

Temperature

Thrust

y^ = - 1847.633 + 3.653x

(b)

P

i

xi = 14, 292,

P

i

yi = 35, 578,

P

i

x2

i = 22, 954, 054,

P

i

xiyi = 57, 441, 610, n = 9.

Therefore,

b =

(9)(57, 441, 610) − (14, 292)(35, 578)

(9)(22, 954, 054) − (14, 292)2 = 3.6529,

a =

35, 578 − (3.6529)(14, 292)

9

= −1847.69.

Hence ˆy = −1847.69 + 3.6529x.

11.12 (a) The scatter plot and the regression line are shown here.

154 Chapter 11 Simple Linear Regression and Correlation

30 40 50 60 70

250 260 270 280 290 300 310 320

Temperature

Power Consumed

y^ = 218.255 + 1.384x

(b)

P

i

xi = 401,

P

i

yi = 2301,

P

i

x2

i = 22, 495,

P

i

xiyi = 118, 652, n = 8. Therefore,

b =

(8)(118, 652) − (401)(2301))

(8)(22, 495) − (401)2 = 1.3839,

a =

2301 − (1.3839)(401)

8

= 218.26.

Hence ˆy = 218.26 + 1.3839x.

(c) For x = 65◦F, ˆy = 218.26 + (1.3839)(65) = 308.21.

11.13 (a) The scatter plot and the regression line are shown here. A simple linear model

seems suitable for the data.

30 40 50 60 70

250 260 270 280 290 300 310 320

Temperature

Power Consumed

y^ = 218.255 + 1.384x

(b)

P

i

xi = 999,

P

i

yi = 670,

P

i

x2

i = 119, 969,

P

i

xiyi = 74, 058, n = 10. Therefore,

b =

(10)(74, 058) − (999)(670)

(10)(119, 969) − (999)2 = 0.3533,

a =

670 − (0.3533)(999)

10

= 31.71.

Hence ˆy = 31.71 + 0.3533x.

(c) See (a).

Solutions for Exercises in Chapter 11 155

11.14 From the data summary, we obtain

b =

(12)(318) − [(4)(12)][(12)(12)]

(12)(232) − [(4)(12)]2 = −6.45,

a = 12 − (−6.45)(4) = 37.8.

Hence, ˆy = 37.8 − 6.45x. It appears that attending professional meetings would not

result in publishing more papers.

11.15 The least squares estimator A of α is a linear combination of normally distributed

random variables and is thus normal as well.

E(A) = E( ¯ Y − B¯x) = E( ¯ Y ) − ¯xE(B) = α + β¯x − β¯x = α,

σ2

A = σ2

¯ Y −B¯x = σ2

¯ Y + ¯x2σ2

B − 2¯xσ¯ Y B =

σ2

n

+

¯x2σ2

Pn

i=1

(xi − ¯x)2

, since σ¯ Y B = 0.

Hence

σ2

A =

Pn

i=1

x2

i

n

Pn

i=1

(xi − ¯x)2

σ2.

11.16 We have the following:

Cov( ¯ Y ,B) = E

1

n

Xn

i=1

Yi −

1

n

Xn

i=1

μYi

!

Pn

i=1

(xi − ¯x)Yi

Pn

i=1

(xi − ¯x)2 −

Pn

i=1

(xi − ¯x)μYi

Pn

i=1

(xi − ¯x)2

=

Pn

i=1

(xi − ¯x)E(Yi − μYi)2 +

P

i6=j

(xi − ¯x)E(Yi − μYi)(Yj − μYj )

n

Pn

i=1

(xi − ¯x)2

=

Pn

i=1

(xi − ¯x)σ2

Yi +

P

i6=j

Cov(Yi, Yj)

n

Pn

i=1

(xi − ¯x)2

.

Now, σ2

Yi = σ2 for all i, and Cov(Yi, Yj) = 0 for i 6= j. Therefore,

Cov( ¯ Y ,B) =

σ2

Pn

i=1

(xi − ¯x)

n

Pn

i=1

(xi − ¯x)2

= 0.

156 Chapter 11 Simple Linear Regression and Correlation

11.17 Sxx = 26, 591.63 − 778.72/25 = 2336.6824, Syy = 172, 891.46 − 20502/25 = 4791.46,

Sxy = 65, 164.04 − (778.7)(2050)/25 = 1310.64, and b = 0.5609.

(a) s2 = 4791.46−(0.5609)(1310.64)

23 = 176.362.

(b) The hypotheses are

H0 : β = 0,

H1 : β 6= 0.

α = 0.05.

Critical region: t < −2.069 or t > 2.069.

Computation: t = 0.5609 √176.362/2336.6824

= 2.04.

Decision: Do not reject H0.

11.18 Sxx = 57, 557 − 7072/9 = 2018.2222, Syy = 51, 980 − 6582/9 = 3872.8889, Sxy =

53, 258 − (707)(658)/9 = 1568.4444, a = 12.0623 and b = 0.7771.

(a) s2 = 3872.8889−(0.7771)(1568.4444)

7 = 379.150.

(b) Since s = 19.472 and t0.025 = 2.365 for 7 degrees of freedom, then a 95% confidence

interval is

12.0623 ± (2.365)

s

(379.150)(57, 557)

(9)(2018.222)

= 12.0623 ± 81.975,

which implies −69.91 < α < 94.04.

(c) 0.7771 ± (2.365)

q

379.150

2018.2222 implies −0.25 < β < 1.80.

11.19 Sxx = 25.85 − 16.52/11 = 1.1, Syy = 923.58 − 100.42/11 = 7.2018, Sxy = 152.59 −

(165)(100.4)/11 = 1.99, a = 6.4136 and b = 1.8091.

(a) s2 = 7.2018−(1.8091)(1.99)

9 = 0.40.

(b) Since s = 0.632 and t0.025 = 2.262 for 9 degrees of freedom, then a 95% confidence

interval is

6.4136 ± (2.262)(0.632)

s

25.85

(11)(1.1)

= 6.4136 ± 2.0895,

which implies 4.324 < α < 8.503.

(c) 1.8091 ± (2.262)(0.632)/√1.1 implies 0.446 < β < 3.172.

11.20 Sxx = 8134.26 − 311.62/12 = 43.0467, Syy = 7407.80 − 297.22/12 = 47.1467, Sxy =

7687.76 − (311.6)(297.2)/12 = −29.5333, a = 42.5818 and b = −0.6861.

(a) s2 = 47.1467−(−0.6861)(−29.5333)

10 = 2.688.

Solutions for Exercises in Chapter 11 157

(b) Since s = 1.640 and t0.005 = 3.169 for 10 degrees of freedom, then a 99% confidence

interval is

42.5818 ± (3.169)(1.640)

s

8134.26

(12)(43.0467)

= 42.5818 ± 20.6236,

which implies 21.958 < α < 63.205.

(c) −0.6861 ± (3.169)(1.640)/√43.0467 implies −1.478 < β < 0.106.

11.21 Sxx = 37, 125 − 6752/18 = 11, 812.5, Syy = 17, 142 − 4882/18 = 3911.7778, Sxy =

25, 005 − (675)(488)/18 = 6705, a = 5.8254 and b = 0.5676.

(a) s2 = 3911.7778−(0.5676)(6705)

16 = 6.626.

(b) Since s = 2.574 and t0.005 = 2.921 for 16 degrees of freedom, then a 99% confidence

interval is

5.8261 ± (2.921)(2.574)

s

37, 125

(18)(11, 812.5)

= 5.8261 ± 3.1417,

which implies 2.686 < α < 8.968.

(c) 0.5676 ± (2.921)(2.574)/√11, 812.5 implies 0.498 < β < 0.637.

11.22 The hypotheses are

H0 : α = 10,

H1 : α > 10.

α = 0.05.

Critical region: t > 1.734.

Computations: Sxx = 67, 100− 11102/20 = 5495, Syy = 74, 725 − 11732/20 = 5928.55,

Sxy = 67, 690 − (1110)(1173)/20 = 2588.5, s2 = 5928.55−(0.4711)(2588.5)

18 = 261.617 and

then s = 16.175. Now

t =

32.51 − 10

16.175

p

67, 100/(20)(5495)

= 1.78.

Decision: Reject H0 and claim α > 10.

11.23 The hypotheses are

H0 : β = 6,

H1 : β < 6.

α = 0.025.

Critical region: t = −2.228.

158 Chapter 11 Simple Linear Regression and Correlation

Computations: Sxx = 15, 650 − 4102/12 = 1641.667, Syy = 2, 512.925 − 54452/12 =

42, 256.25, Sxy = 191, 325 − (410)(5445)/12 = 5, 287.5, s2 = 42,256.25−(3,221)(5,287.5)

10 =

2, 522.521 and then s = 50.225. Now

t =

3.221 − 6

50.225/√1641.667

= −2.24.

Decision: Reject H0 and claim β < 6.

11.24 Using the value s = 19.472 from Exercise 11.18(a) and the fact that ¯y0 = 74.230 when

x0 = 80, and ¯x = 78.556, we have

74.230 ± (2.365)(19.472)

r

1

9

+

1.4442

2018.222

= 74.230 ± 15.4216.

Simplifying it we get 58.808 < μY | 80 < 89.652.

11.25 Using the value s = 1.64 from Exercise 11.20(a) and the fact that y0 = 25.7724 when

x0 = 24.5, and ¯x = 25.9667, we have

(a) 25.7724 ± (2.228)(1.640)

q

1

12 + (−1.4667)2

43.0467 = 25.7724 ± 1.3341 implies 24.438 <

μY | 24.5 < 27.106.

(b) 25.7724±(2.228)(1.640)

q

1 + 1

12 + (−1.4667)2

43.0467 = 25.7724±3.8898 implies 21.883 <

y0 < 29.662.

11.26 95% confidence bands are obtained by plotting the limits

(6.4136 + 1.809x) ± (2.262)(0.632)

r

1

11

+

(x − 1.5)2

1.1

.

1.0 1.2 1.4 1.6 1.8 2.0

8.0 8.5 9.0 9.5 10.0 10.5

Temperature

Converted Sugar

11.27 Using the value s = 0.632 from Exercise 11.19(a) and the fact that y0 = 9.308 when

x0 = 1.6, and ¯x = 1.5, we have

9.308 ± (2.262)(0.632)

r

1 +

1

11

+

0.12

1.1

= 9.308 ± 1.4994

implies 7.809 < y0 < 10.807.

Solutions for Exercises in Chapter 11 159

11.28 sing the value s = 2.574 from Exercise 11.21(a) and the fact that y0 = 34.205 when

x0 = 50, and ¯x = 37.5, we have

(a) 34.205±(2.921)(2.574)

q

1

18 + 12.52

11,812.5 = 34.205±1.9719 implies 32.23 < μY | 50 <

36.18.

(b) 34.205 ± (2.921)(2.574)

q

1 + 1

18 + 12.52

11,812.5 = 34.205 ± 7.7729 implies 26.43 < y0 <

41.98.

11.29 (a) 17.1812.

(b) The goal of 30 mpg is unlikely based on the confidence interval for mean mpg,

(27.95, 29.60).

(c) Based on the prediction interval, the Lexus ES300 should exceed 18 mpg.

11.30 It is easy to see that

Xn

i=1

(yi − ˆyi) =

Xn

i=1

(yi − a − bxi) =

Xn

i=1

[yi − (¯y − b¯x) − bxi)

=

Xn

i=1

(yi − ¯yi) − b

Xn

i=1

(xi − ¯x) = 0,

since a = ¯y − b¯x.

11.31 When there are only two data points x1 6= x2, using Exercise 11.30 we know that

(y1 − ˆy1) + (y2 − ˆy2) = 0. On the other hand, by the method of least squares on page

395, we also know that x1(y1 − ˆy1) + x2(y2 − ˆy2) = 0. Both of these equations yield

(x2 − x1)(y2 − ˆy2) = 0 and hence y2 − ˆy2 = 0. Therefore, y1 − ˆy1 = 0. So,

SSE = (y1 − ˆy1)2 + (y2 − ˆy2)2 = 0.

Since R2 = 1 − SSE

SST , we have R2 = 1.

11.32 (a) Suppose that the fitted model is ˆy = bx. Then

SSE =

Xn

i=1

(yi − ˆyi)2 =

Xn

i=1

(yi − bxi)2.

Taking derivative of the above with respect to b and setting the derivative to zero,

we have −2

Pn

i=1

xi(yi − bxi) = 0, which implies b =

n P

i

=

1

xiyi

n P

i

=

1

x2

i

.

(b) σ2

B =

V ar„ n

Pi=1

xiYi«

(

n P

i

=

1

x2

i )2

=

n P

i

=

1

x2

i 2

Y i

(

n P

i

=

1

x2

i )2

= 2

n P

i

=

1

x2

i

, since Yi’s are independent.

160 Chapter 11 Simple Linear Regression and Correlation

(c) E(B) =

E„ n

Pi=1

xiYi«

n P

i

=

1

x2

i

=

n P

i

=

1

xi( xi)

n P

i

=

1

x2

i

= β.

11.33 (a) The scatter plot of the data is shown next.

5 10 15 20 25 30

20 40 60 80 100

x

y

y^ = 3.416x

(b)

Pn

i=1

x2

i = 1629 and

Pn

i=1

xiyi = 5564. Hence b = 5564

1629 = 3.4156. So, ˆy = 3.4156x.

(c) See (a).

(d) Since there is only one regression coefficient, β, to be estimated, the degrees of

freedom in estimating σ2 is n − 1. So,

ˆσ2 = s2 =

SSE

n − 1

=

Pn

i=1

(yi − bxi)2

n − 1

.

(e) V ar(ˆyi) = V ar(Bxi) = x2

i V ar(B) = x2

i 2

n P

i

=

1

x2

i

.

(f) The plot is shown next.

5 10 15 20 25 30

20 40 60 80 100

x

y

11.34 Using part (e) of Exercise 11.33, we can see that the variance of a prediction y0 at x0

Solutions for Exercises in Chapter 11 161

is σ2

y0 = σ2

1 + x20

n P

i

=

1

x2

i

. Hence the 95% prediction limits are given as

(3.4145)(25) ± (2.776)√11.16132

r

1 +

252

1629

= 85.3625 ± 10.9092,

which implies 74.45 < y0 < 96.27.

11.35 (a) As shown in Exercise 11.32, the least squares estimator of β is b =

n P

i

=

1

xiyi

n P

i

=

1

x2

i

.

(b) Since

Pn

i=1

xiyi = 197.59, and

Pn

i=1

x2

i = 98.64, then b = 197.59

98.64 = 2.003 and ˆy = 2.003x.

11.36 It can be calculated that b = 1.929 and a = 0.349 and hence ˆy = 0.349 + 1.929x when

intercept is in the model. To test the hypotheses

H0 : α = 0,

H1 : α 6= 0,

with 0.10 level of significance, we have the critical regions as t < −2.132 or t > 2.132.

Computations: s2 = 0.0957 and t = 0.349 √(0.0957)(98.64)/(6)(25.14)

= 1.40.

Decision: Fail to reject H0; the intercept appears to be zero.

11.37 Now since the true model has been changed,

E(B) =

Pn

i=1

(x1i − ¯x1)E(Yi)

Pn

i=1

(x1i − ¯xi)2

=

Pn

i=1

(x1i − ¯x1)(α + βx1i + γx2i)

Pn

i=1

(x1i − ¯x1)2

=

β

Pn

i=1

(x1i − ¯x1)2 + γ

Pn

i=1

(x1i − ¯x1)x2i

Pn

i=1

(x1i − ¯x1)2

= β + γ

Pn

i=1

(x1i − ¯x1)x2i

Pn

i=1

(x1i − ¯x1)2

.

11.38 The hypotheses are

H0 : β = 0,

H1 : β 6= 0.

Level of significance: 0.05.

Critical regions: f > 5.12.

Computations: SSR = bSxy = 1.8091

1.99 = 3.60 and SSE = Syy − SSR = 7.20 − 3.60 =

3.60.

162 Chapter 11 Simple Linear Regression and Correlation

Source of Sum of Degrees of Mean Computed

Variation Squares Freedom Square f

Regression 3.60 1 3.60 9.00

Error 3.60 9 0.40

Total 7.20 10

Decision: Reject H0.

11.39 (a) Sxx = 1058, Syy = 198.76, Sxy = −363.63, b = Sxy

Sxx

= −0.34370, and a =

210−(−0.34370)(172.5)

25 = 10.81153.

(b) The hypotheses are

H0 : The regression is linear in x,

H1 : The regression is nonlinear in x.

α = 0.05.

Critical regions: f > 3.10 with 3 and 20 degrees of freedom.

Computations: SST = Syy = 198.76, SSR = bSxy = 124.98 and

SSE = Syy − SSR = 73.98. Since

T1. = 51.1, T2. = 51.5, T3. = 49.3, T4. = 37.0 and T5. = 22.1,

then

SSE(pure) =

X5

i=1

X5

j=1

y2

ij −

X5

i=1

T2

i.

5

= 1979.60 − 1910.272 = 69.33.

Hence the “Lack-of-fit SS” is 73.78 − 69.33 = 4.45.

Source of Sum of Degrees of Mean Computed

Variation Squares Freedom Square f

Regression

E rror

Lack of fit

Pure error

124.98

73.98

4.45

69.33

1

23

3

20

124.98

3.22

1.48

3.47

0.43

Total 198.76 24

Decision: Do not reject H0.

11.40 The hypotheses are

H0 : The regression is linear in x,

H1 : The regression is nonlinear in x.

α = 0.05.

Critical regions: f > 3.26 with 4 and 12 degrees of freedom.

Solutions for Exercises in Chapter 11 163

Computations: SST = Syy = 3911.78, SSR = bSxy = 3805.89 and SSE = Syy −

SSR = 105.89. SSE(pure) =

P6

i=1

P3

j=1

y2

ij −

P6

i=1

T2

i.

3 = 69.33, and the “Lack-of-fit SS” is

105.89 − 69.33 = 36.56.

Source of Sum of Degrees of Mean Computed

Variation Squares Freedom Square f

Regression

E rror

Lack of fit

Pure error

3805.89

105.89

36.56

69.33

1

16

4

12

3805.89

6.62

9.14

5.78

1.58

Total 3911.78 17

Decision: Do not reject H0; the lack-of-fit test is insignificant.

11.41 The hypotheses are

H0 : The regression is linear in x,

H1 : The regression is nonlinear in x.

α = 0.05.

Critical regions: f > 3.00 with 6 and 12 degrees of freedom.

Computations: SST = Syy = 5928.55, SSR = bSxy = 1219.35 and SSE = Syy −

SSR = 4709.20. SSE(pure) =

P8

i=1

Pni

j=1

y2

ij −

P8

i=1

T2

i.

ni

= 3020.67, and the “Lack-of-fit SS”

is 4709.20 − 3020.67 = 1688.53.

Source of Sum of Degrees of Mean Computed

Variation Squares Freedom Square f

Regression

E rror

Lack of fit

Pure error

1219.35

4709.20

1688.53

3020.67

1

18

6

12

1219.35

261.62

281.42

251.72

1.12

Total 5928.55 19

Decision: Do not reject H0; the lack-of-fit test is insignificant.

11.42 (a) t = 2.679 and 0.01 < P(T > 2.679) < 0.015, hence 0.02 < P-value < 0.03. There

is a strong evidence that the slope is not 0. Hence emitter drive-in time influences

gain in a positive linear fashion.

(b) f = 56.41 which results in a strong evidence that the lack-of-fit test is significant.

Hence the linear model is not adequate.

(c) Emitter does does not influence gain in a linear fashion. A better model is a

quadratic one using emitter drive-in time to explain the variability in gain.

164 Chapter 11 Simple Linear Regression and Correlation

11.43 ˆy = −21.0280 + 0.4072x; fLOF = 1.71 with a P-value = 0.2517. Hence, lack-of-fit test

is insignificant and the linear model is adequate.

11.44 (a) ˆy = 0.011571 + 0.006462x with t = 7.532 and P(T > 7.532) < 0.0005 Hence

P-value < 0.001; the slope is significantly different from 0 in the linear regression

model.

(b) fLOF = 14.02 with P-value < 0.0001. The lack-of-fit test is significant and the

linear model does not appear to be the best model.

11.45 (a) ˆy = −11.3251 − 0.0449 temperature.

(b) Yes.

(c) 0.9355.

(d) The proportion of impurities does depend on temperature.

−270 −265 −260

0.2 0.4 0.6 0.8 1.0

Temperature

Proportion of Impurity

However, based on the plot, it does not appear that the dependence is in linear

fashion. If there were replicates, a lack-of-fit test could be performed.

11.46 (a) ˆy = 125.9729 + 1.7337 population; P-value for the regression is 0.0023.

(b) f6,2 = 0.49 and P-value = 0.7912; the linear model appears to be adequate based

on the lack-of-fit test.

(c) f1,2 = 11.96 and P-value = 0.0744. The results do not change. The pure error

test is not as sensitive because the loss of error degrees of freedom.

11.47 (a) The figure is shown next.

(b) ˆy = −175.9025 + 0.0902 year; R2 = 0.3322.

(c) There is definitely a relationship between year and nitrogen oxide. It does not

appear to be linear.

Solutions for Exercises in Chapter 11 165

700 750 800 850 900 950 1000

−5 0 5 10

Time

Residual

11.48 The ANOVA model is:

Source of Sum of Degrees of Mean Computed

Variation Squares Freedom Square f

Regression

E rror

Lack of fit

Pure error

135.2000

10.4700

6.5150

3.9550

1

14

2

12

135.2000

0.7479

3.2575

0.3296

9.88

Total 145.6700 15

The P-value = 0.0029 with f = 9.88.

Decision: Reject H0; the lack-of-fit test is significant.

11.49 Sxx = 36, 354 − 35, 882.667 = 471.333, Syy = 38, 254 − 37, 762.667 = 491.333, and

Sxy = 36, 926 − 36, 810.667 = 115.333. So, r = 115 √(471.333)(491.333)

= 0.240.

11.50 The hypotheses are

H0 : ρ = 0,

H1 : ρ 6= 0.

α = 0.05.

Critical regions: t < −2.776 or t > 2.776.

Computations: t = 0.240√4 √1−0.2402 = 0.51.

Decision: Do not reject H0.

11.51 Since b = Sxy

Sxx

, we can write s2 = Syy−bSxy

n−2 = Syy−b2Sxx

n−2 . Also, b = r

q

Syy

Sxx

so that

s2 = Syy−r2Syy

n−2 = (1−r2)Syy

n−2 , and hence

t =

b

s√Sxx

=

r

p

p Syy/Sxx

SyySxx(1 − r2)/(n − 2)

=

r√n − 2

√1 − r2

.

166 Chapter 11 Simple Linear Regression and Correlation

11.52 (a) Sxx = 128.6602 − 32.682/9 = 9.9955, Syy = 7980.83 − 266.72/9 = 77.62, and

Sxy = 990.268 − (32.68)(266.7)/9 = 21.8507. So, r = 21.8507 √(9.9955)(77.62)

= 0.784.

(b) The hypotheses are

H0 : ρ = 0,

H1 : ρ > 0.

α = 0.01.

Critical regions: t > 2.998.

Computations: t = 0.784√7 √1−0.7842 = 3.34.

Decision: Reject H0; ρ > 0.

(c) (0.784)2(100%) = 61.5%.

11.53 (a) From the data of Exercise 11.1 we can calculate

Sxx = 26, 591.63 − (778.7)2/25 = 2336.6824,

Syy = 172, 891.46 − (2050)2/25 = 4791.46,

Sxy = 65, 164.04 − (778.7)(2050)/25 = 1310.64.

Therefore, r = 1310.64 √(2236.6824)(4791.46)

= 0.392.

(b) The hypotheses are

H0 : ρ = 0,

H1 : ρ 6= 0.

α = 0.05.

Critical regions: t < −2.069 or t > 2.069.

Computations: t = 0.392√23 √1−0.3922 = 2.04.

Decision: Fail to reject H0 at level 0.05. However, the P-value = 0.0530 which is

marginal.

11.54 (a) From the data of Exercise 11.9 we find Sxx = 244.26 − 452/9 = 19.26, Syy =

133, 786 − 10942/9 = 804.2222, and Sxy = 5348.2 − (45)(1094)/9 = −121.8. So,

r = −121.8 √(19.26)(804.2222)

= −0.979.

(b) The hypotheses are

H0 : ρ = −0.5,

H1 : ρ < −0.5.

α = 0.025.

Critical regions: z < −1.96.

Computations: z =

√6

2 ln

h

(0.021)(1.5)

(1.979)(0.5)

i

= −4.22.

Decision: Reject H0; ρ < −0.5.

Solutions for Exercises in Chapter 11 167

(c) (−0.979)2(100%) = 95.8%.

11.55 Using the value s = 16.175 from Exercise 11.6 and the fact that ˆy0 = 48.994 when

x0 = 35, and ¯x = 55.5, we have

(a) 48.994±(2.101)(16.175)

p

1/20 + (−20.5)2/5495 which implies to 36.908 < μY | 35 <

61.080.

(b) 48.994 ± (2.101)(16.175)

p

1 + 1/20 + (−20.5)2/5495 which implies to 12.925 <

y0 < 85.063.

11.56 The fitted model can be derived as ˆy = 3667.3968 − 47.3289x.

The hypotheses are

H0 : β = 0,

H1 : β 6= 0.

t = −0.30 with P-value = 0.77. Hence H0 cannot be rejected.

11.57 (a) Sxx = 729.18 − 118.62/20 = 25.882, Sxy = 1714.62 − (118.6)(281.1)/20 = 47.697,

so b = Sxy

Sxx

= 1.8429, and a = ¯y − b¯x = 3.1266. Hence ˆy = 3.1266 + 1.8429x.

(b) The hypotheses are

H0 : the regression is linear in x,

H1 : the regression is not linear in x.

α = 0.05.

Critical region: f > 3.07 with 8 and 10 degrees of freedom.

Computations: SST = 13.3695, SSR = 87.9008, SSE = 50.4687, SSE(pure) =

16.375, and Lack-of-fit SS = 34.0937.

Source of Sum of Degrees of Mean Computed

Variation Squares Freedom Square f

Regression

E rror

Lack of fit

Pure error

87.9008

50.4687

34.0937

16.375

1

18

8

10

87.9008

2.8038

4.2617

1.6375

2.60

Total 138.3695 19

The P-value = 0.0791. The linear model is adequate at the level 0.05.

11.58 Using the value s = 50.225 and the fact that ˆy0 = $448.644 when x0 = $45, and

¯x = $34.167, we have

(a) 488.644 ± (1.812)(50.225)

q

1/12 + 10.8332

1641.667 , which implies 452.835 < μY | 45 <

524.453.

168 Chapter 11 Simple Linear Regression and Correlation

(b) 488.644 ± (1.812)(50.225)

q

1 + 1/12 + 10.8332

1641.667 , which implies 390.845 < y0 <

586.443.

11.59 (a) ˆy = 7.3598 + 135.4034x.

(b) SS(Pure Error) = 52, 941.06; fLOF = 0.46 with P-value = 0.64. The lack-of-fit

test is insignificant.

(c) No.

11.60 (a) Sxx = 672.9167, Syy = 728.25, Sxy = 603.75 and r = 603.75 √(672.9167)(728.25)

= 0.862,

which means that (0.862)2(100%) = 74.3% of the total variation of the values of

Y in our sample is accounted for by a linear relationship with the values of X.

(b) To estimate and test hypotheses on ρ, X and Y are assumed to be random

variables from a bivariate normal distribution.

(c) The hypotheses are

H0 : ρ = 0.5,

H1 : ρ > 0.5.

α = 0.01.

Critical regions: z > 2.33.

Computations: z =

√9

2 ln

h

(1.862)(0.5)

(0.138)(1.5)

i

= 2.26.

Decision: Reject H0; ρ > 0.5.

11.61 s2 =

n P

i

=

1

(yi−ˆyi)2

n−2 . Using the centered model, ˆyi = ¯y + b(xi − ¯x) + ǫi.

(n − 2)E(S2) = E

Xn

i=1

[α + β(xi − ¯x) + ǫi − (¯y + b(xi − ¯x))]2

=

Xn

i=1

E

(α − ¯y)2 + (β − b)2(xi − ¯x)2 + ǫ2

i − 2b(xi − ¯x)ǫi − 2¯yǫi

,

(other cross product terms go to 0)

=

nσ2

n

+

σ2Sxx

Sxx

+ nσ2 − 2

σ2Sxx

Sxx − 2

nσ2

n

= (n − 2)σ2.

11.62 (a) The confidence interval is an interval on the mean sale price for a given buyer’s

bid. The prediction interval is an interval on a future observed sale price for a

given buyer’s bid.

(b) The standard errors of the prediction of sale price depend on the value of the

buyer’s bid.

Solutions for Exercises in Chapter 11 169

(c) Observations 4, 9, 10, and 17 have the lowest standard errors of prediction. These

observations have buyer’s bids very close to the mean.

11.63 (a) The residual plot appears to have a pattern and not random scatter. The R2 is

only 0.82.

(b) The log model has an R2 of 0.84. There is still a pattern in the residuals.

(c) The model using gallons per 100 miles has the best R2 with a 0.85. The residuals

appear to be more random. This model is the best of the three models attempted.

Perhaps a better model could be found.

11.64 (a) The plot of the data and an added least squares fitted line are given here.

150 200 250 300

75 80 85 90 95

Temperature

Yield

(b) Yes.

(c) ˆy = 61.5133 + 0.1139x.

Source of Sum of Degrees of Mean Computed

Variation Squares Freedom Square f

Regression

E rror

Lack of fit

Pure error

486.21

24.80

3.61

21.19

1

10

2

8

486.21

2.48

1.81

2.65

0.68

Total 511.01 11

The P-value = 0.533.

(d) The results in (c) show that the linear model is adequate.

11.65 (a) ˆy = 90.8904 − 0.0513x.

(b) The t-value in testing H0 : β = 0 is −6.533 which results in a P-value < 0.0001.

Hence, the time it takes to run two miles has a significant influence on maximum

oxygen uptake.

(c) The residual graph shows that there may be some systematic behavior of the

residuals and hence the residuals are not completely random

170 Chapter 11 Simple Linear Regression and Correlation

700 750 800 850 900 950 1000

−5 0 5 10

Time

Residual

11.66 Let Y ∗ i = Yi − α, for i = 1, 2, . . . , n. The model Yi = α + βxi + ǫi is equivalent to

Y ∗ i = βxi + ǫi. This is a “regression through the origin” model that is studied in

Exercise 11.32.

(a) Using the result from Exercise 11.32(a), we have

b =

Pn

i=1

xi(yi − α)

Pn

i=1

x2

i

=

Pn

i=1

xiyi − n¯xα

Pn

i=1

x2

i

.

(b) Also from Exercise 11.32(b) we have σ2

B = 2

n P

i

=

1

x2

i

.

11.67 SSE =

Pn

i=1

(yi − βxi)2. Taking derivative with respect to β and setting this as 0, we

get

Pn

i=1

xi(yi − bxi) = 0, or

Pn

i=1

xi(yi − ˆyi) = 0. This is the only equation we can get

using the least squares method. Hence in general,

Pn

i=1

(yi − ˆyi) = 0 does not hold for a

regression model with zero intercept.

11.68 No solution is provided.

Chapter 12

Multiple Linear Regression and

Certain Nonlinear Regression Models

12.1 (a) by = 27.5467 + 0.9217x1 + 0.2842x2.

(b) When x1 = 60 and x2 = 4, the predicted value of the chemistry grade is

ˆy = 27.5467 + (0.9217)(60) + (0.2842)(4) = 84.

12.2 ˆy = −3.3727 + 0.0036x1 + 0.9476x2.

12.3 ˆy = 0.7800 + 2.7122x1 + 2.0497x2.

12.4 (a) ˆy = −22.99316 + 1.39567x1 + 0.21761x2.

(b) ˆy = −22.99316 + (1.39567)(35) + (0.21761)(250) = 80.25874.

12.5 (a) ˆy = 56.46333 + 0.15253x − 0.00008x2.

(b) ˆy = 56.46333 + (0.15253)(225)− (0.00008)(225)2 = 86.73333%.

12.6 (a) ˆ d = 13.35875 − 0.33944v − 0.01183v2.

(b) ˆ d = 13.35875 − (−0.33944)(70) − (0.01183)(70)2 = 47.54206.

12.7 ˆy = 141.61178 − 0.28193x + 0.00031x2.

12.8 (a) ˆy = 19.03333 + 1.0086x − 0.02038x2.

(b) SSE = 24.47619 with 12 degrees of freedom and SS(pure error) = 24.36667

with 10 degrees of freedom. So, SSLOF = 24.47619 − 24.36667 = 0.10952 with

2 degrees of freedom. Hence f = 0.10952/2

24.36667/10 = 0.02 with a P-value of 0.9778.

Therefore, there is no lack of fit and the quadratic model fits the data well.

12.9 (a) ˆy = −102.71324 + 0.60537x1 + 8.92364x2 + 1.43746x3 + 0.01361x4.

(b) ˆy = −102.71324+(0.60537)(75)+(8.92364)(24)+(1.43746)(90)+(0.01361)(98) =

287.56183.

171

172 Chapter 12 Multiple Linear Regression and Certain Nonlinear Regression Models

12.10 (a) ˆy = 1.07143 + 4.60317x − 1.84524x2 + 0.19444x3.

(b) ˆy = 1.07143 + (4.60317)(2) − (1.84524)(2)2 + (0.19444)(2)3 = 4.45238.

12.11 ˆy = 3.3205 + 0.42105x1 − 0.29578x2 + 0.01638x3 + 0.12465x4.

12.12 ˆy = 1, 962.94816 − 15.85168x1 + 0.05593x2 + 1.58962x3 − 4.21867x4 − 394.31412x5.

12.13 ˆy = −6.51221 + 1.99941x1 − 3.67510x2 + 2.52449x3 + 5.15808x4 + 14.40116x5.

12.14 ˆy = −884.667 − 3 − 0.83813x1 + 4.90661x2 + 1.33113x3 + 11.93129x4.

12.15 (a) ˆy = 350.99427 − 1.27199x1 − 0.15390x2.

(b) ˆy = 350.99427 − (1.27199)(20) − (0.15390)(1200) = 140.86930.

12.16 (a) ˆy = −21.46964 − 3.32434x1 + 0.24649x2 + 20.34481x3.

(b) ˆy = −21.46964 − (3.32434)(14) + (0.24649)(220) + (20.34481)(5) = 87.94123.

12.17 s2 = 0.16508.

12.18 s2 = 0.43161.

12.19 s2 = 242.71561.

12.20 Using SAS output, we obtain

ˆσ2

b1 = 3.747 × 10−7, ˆσ2

b2 = 0.13024, ˆσb1b2 = −4.165 × 10−7.

12.21 Using SAS output, we obtain

(a) ˆσ2

b2 = 28.09554.

(b) ˆσb1b4 = −0.00958.

12.22 Using SAS output, we obtain

0.4516 < μY |x1=900,x2=1 < 1.2083, and −0.1640 < y0 < 1.8239.

12.23 Using SAS output, we obtain a 90% confidence interval for the mean response when

x = 19.5 as 29.9284 < μY |x=19.5 < 31.9729.

12.24 Using SAS output, we obtain

263.7879 < μY |x1=75,x2=24,x3=90,x4=98 < 311.3357, and 243.7175 < y0 < 331.4062.

12.25 The hypotheses are

H0 : β2 = 0,

H1 : β2 6= 0.

The test statistic value is t = 2.86 with a P-value = 0.0145. Hence, we reject H0 and

conclude β2 6= 0.

Solutions for Exercises in Chapter 12 173

12.26 The test statistic is t = 0.00362

0.000612 = 5.91 with P-value = 0.0002. Reject H0 and claim

that β1 6= 0.

12.27 The hypotheses are

H0 : β1 = 2,

H1 : β1 6= 2.

The test statistics is t = 2.71224−2

0.20209 = 3.524 with P-value = 0.0097. Reject H0 and

conclude that β1 6= 2.

12.28 Using SAS output, we obtain

(a) s2 = 650.1408.

(b) ˆy = 171.6501, 135.8735 < μY |x1=20,x2=1000 < 207.4268, and

82.9677 < y0 < 260.3326.

12.29 (a) P-value = 0.3562. Hence, fail to reject H0.

(b) P-value = 0.1841. Again, fail to reject H0.

(c) There is not sufficient evidence that the regressors x1 and x2 significantly influence

the response with the described linear model.

12.30 (a) s2 = 17.22858.

(b) ˆy = 104.9617 and 95.5660 < y0 < 114.3574.

12.31 R2 = SSR

SST = 10953

10956 = 99.97%. Hence, 99.97% of the variation in the response Y in our

sample can be explained by the linear model.

12.32 The hypotheses are:

H0 : β1 = β2 = 0,

H1 : At least one of the βi’s is not zero, for i = 1, 2.

Using the f-test, we obtain that f = MSR

MSE = 5476.60129

0.43161 = 12688.7 with P-value <

0.0001. Hence, we reject H0. The regression explained by the model is significant.

12.33 f = 5.11 with P-value = 0.0303. At level of 0.01, we fail to reject H0 and we cannot

claim that the regression is significant.

12.34 The hypotheses are:

H0 : β1 = β2 = 0,

H1 : At least one of the βi’s is not zero, for i = 1, 2.

174 Chapter 12 Multiple Linear Regression and Certain Nonlinear Regression Models

The partial f-test statistic is

f =

(160.93598 − 145.88354)/2

1.49331

= 5.04, with 2 and 7 degrees of freedom.

The resulting P-value = 0.0441. Therefore, we reject H0 and claim that at least one

of β1 and β2 is not zero.

12.35 f = (6.90079−1.13811)/1

0.16508 = 34.91 with 1 and 9 degrees of freedom. The P-value = 0.0002

which implies that H0 is rejected.

12.36 (a) ˆy = 0.900 + 0.575x1 + 0.550x2 + 1.150x3.

(b) For the model in (a), SSR = 15.645, SSE = 1.375 and SST = 17.020. The

ANOVA table for all these single-degree-of-freedom components can be displayed

as:

Source of Degrees of Mean Computed

Variation Freedom Square f P-value

x1

x2

x3

Error

1

1

1

4

2.645

2.420

10.580

0.34375

7.69

7.04

30.78

0.0501

0.0568

0.0052

Total 7

β3 is found to be significant at the 0.01 level and β1 and β2 are not significant.

12.37 The hypotheses are:

H0 : β1 = β2 = 0,

H1 : At least one of the βi’s is not zero, for i = 1, 2.

The partial f-test statistic is

f =

(4957.24074 − 17.02338)/2

242.71561

= 10.18, with 2 and 7 degrees of freedom.

The resulting P-value = 0.0085. Therefore, we reject H0 and claim that at least one

of β1 and β2 is not zero.

12.38 Using computer software, we obtain the following.

R(β1 | β0) = 2.645,

R(β1 | β0, β2, β3) = R(β0, β1, β2, β3) − R(β0, β2, β3) = 15.645 − 13.000 = 2.645.

R(β2 | β0, β1) = R(β0, β1, β2) − R(β0, β1) = 5.065 − 2.645 = 2.420,

R(β2 | β0, β1, β3) = R(β0, β1, β2, β3) − R(β0, β1, β3) = 15.645 − 13.225 = 2.420,

R(β3 | β0, β1, β2) = R(β0, β1, β2, β3) − R(β0, β1, β2) = 15.645 − 5.065 = 10.580.

Solutions for Exercises in Chapter 12 175

12.39 The following is the summary.

s2 R2 R2

adj

The model using weight alone 8.12709 0.8155 0.8104

The model using weight and drive ratio 4.78022 0.8945 0.8885

The above favor the model using both explanatory variables. Furthermore, in the

model with two independent variables, the t-test for β2, the coefficient of drive ratio,

shows P-value < 0.0001. Hence, the drive ratio variable is important.

12.40 The following is the summary:

s2 C.V. R2

adj Average Length of the CIs

The model with x3 4.29738 7.13885 0.8823 5.03528

The model without x3 4.00063 6.88796 0.8904 4.11769

These numbers favor the model without using x3. Hence, variable x3 appears to be

unimportant.

12.41 The following is the summary:

s2 C.V. R2

adj

The model with 3 terms 0.41966 4.62867 0.9807

The model without 3 terms 1.60019 9.03847 0.9266

Furthermore, to test β11 = β12 = β22 = 0 using the full model, f = 15.07 with

P-value = 0.0002. Hence, the model with interaction and pure quadratic terms is

better.

12.42 (a) Full model: ˆy = 121.75 + 2.50x1 + 14.75x2 + 21.75x3, with R2

adj = 0.9714.

Reduced model: ˆy = 121.75 + 14.75x2 + 21.75x3, with R2

adj = 0.9648.

There appears to be little advantage using the full model.

(b) The average prediction interval widths are:

full model: 32.70; and reduced model: 32.18. Hence, the model without using x1

is very competitive.

12.43 The following is the summary:

s2 C.V. R2

adj Average Length of the CIs

x1, x2 650.14075 16.55705 0.7696 106.60577

x1 967.90773 20.20209 0.6571 94.31092

x2 679.99655 16.93295 0.7591 78.81977

In addition, in the full model when the individual coefficients are tested, we obtain

P-value = 0.3562 for testing β1 = 0 and P-value = 0.1841 for testing β2 = 0.

In comparing the three models, it appears that the model with x2 only is slightly

better.

176 Chapter 12 Multiple Linear Regression and Certain Nonlinear Regression Models

12.44 Here is the summary for all four models (including the full model)

s2 C.V. R2

adj

x1, x2, x3 17.22858 3.78513 0.9899

x1, x2 297.97747 15.74156 0.8250

x1, x3 17.01876 3.76201 0.9900

x2, x3 17.07575 3.76830 0.9900

It appears that a two-variable model is very competitive with the full model as long as

the model contains x3.

12.45 (a) ˆy = 5.95931−0.00003773 odometer +0.33735 octane −12.62656 van −12.98455 suv.

(b) Since the coefficients of van and suv are both negative, sedan should have the best

gas mileage.

(c) The parameter estimates (standard errors) for van and suv are −12.63 (1.07) and

−12, 98 (1.11), respectively. So, the difference between the estimates are smaller

than one standard error of each. So, no significant difference in a van and an suv

in terms of gas mileage performance.

12.46 The parameter estimates are given here.

Variable DF Estimate Standar Error t P-value

Intercept

Income

Family

Female

1

1

1

1

−206.64625

0.00543

−49.24044

236.72576

163.70943

0.00274

51.95786

110.57158

−1.26

1.98

−0.95

2.14

0.2249

0.0649

0.3574

0.0480

(a) ˆy = −206.64625 + 0.00543Income − 49.24044Family + 236/72576Female. The

company would prefer female customers.

(b) Since the P-value = 0.0649 for the coefficient of the “Income,” it is at least

marginally important. Note that the R2 = 0.3075 which is not very high. Perhaps

other variables need to be considered.

12.47 (a) Han\g Time = 1.10765 + 0.01370 LLS + 0.00429 Power.

(b) Han\g Time = 1.10765 + (0.01370)(180) + (0.00429)(260) = 4.6900.

(c) 4.4502 < μHang Time | LLS=180, Power=260 < 4.9299.

12.48 (a) For forward selection, variable x1 is entered first, and no other variables are entered

at 0.05 level. Hence the final model is ˆy = −6.33592 + 0.33738x1.

(b) For the backward elimination, variable x3 is eliminated first, then variable x4 and

then variable x2, all at 0.05 level of significance. Hence only x1 remains in the

model and the final model is the same one as in (a).

Solutions for Exercises in Chapter 12 177

(c) For the stepwise regression, after x1 is entered, no other variables are entered.

Hence the final model is still the same one as in (a) and (b).

12.49 Using computer output, with α = 0.05, x4 was removed first, and then x1. Neither x2

nor x3 were removed and the final model is ˆy = 2.18332 + 0.95758x2 + 3.32533x3.

12.50 (a) ˆy = −29.59244 + 0.27872x1 + 0.06967x2 + 1.24195x3 − 0.39554x4 + 0.22365x5.

(b) The variables x3 and x5 were entered consecutively and the final model is ˆy =

−56.94371 + 1.63447x3 + 0.24859x5.

(c) We have a summary table displayed next.

Model s2 PRESS R2 P

i |δi|

x2x5

x1x5

x1x3x5

x3x5

x3x4x5

x2x4x5

x2x3x5

x3x4

x1x2x5

x5

x3

x2x3x4x5

x1

x2x3

x1x3

x1x3x4x5

x2

x4x5

x1x2x3x5

x2x3x4

x1x2

x1x4x5

x1x3x4

x2x4

x1x2x4x5

x1x2x3x4x5

x1x4

x1x2x3

x4

x1x2x3x4

x1x2x4

176.735

174.398

174.600

194.369

192.006

196.211

186.096

249.165

184.446

269.355

257.352

197.782

274.853

264.670

226.777

188.333

328.434

289.633

195.344

269.800

297.294

192.822

240.828

352.781

207.477

214.602

287.794

249.038

613.411

266.542

317.783

2949.13

3022.18

3207.34

3563.40

3637.70

3694.97

3702.90

3803.00

3956.41

3998.77

4086.47

4131.88

4558.34

4721.55

4736.02

4873.16

4998.07

5136.91

5394.56

5563.87

5784.20

5824.58

6564.79

6902.14

7675.70

7691.30

7714.86

7752.69

8445.98

10089.94

10591.58

0.7816

0.7845

0.8058

0.7598

0.7865

0.7818

0.7931

0.6921

0.7949

0.6339

0.6502

0.8045

0.6264

0.6730

0.7198

0.8138

0.5536

0.6421

0.8069

0.7000

0.6327

0.7856

0.7322

0.5641

0.7949

0.8144

0.6444

0.7231

0.1663

0.7365

0.6466

151.681

166.223

174.819

189.881

190.564

170.847

184.285

192.172

189.107

189.373

199.520

192.000

202.533

210.853

219.630

207.542

217.814

209.232

216.934

234.565

231.374

216.583

248.123

248.621

249.604

257.732

249.221

264.324

259.968

297.640

294.044

178 Chapter 12 Multiple Linear Regression and Certain Nonlinear Regression Models

(d) It appears that the model with x2 = LLS and x5 = Power is the best in terms of

PRESS, s2, and

P

i |δi|.

12.51 (a) ˆy = −587.21085 + 428.43313x.

(b) ˆy = 1180.00032 − 192.69121x + 35.20945x2.

(c) The summary of the two models are given as:

Model s2 R2 PRESS

μY = β0 + β1x 1,105,054 0.8378 18,811,057.08

μY = β0 + β1x + β11x2 430,712 0.9421 8,706,973.57

It appears that the model with a quadratic term is preferable.

12.52 The parameter estimate for β4 is 0.22365 with a standard error of 0.13052. Hence,

t = 1.71 with P-value = 0.6117. Fail to reject H0.

12.53 ˆσ2

b1 = 20, 588.038, ˆσ2

b11 = 62.650, and ˆσb1b11 = −1, 103.423.

12.54 (a) The following is the summary of the models.

Model s2 R2 PRESS Cp

x2x3

x2

x1x2

x1x2x3

x3

x1

x1x3

8094.15

8240.05

8392.51

8363.55

8584.27

8727.47

8632.45

0.51235

0.48702

0.49438

0.51292

0.46559

0.45667

0.47992

282194.34

282275.98

289650.65

294620.94

297242.74

304663.57

306820.37

2.0337

1.5422

3.1039

4.0000

2.8181

3.3489

3.9645

(b) The model with ln(x2) appears to have the smallest Cp with a small PRESS.

Also, the model ln(x2) and ln(x3) has the smallest PRESS. Both models appear

to better than the full model.

12.55 (a) There are many models here so the model summary is not displayed. By using

MSE criterion, the best model, contains variables x1 and x3 with s2 = 313.491.

If PRESS criterion is used, the best model contains only the constant term with

s2 = 317.51. When the Cp method is used, the best model is model with the

constant term.

(b) The normal probability plot, for the model using intercept only, is shown next.

We do not appear to have the normality.

−2 −1 0 1 2

−2.0 −1.5 −1.0 −0.5 0.0 0.5 1.0

Normal Q−Q Plot

Theoretical Quantiles

Sample Quantiles

Solutions for Exercises in Chapter 12 179

12.56 (a) [Volt = −1.64129 + 0.000556 Speed − 67.39589 Extension.

(b) P-values for the t-tests of the coefficients are all < 0.0001.

(c) The R2 = 0.9607 and the model appears to have a good fit. The residual plot

and a normal probability plot are given here.

5500 6000 6500 7000 7500 8000 8500

−400 −200 0 200 400

y^

Residual

−2 −1 0 1 2

−400 −200 0 200 400

Normal Q−Q Plot

Theoretical Quantiles

Sample Quantiles

12.57 (a) ˆy = 3.13682 + 0.64443x1 − 0.01042x2 + 0.50465x3 − 0.11967x4 − 2.46177x5 +

1.50441x6.

(b) The final model using the stepwise regression is

ˆy = 4.65631 + 0.51133x3 − 0.12418x4.

(c) Using Cp criterion (smaller the better), the best model is still the model stated in

(b) with s2 = 0.73173 and R2 = 0.64758. Using the s2 criterion, the model with

x1, x3 and x4 has the smallest value of 0.72507 and R2 = 0.67262. These two

models are quite competitive. However, the model with two variables has one less

variable, and thus may be more appealing.

(d) Using the model in part (b), displayed next is the Studentized residual plot. Note

that observations 2 and 14 are beyond the 2 standard deviation lines. Both of

those observations may need to be checked.

8 9 10 11 12

−2 −1 0 1 2

y^

Studentized Residual

1

2

3

4

5

6

7

8

9

10 11

12 13

14

15

16

17

18

19

12.58 The partial F-test shows a value of 0.82, with 2 and 12 degrees of freedom. Consequently,

the P-value = 0.4622, which indicates that variables x1 and x6 can be excluded

from the model.

180 Chapter 12 Multiple Linear Regression and Certain Nonlinear Regression Models

12.59 (a) ˆy = 125.86555 + 7.75864x1 + 0.09430x2 − 0.00919x1x2.

(b) The following is the summary of the models.

Model s2 R2 PRESS Cp

x2

x1

x1x2

x1x2x3

680.00

967.91

650.14

561.28

0.80726

0.72565

0.86179

0.92045

7624.66

12310.33

12696.66

15556.11

2.8460

4.8978

3.4749

4.0000

It appears that the model with x2 alone is the best.

12.60 (a) The fitted model is ˆy = 85.75037−15.93334x1+2.42280x2+1.82754x3+3.07379x4.

(b) The summary of the models are given next.

Model s2 PRESS R2 Cp

x1x2x4

x4

x3x4

x1x2x3x4

x1x4

x2x4

x2x3x4

x1x3x4

x2

x3

x2x3

x1

x1x3

x1x2

x1x2x3

9148.76

19170.97

21745.08

10341.20

10578.94

21630.42

25160.18

12341.87

160756.81

171264.68

183701.86

95574.16

107287.63

109137.20

125126.59

447, 884.34

453, 304.54

474, 992.22

482, 210.53

488, 928.91

512, 749.78

532, 065.42

614, 553.42

1, 658, 507.38

1, 888, 447.43

1, 896, 221.30

2, 213, 985.42

2, 261, 725.49

2, 456, 103.03

2, 744, 659.14

0.9603

0.8890

0.8899

0.9626

0.9464

0.8905

0.8908

0.9464

0.0695

0.0087

0.0696

0.4468

0.4566

0.4473

0.4568

3.308

8.831

10.719

5.000

3.161

10.642

12.598

5.161

118.362

126.491

120.349

67.937

68.623

69.875

70.599

When using PRESS as well as the s2 criterion, a model with x1, x2 and x4 appears

to be the best, while when using the Cp criterion, the model with x1 and x4 is the

best. When using the model with x1, x2 and x4, we find out that the P-value for

testing β2 = 0 is 0.1980 which implies that perhaps x2 can be excluded from the

model.

(c) The model in part (b) has smaller Cp as well as competitive PRESS in comparison

to the full model.

12.61 Since H = X(X‘X)−1X‘, and

Pn

i=1

hii = tr(H), we have

Xn

i=1

hii = tr(X(X‘X)−1X‘) = tr(X‘X(X‘X)−1) = tr(Ip) = p,

where Ip is the p × p identity matrix. Here we use the property of tr(AB) = tr(BA)

in linear algebra.

Solutions for Exercises in Chapter 12 181

12.62 (a) ˆy = 9.9375 + 0.6125x1 + 1.3125x2 + 1.4625x3.

(b) The ANOVA table for all these single-degree-of-freedom components can be displayed

as:

Source of Degrees of Mean Computed

Variation Freedom Square f P-value

x1

x2

x3

Error

1

1

1

4

3.00125

13.78125

17.11125

3.15625

0.95

4.37

5.42

0.3847

0.1049

0.0804

Total 7

Only x3 is near significant.

12.63 (a) For the completed second-order model, we have

PRESS = 9, 657, 641.55,

Xn

i=1

|yi − ˆyi,−i| = 5, 211.37.

(b) When the model does not include any term involving x4,

PRESS = 6, 954.49,

Xn

i=1

|yi − ˆyi,−i| = 277.292.

Apparently, the model without x4 is much better.

(c) For the model with x4:

PRESS = 312, 963.71,

Xn

i=1

|yi − ˆyi,−i| = 762.57.

For the model without x4:

PRESS = 3, 879.89,

Xn

i=1

|yi − ˆyi,−i| = 220.12

Once again, the model without x4 performs better in terms of PRESS and

Pn

i=1 |yi−

ˆyi,−i|.

12.64 (a) The stepwise regression results in the following fitted model:

ˆy = 102.20428 − 0.21962x1 − 0.0723 − x2 − 2.68252x3 − 0.37340x5 + 0.30491x6.

(b) Using the Cp criterion, the best model is the same as the one in (a).

12.65 (a) Yes. The orthogonality is maintained with the use of interaction terms.

182 Chapter 12 Multiple Linear Regression and Certain Nonlinear Regression Models

(b) No. There are no degrees of freedom available for computing the standard error.

12.66 The fitted model is ˆy = 26.19333+0.04772x1+0.76011x2−0.00001334x11−0.00687x22+

0.00011333x12. The t-tests for each coefficient show that x12 and x22 may be eliminated.

So, we ran a test for β12 = β22 = 0 which yields P-value = 0.2222. Therefore, both x12

and x22 may be dropped out from the model.

12.67 (a) The fitted model is ˆy = −0.26891 + 0.07755x1 + 0.02532x2 − 0.03575x3. The

f-value of the test for H0 : β1 = β2 = β3 = 0 is 35,28 with P-value < 0.0001.

Hence, we reject H0.

(b) The residual plots are shown below and they all display random residuals.

7.5 8.0 8.5 9.0 9.5 10.0

−2 −1 0 1 2

x1

Studentized Residual

0 1 2 3 4 5

−2 −1 0 1 2

x2

Studentized Residual

0 2 4 6 8

−2 −1 0 1 2

x3

Studentized Residual

(c) The following is the summary of these three models.

Model PRESS Cp

x1, x2, x3 0.091748 24.7365

x1, x2, x3, x2

1, x2

2 , x2

3 0.08446 12.3997

x1, x2, x3, x2

1 , x2

2 , x2

3, x12, x13, x23 0.088065 10

It is difficult to decide which of the three models is the best. Model I contains

all the significant variables while models II and III contain insignificant variables.

However, the Cp value and PRESS for model are not so satisfactory. Therefore,

some other models may be explored.

12.68 Denote by Z1 = 1 when Group=1, and Z1 = 0 otherwise;

Denote by Z2 = 1 when Group=2, and Z2 = 0 otherwise;

Denote by Z3 = 1 when Group=3, and Z3 = 0 otherwise;

Solutions for Exercises in Chapter 12 183

(a) The parameter estimates are:

Variable DF Parameter Estimate P-value

Intercept

BMI

z1

z2

1

1

1

1

46.34694

−1.79090

−23.84705

−17.46248

0.0525

0.0515

0.0018

0.0109

Yes, Group I has a mean change in blood pressure that was significantly lower

than the control group. It is about 23.85 points lower.

(b) The parameter estimates are:

Variable DF Parameter Estimate P-value

Intercept

BMI

z1

z3

1

1

1

1

28.88446

−1.79090

−6.38457

17.46248

0.1732

0.0515

0.2660

0.0109

Although Group I has a mean change in blood pressure that was 6.38 points lower

than that of Group II, the difference is not very significant due to a high P-value.

12.69 (b) All possible regressions should be run. R2 = 0.9908 and there is only one significant

variable.

(c) The model including x2, x3 and x5 is the best in terms of Cp, PRESS and has all

variables significant.

12.70 Using the formula of R2

adj on page 467, we have

R2

adj = 1 −

SSE/(n − k − 1)

SST/(n − 1)

= 1 −

MSE

MST

.

Since MST is fixed, maximizing R2

adj is thus equivalent to minimizing MSE.

12.71 (a) The fitted model is ˆp = 1

1+e2.7620−0.0308x .

(b) The χ2-values for testing b0 = 0 and b1 = 0 are 471.4872 and 243.4111, respectively.

Their corresponding P-values are < 0.0001. Hence, both coefficients are

significant.

(c) ED50 = −−2.7620

0.0308 = 89.675.

12.72 (a) The fitted model is ˆp = 1

1+e2.9949−0.0308x .

(b) The increase in odds of failure that results by increasing the load by 20 lb/in.2 is

e(20)(0.0308) = 1.8515.

Chapter 13

One-Factor Experiments: General

13.1 Using the formula of SSE, we have

SSE =

Xk

i=1

Xn

j=1

(yij − ¯yi.)2 =

Xk

i=1

Xn

j=1

(ǫij − ¯ǫi.)2 =

Xk

i=1

"

Xn

j=1

ǫ2

ij − n¯ǫ2

i.

#

.

Hence

E(SSE) =

Xk

i=1

"

Xn

j=1

E(ǫ2

ij) − nE(¯ǫ2

i.)

#

=

Xk

i=1

nσ2 − n

σ2

n

= k(n − 1)σ2.

Thus E

h

SSE

k(n−1)

i

= k(n−1) 2

k(n−1) = σ2.

13.2 Since SSA = n

Pk

i=1

(¯yi. − ¯y..)2 = n

Pk

i=1

¯y2

i. − kn¯y2

.., yij ∼ n(y; μ + αi, σ2), and hence

¯yi. ∼ n

y; μ + αi, √n

and ¯y.. ∼ n

μ + ¯α, √kn

, then

E(¯y2

i.) = V ar(¯yi.) + [E(¯yi.)]2 =

σ2

n

+ (μ + αi)2,

and

E[¯y2

..] =

σ2

kn

+ (μ + ¯α)2 =

σ2

kn

+ μ2,

due to the constraint on the α’s. Therefore,

E(SSA) = n

Xk

i=1

E(¯y2

i.) − knE(¯y2

..) = kσ2 + n

Xk

i=1

(μ + αi)2 − (σ2 + knμ2)

= (k − 1)σ2 + n

Xk

i=1

α2

i .

185

186 Chapter 13 One-Factor Experiments: General

13.3 The hypotheses are

H0 : μ1 = μ2 = · · · = μ6,

H1 : At least two of the means are not equal.

α = 0.05.

Critical region: f > 2.77 with v1 = 5 and v2 = 18 degrees of freedom.

Computation:

Source of Sum of Degrees of Mean Computed

Variation Squares Freedom Square f

Treatment

Error

5.34

62.64

5

18

1.07

3.48

0.31

Total 67.98 23

with P-value=0.9024.

Decision: The treatment means do not differ significantly.

13.4 The hypotheses are

H0 : μ1 = μ2 = · · · = μ5,

H1 : At least two of the means are not equal.

α = 0.05.

Critical region: f > 2.87 with v1 = 4 and v2 = 20 degrees of freedom.

Computation:

Source of Sum of Degrees of Mean Computed

Variation Squares Freedom Square f

Tablets

Error

78.422

59.532

4

20

19.605

2.977

6.59

Total 137.954 24

with P-value=0.0015.

Decision: Reject H0. The mean number of hours of relief differ significantly.

13.5 The hypotheses are

H0 : μ1 = μ2 = μ3,

H1 : At least two of the means are not equal.

α = 0.01.

Computation:

Solutions for Exercises in Chapter 13 187

Source of Sum of Degrees of Mean Computed

Variation Squares Freedom Square f

Shelf Height

Error

399.3

288.8

2

21

199.63

13.75

14.52

Total 688.0 23

with P-value=0.0001.

Decision: Reject H0. The amount of money spent on dog food differs with the shelf

height of the display.

13.6 The hypotheses are

H0 : μA = μB = μC,

H1 : At least two of the means are not equal.

α = 0.01.

Computation:

Source of Sum of Degrees of Mean Computed

Variation Squares Freedom Square f

Drugs

Error

158.867

393.000

2

27

79.433

14.556

5.46

Total 551.867 29

with P-value=0.0102.

Decision: Since α = 0.01, we fail to reject H0. However, this decision is very marginal

since the P-value is very close to the significance level.

13.7 The hypotheses are

H0 : μ1 = μ2 = μ3 = μ4,

H1 : At least two of the means are not equal.

α = 0.05.

Computation:

Source of Sum of Degrees of Mean Computed

Variation Squares Freedom Square f

Treatments

Error

119.787

638.248

3

36

39.929

17.729

2.25

Total 758.035 39

with P-value=0.0989.

Decision: Fail to reject H0 at level α = 0.05.

188 Chapter 13 One-Factor Experiments: General

13.8 The hypotheses are

H0 : μ1 = μ2 = μ3,

H1 : At least two of the means are not equal.

Computation:

Source of Sum of Degrees of Mean Computed

Variation Squares Freedom Square f

Solvents

Error

3.3054

1.9553

2

29

1.6527

0.0674

24.51

Total 5.2608 31

with P-value< 0.0001.

Decision: There is significant difference in the mean sorption rate for the three solvents.

The mean sorption for the solvent Chloroalkanes is the highest. We know that it is

significantly higher than the rate of Esters.

13.9 The hypotheses are

H0 : μ1 = μ2 = μ3 = μ4,

H1 : At least two of the means are not equal.

α = 0.01.

Computation:

Source of Sum of Degrees of Mean Computed

Variation Squares Freedom Square f

Treatments

Error

27.5506

18.6360

3

17

9.1835

1.0962

8.38

Total 46.1865 20

with P-value= 0.0012.

Decision: Reject H0. Average specific activities differ.

13.10 s50 = 3.2098, s100 = 4.5253, s200 = 5.1788, and s400 = 3.6490. Since the sample sizes

are all the same,

s2

p =

1

4

X4

i=1

s2

i = 17.7291.

Therefore, the Bartlett’s statistic is

b =

Q4

i=1

s2

i

1/4

s2

p

= 0.9335.

Solutions for Exercises in Chapter 13 189

Using Table A.10, the critical value of the Bartlett’s test with k = 4 and α = 0.05 is

0.7970. Since b > 0.7970, we fail to reject H0 and hence the variances can be assumed

equal.

13.11 Computation:

Source of Sum of Degrees of Mean Computed

Variation Squares Freedom Square f

B vs. A, C, D

C vs. A, D

A vs. D

Error

30.6735

49.9230

5.3290

34.3800

1

1

1

16

30.6735

49.9230

5.3290

2.1488

14.28

23.23

2.48

(a) P-value=0.0016. B is significantly different from the average of A, C, and D.

(b) P-value=0.0002. C is significantly different from the average of A and D.

(c) P-value=0.1349. A can not be shown to differ significantly from D.

13.12 (a) The hypotheses are

H0 : μ29 = μ54 = μ84,

H1 : At least two of the means are not equal.

Source of Sum of Degrees of Mean Computed

Variation Squares Freedom Square f

Protein Levels

Error

32, 974.87

28, 815.80

2

9

16, 487.43

3, 201.76

5.15

Total 61, 790.67 11

with P-value= 0.0323.

Decision: Reject H0. The mean nitrogen loss was significantly different for the

three protein levels.

(b) For testing the contrast L = 2μ29 − μ54 − μ84 at level α = 0.05, we have SSw =

31, 576.42 and f = 9.86, with P-value=0.0119. Hence, the mean nitrogen loss

for 29 grams of protein was different from the average of the two higher protein

levels.

13.13 (a) The hypotheses are

H0 : μ1 = μ2 = μ3 = μ4,

H1 : At least two of the means are not equal.

Source of Sum of Degrees of Mean Computed

Variation Squares Freedom Square f

Treatments

Error

1083.60

1177.68

3

44

361.20

26.77

13.50

Total 2261.28 47

190 Chapter 13 One-Factor Experiments: General

with P-value< 0.0001.

Decision: Reject H0. The treatment means are different.

(b) For testing two contrasts L1 = μ1 − μ2 and L2 = μ3 − μ4 at level α = 0.01, we

have the following

Contrast Sum of Squares Computed f P-value

1 vs. 2

3 vs. 4

785.47

96.00

29.35

3.59

< 0.0001

0.0648

Hence, Bath I and Bath II were significantly different for 5 launderings, and Bath

I and Bath II were not different for 10 launderings.

13.14 The means of the treatments are:

¯y1. = 5.44, ¯y2. = 7.90, ¯y3. = 4.30, ¯y4. = 2.98, and ¯y5. = 6.96.

Since q(0.05, 5, 20) = 4.24, the critical difference is (4.24)

q

2.9766

5 = 3.27. Therefore,

the Tukey’s result may be summarized as follows:

¯y5. ¯y3. ¯y1. ¯y5. ¯y2.

2.98 4.30 5.44 6.96 7.90

13.15 Since q(0.05, 4, 16) = 4.05, the critical difference is (4.05)

q

2.14875

5 = 2.655. Hence

¯y3. ¯y1. ¯y4. ¯y2.

56.52 59.66 61.12 61.96

13.16 (a) The hypotheses are

H0 : μ1 = μ2 = μ3 = μ4,

H1 : At least two of the means are not equal.

α = 0.05.

Computation:

Source of Sum of Degrees of Mean Computed

Variation Squares Freedom Square f

Blends

Error

119.649

44.920

3

8

39.883

5.615

7.10

Total 164.569 11

with P-value= 0.0121.

Decision: Reject H0. There is a significant difference in mean yield reduction for

the 4 preselected blends.

Solutions for Exercises in Chapter 13 191

(b) Since

p

s2/3 = 1.368 we get

p 2 3 4

rp 3.261 3.399 3.475

Rp 4.46 4.65 4.75

Therefore,

¯y3. ¯y1. ¯y2. ¯y4.

23.23 25.93 26.17 31.90

(c) Since q(0.05, 4, 8) = 4.53, the critical difference is 6.20. Hence

¯y3. ¯y1. ¯y2. ¯y4.

23.23 25.93 26.17 31.90

13.17 (a) The hypotheses are

H0 : μ1 = μ2 = · · · = μ5,

H1 : At least two of the means are not equal.

Computation:

Source of Sum of Degrees of Mean Computed

Variation Squares Freedom Square f

Procedures

Error

7828.30

3256.50

4

15

1957.08

217.10

9.01

Total 11084.80 19

with P-value= 0.0006.

Decision: Reject H0. There is a significant difference in the average species count

for the different procedures.

(b) Since q(0.05, 5, 15) = 4.373 and

q

217.10

4 = 7.367, the critical difference is 32.2.

Hence

¯yK ¯yS ¯ySub ¯yM ¯yD

12.50 24.25 26.50 55.50 64.25

13.18 The hypotheses are

H0 : μ1 = μ2 = · · · = μ5,

H1 : At least two of the means are not equal.

Computation:

192 Chapter 13 One-Factor Experiments: General

Source of Sum of Degrees of Mean Computed

Variation Squares Freedom Square f

Angles

Error

99.024

23.136

4

20

24.756

1.157

21.40

Total 122.160 24

with P-value< 0.0001.

Decision: Reject H0. There is a significant difference in mean pressure for the different

angles.

13.19 When we obtain the ANOVA table, we derive s2 = 0.2174. Hence

p

2s2/n =

p

(2)(0.2174)/5 = 0.2949.

The sample means for each treatment levels are

¯yC = 6.88, ¯y1. = 8.82, ¯y2. = 8.16, ¯y3. = 6.82, ¯y4. = 6.14.

Hence

d1 =

8.82 − 6.88

0.2949

= 6.579, d2 =

8.16 − 6.88

0.2949

= 4.340,

d3 =

6.82 − 6.88

0.2949

= −0.203, d4 =

6.14 − 6.88

0.2949

= −2.509.

From Table A.14, we have d0.025(4, 20) = 2.65. Therefore, concentrations 1 and 2 are

significantly different from the control.

13.20 The ANOVA table can be obtained as follows:

Source of Sum of Degrees of Mean Computed

Variation Squares Freedom Square f

Cables

Error

1924.296

2626.917

8

99

240.537

26.535

9.07

Total 4551.213 107

with P-value< 0.0001.

The results from Tukey’s procedure can be obtained as follows:

¯y2. ¯y3. ¯y1. ¯y4. ¯y6. ¯y7. ¯y5. ¯y8. ¯y9.

−7.000 −6.083 −4.083 −2.667 0.833 0.917 1.917 3.333 6.250

The cables are significantly different:

9 is different from 4, 1, 2, 3

8 is different from 1, 3, 2

5, 7, 6 are different from 3, 2.

Solutions for Exercises in Chapter 13 193

13.21 Aggregate 4 has a significantly lower absorption rate than the other aggregates.

13.22 (a) The hypotheses are

H0 : μC = μL = μM = μH,

H1 : At least two of the means are not equal.

Computation:

Source of Sum of Degrees of Mean Computed

Variation Squares Freedom Square f

Finance Leverages

Error

80.7683

100.8700

3

20

26.9228

5.0435

5.34

Total 181.6383 23

with P-value= 0.0073.

Decision: Reject H0. The means are not all equal for the different financial

leverages.

(b)

p

2s2/n =

p

(2)(5.0435)/6 = 1.2966. The sample means for each treatment levels

are

¯yC = 4.3833, ¯yL = 5.1000, ¯yH = 8.3333, ¯yM = 8.4167.

Hence

dL =

5.1000 − 4.3833

1.2966

= 0.5528, dM =

8.4167 − 4.3833

1.2966

= 3.1108,

dL =

8.333 − 4.3833

1.2966

= 3.0464.

From Table A.14, we have d0.025(3, 20) = 2.54. Therefore, the mean rate of return

are significantly higher for median and high financial leverage than for control.

13.23 The ANOVA table can be obtained as follows:

Source of Sum of Degrees of Mean Computed

Variation Squares Freedom Square f

Temperatures

Error

1268.5333

112.8333

4

25

317.1333

4.5133

70.27

Total 1381.3667 29

with P-value< 0.0001.

The results from Tukey’s procedure can be obtained as follows:

¯y0 ¯y25 ¯y100 ¯y75 ¯y50

55.167 60.167 64.167 70.500 72.833

194 Chapter 13 One-Factor Experiments: General

The batteries activated at temperature 50 and 75 have significantly longer activated

life.

13.24 The Duncan’s procedure shows the following results:

¯yE ¯yA ¯yC

0.3300 0.9422 1.0063

Hence, the sorption rate using the Esters is significantly lower than the sorption rate

using the Aromatics or the Chloroalkanes.

13.25 Based on the definition, we have the following.

SSB = k

Xb

j=1

(¯y.j − ¯y..)2 = k

Xb

j=1

T.j

k −

T..

bk

2

=

Xb

j=1

T2

.j

k − 2

T2

..

bk

+

T2

..

bk

=

Xb

j=1

T2

.j

k −

T2

..

bk

.

13.26 From the model

yij = μ + αi + βj + ǫij ,

and the constraints

Xk

i=1

αi = 0 and

Xb

j=1

βj = 0,

we obtain

¯y.j = μ + βj + ¯ǫ.j and ¯y.. = μ + ¯ǫ...

Hence

SSB = k

Xb

j=1

(¯y.j − ¯y..)2 = k

Xb

j=1

(βj + ¯ǫ.j − ¯ǫ..)2.

Since E(¯ǫ.j) = 0 and E(¯ǫ..) = 0, we obtain

E(SSB) = k

Xb

j=1

β2

j + k

Xb

j=1

E(¯ǫ2

. j) − bkE(¯ǫ2

..).

We know that E(¯ǫ2

.j) = 2

k and E(¯ǫ2

..) = 2

bk . Then

E(SSB) = k

Xb

j=1

β2

j + bσ2 − σ2 = (b − 1)σ2 + k

Xb

j=1

β2

j .

13.27 (a) The hypotheses are

H0 : α1 = α2 = α3 = α4 = 0, fertilizer effects are zero

H1 : At least one of the αi’s is not equal to zero.

α = 0.05.

Critical region: f > 4.76.

Computation:

Solutions for Exercises in Chapter 13 195

Source of Sum of Degrees of Mean Computed

Variation Squares Freedom Square f

Fertilizers

Blocks

Error

218.1933

197.6317

71.4017

3

2

6

72.7311

98.8158

11.9003

6.11

Total 487.2267 11

P-value= 0.0296. Decision: Reject H0. The means are not all equal.

(b) The results of testing the contrasts are shown as:

Source of Sum of Degrees of Mean Computed

Variation Squares Freedom Square f

(f1, f3) vs (f2. f4)

f1 vs f3

Error

206.6700

11.4817

71.4017

1

1

6

206.6700

11.4817

11.9003

17.37

0.96

The corresponding P-values for the above contrast tests are 0.0059 and 0.3639,

respectively. Hence, for the first contrast, the test is significant and the for the

second contrast, the test is insignificant.

13.28 The hypotheses are

H0 : α1 = α2 = α3 = 0, no differences in the varieties

H1 : At least one of the αi’s is not equal to zero.

α = 0.05.

Critical region: f > 5.14.

Computation:

Source of Sum of Degrees of Mean Computed

Variation Squares Freedom Square f

Treatments

Blocks

Error

24.500

171.333

42.167

2

3

6

12.250

57.111

7.028

1.74

Total 238.000 11

P-value=0.2535. Decision: Do not reject H0; could not show that the varieties of

potatoes differ in field.

13.29 The hypotheses are

H0 : α1 = α2 = α3 = 0, brand effects are zero

H1 : At least one of the αi’s is not equal to zero.

α = 0.05.

Critical region: f > 3.84.

Computation:

196 Chapter 13 One-Factor Experiments: General

Source of Sum of Degrees of Mean Computed

Variation Squares Freedom Square f

Treatments

Blocks

Error

27.797

16.536

18.556

2

4

8

13.899

4.134

2.320

5.99

Total 62.889 14

P-value=0.0257. Decision: Reject H0; mean percent of foreign additives is not the

same for all three brand of jam. The means are:

Jam A: 2.36, Jam B: 3.48, Jam C: 5.64.

Based on the means, Jam A appears to have the smallest amount of foreign additives.

13.30 The hypotheses are

H0 : α1 = α2 = α3 = α4 = 0, courses are equal difficulty

H1 : At least one of the αi’s is not equal to zero.

α = 0.05.

Computation:

Source of Sum of Degrees of Mean Computed

Variation Squares Freedom Square f

Subjects

Students

Error

42.150

1618.700

1112.100

3

4

12

14.050

404.675

92.675

0.15

Total 2772.950 19

P-value=0.9267. Decision: Fail to reject H0; there is no significant evidence to conclude

that courses are of different difficulty.

13.31 The hypotheses are

H0 : α1 = α2 = · · · = α6 = 0, station effects are zero

H1 : At least one of the αi’s is not equal to zero.

α = 0.01.

Computation:

Source of Sum of Degrees of Mean Computed

Variation Squares Freedom Square f

Stations

Dates

Error

230.127

3.259

44.018

5

5

25

46.025

0.652

1.761

26.14

Total 277.405 35

Solutions for Exercises in Chapter 13 197

P-value< 0.0001. Decision: Reject H0; the mean concentration is different at the

different stations.

13.32 The hypotheses are

H0 : α1 = α2 = α3 = 0, station effects are zero

H1 : At least one of the αi’s is not equal to zero.

α = 0.05.

Computation:

Source of Sum of Degrees of Mean Computed

Variation Squares Freedom Square f

Stations

Months

Error

10.115

537.030

744.416

2

11

22

5.057

48.821

33.837

0.15

Total 1291.561 35

P-value= 0.8620. Decision: Do not reject H0; the treatment means do not differ

significantly.

13.33 The hypotheses are

H0 : α1 = α2 = α3 = 0, diet effects are zero

H1 : At least one of the αi’s is not equal to zero.

α = 0.01.

Computation:

Source of Sum of Degrees of Mean Computed

Variation Squares Freedom Square f

Diets

Subjects

Error

4297.000

6033.333

1811.667

2

5

10

2148.500

1206.667

181.167

11.86

Total 12142.000 17

P-value= 0.0023. Decision: Reject H0; differences among the diets are significant.

13.34 The hypotheses are

H0 : α1 = α2 = α3 = 0, analyst effects are zero

H1 : At least one of the αi’s is not equal to zero.

α = 0.05.

Computation:

198 Chapter 13 One-Factor Experiments: General

Source of Sum of Degrees of Mean Computed

Variation Squares Freedom Square f

Analysts

Individuals

Error

0.001400

0.021225

0.001400

2

3

6

0.000700

0.007075

0.000233

3.00

Total 0.024025 11

P-value= 0.1250. Decision: Do not reject H0; cannot show that the analysts differ

significantly.

13.35 The hypotheses are

H0 : α1 = α2 = α3 = α4 = α5 = 0, treatment effects are zero

H1 : At least one of the αi’s is not equal to zero.

α = 0.01.

Computation:

Source of Sum of Degrees of Mean Computed

Variation Squares Freedom Square f

Treatments

Locations

Error

79630.133

634334.667

689106.667

4

5

20

19907.533

126866.933

34455.333

0.58

Total 1403071.467 29

P-value= 0.6821. Decision: Do not reject H0; the treatment means do not differ

significantly.

13.36 The hypotheses are

H0 : α1 = α2 = α3 = 0, treatment effects are zero

H1 : At least one of the αi’s is not equal to zero.

α = 0.01.

Computation:

Source of Sum of Degrees of Mean Computed

Variation Squares Freedom Square f

Treatments

Subjects

Error

203.2792

188.2271

212.8042

2

9

18

101.6396

20.9141

11.8225

8.60

Total 604.3104 29

P-value= 0.0024. Decision: Reject H0; the mean weight losses are different for different

treatments and the therapists had the greatest effect on the weight loss.

Solutions for Exercises in Chapter 13 199

13.37 The total sums of squares can be written as

X

i

X

j

X

k

(yijk − ¯y...)2 =

X

i

X

j

X

k

[(¯yi.. − ¯y...) + (¯y.j. − ¯y...) + (¯y..k − ¯y...)

+ (yijk − ¯yi.. − ¯y.j. − ¯y..k + 2¯y...)]2

=r

X

i

(¯yi.. − ¯y...)2 + r

X

j

(¯y.j. − ¯y...)2 + r

X

k

(¯y..k − ¯y...)2

+

X

i

X

j

X

k

(yijk − ¯yi.. − ¯y.j. − ¯y..k + 2¯y...)2

+ 6 cross-product terms,

and all cross-product terms are equal to zeroes.

13.38 For the model yijk = μ + αi + βj + τk + ǫijk, we have

¯y..k = μ + τk + ¯ǫ..k, and ¯y... = μ + ¯ǫ....

Hence SSTr = r

P

k

(¯y..k − ¯y...)2 = r

P

k

(τk + ¯ǫ..k − ¯ǫ...)2, and

E(SSTr) = r

X

k

τ 2

k + r

X

k

E(¯ǫ2

..k) − r2

X

k

E(¯ǫ2

...)

= r

X

k

τ 2

k + r

X

k

σ2

r − r2 σ2

r2 = r

X

k

τ 2

k + rσ2 − σ2

= (r − 1)σ2 + r

X

k

τ 2

k .

13.39 The hypotheses are

H0 : τ1 = τ2 = τ3 = τ4 = 0, professor effects are zero

H1 : At least one of the τi’s is not equal to zero.

α = 0.05.

Computation:

Source of Sum of Degrees of Mean Computed

Variation Squares Freedom Square f

Time Periods

Courses

Professors

Error

474.50

252.50

723.50

287.50

3

3

3

6

158.17

84.17

241.17

47.92

5.03

Total 1738.00 15

P-value= 0.0446. Decision: Reject H0; grades are affected by different professors.

200 Chapter 13 One-Factor Experiments: General

13.40 The hypotheses are

H0 : τA = τB = τC = τD = τE = 0, color additive effects are zero

H1 : At least one of the τi’s is not equal to zero.

α = 0.05.

Computation:

Source of Sum of Degrees of Mean Computed

Variation Squares Freedom Square f

Workers

Days

Additives

Error

12.4344

14.7944

3.9864

9.2712

4

4

4

12

3.1086

3.6986

0.9966

0.7726

1.29

Total 40.4864 24

P-value= 0.3280. Decision: Do not reject H0; color additives could not be shown to

have an effect on setting time.

13.41 The hypotheses are

H0 : α1 = α2 = α3 = 0, dye effects are zero

H1 : At least one of the αi’s is not equal to zero.

α = 0.05.

Computation:

Source of Sum of Degrees of Mean Computed

Variation Squares Freedom Square f

Amounts

Plants

Error

1238.8825

53.7004

101.2433

2

1

20

619.4413

53.7004

5.0622

122.37

Total 1393.8263 23

P-value< 0.0001. Decision: Reject H0; color densities of fabric differ significantly for

three levels of dyes.

13.42 (a) After a transformation g(y) = √y, we come up with an ANOVA table as follows.

Source of Sum of Degrees of Mean Computed

Variation Squares Freedom Square f

Materials

Error

7.5123

6.2616

2

27

3.7561

0.2319

16.20

Total 13.7739 29

Solutions for Exercises in Chapter 13 201

(b) The P-value< 0.0001. Hence, there is significant difference in flaws among three

materials.

(c) A residual plot is given below and it does show some heterogeneity of the variance

among three treatment levels.

1.0 1.5 2.0 2.5 3.0

−0.5 0.0 0.5 1.0

material

residual

(d) The purpose of the transformation is to stabilize the variances.

(e) One could be the distribution assumption itself. Once the data is transformed, it

is not necessary that the data would follow a normal distribution.

(f) Here the normal probability plot on residuals is shown.

−2 −1 0 1 2

−0.5 0.0 0.5 1.0

Theoretical Quantiles

Sample Quantiles

It appears to be close to a straight

line. So, it is likely that the transformed data are normally distributed.

13.43 (a) The hypotheses are

H0 : σ2

= 0,

H1 : σ2

6= 0

α = 0.05.

Computation:

202 Chapter 13 One-Factor Experiments: General

Source of Sum of Degrees of Mean Computed

Variation Squares Freedom Square f

Operators

Error

371.8719

99.7925

3

12

123.9573

8.3160

14.91

Total 471.6644 15

P-value= 0.0002. Decision: Reject H0; operators are different.

(b) ˆσ2 = 8.316 and ˆσ2

= 123.9573−8.3160

4 = 28.910.

13.44 The model is yij = μ + Ai + Bj + ǫij . Hence

¯y.j = μ + ¯ A. + Bj + ¯ǫ.j , and ¯y.. = μ + ¯ A. + ¯B. + ¯ǫ...

Therefore,

SSB = k

Xb

j=1

(¯y.j − ¯y..)2 = k

Xb

j=1

[(Bj − ¯B.) + (¯ǫ.j − ¯ǫ..)]2,

and

E(SSB) = k

Xb

j=1

E(B2

j ) − kbE(¯B2

. ) + k

Xb

j=1

E(¯ǫ2

.j) − kbE(¯ǫ2

..)

= kbσ2

− kσ2

+ bσ2 − σ2 = (b − 1)σ2 + k(b − 1)σ2

.

13.45 (a) The hypotheses are

H0 : σ2

= 0,

H1 : σ2

6= 0

α = 0.05.

Computation:

Source of Sum of Degrees of Mean Computed

Variation Squares Freedom Square f

Treatments

Blocks

Error

23.238

45.283

27.937

3

4

12

7.746

11.321

2.328

3.33

Total 96.458 19

P-value= 0.0565. Decision: Not able to show a significant difference in the random

treatments at 0.05 level, although the P-value shows marginal significance.

(b) σ2

= 7.746−2.328

5 = 1.084, and σ2

= 11.321−2.328

4 = 2.248.

13.46 From the model

yijk = μ + Ai + Bj + Tk + ǫij ,

Solutions for Exercises in Chapter 13 203

we have

¯y..k = μ + ¯ A. + ¯B. + Tk + ¯ǫ..k, and ¯y... = μ + ¯ A. + ¯B. + ¯ T. + ¯ǫ....

Hence,

SSTr = r

X

k

(¯y..k − ¯y...)2 = r

X

k

[(Tj − ¯ T.) + (¯ǫ..k − ¯ǫ...)]2,

and

E(SSTr) = r

X

k

E(T2

k ) − r2E( ¯ T2

. ) + r

X

k

E(¯ǫ2

. .k) − r2E(¯ǫ2

...)

= r2σ2

− rσ2

+ rσ2 − σ2 = (r − 1)(σ2 + rσ2

).

13.47 (a) The matrix is

A =

bk b b · · · b k k · · · k

b b 0 · · · 0 1 1 · · · 1

b 0 b · · · 0 1 1 · · · 1

...

...

...

. . .

...

...

...

. . .

...

b 0 0 · · · b 1 1 · · · 1

k 1 1 · · · k 0 0 · · · 0

k 1 1 · · · 0 k 0 · · · 0

...

...

...

. . .

...

...

...

. . .

...

k 1 1 · · · 0 0 0 · · · k

,

where b = number of blocks and k = number of treatments. The vectors are

b

′

= (μ, α1, α2, · · · , αk, β1, β2, · · · , βb)′, and

g

′

= (T.., T1., T2., · · · , Tk., T.1, T.2, · · · , T.b)

′

.

(b) Solving the system Ab = g with the constraints

Pk

i=1

αi = 0 and

Pb

j=1

βj = 0, we

have

ˆμ = ¯y..,

ˆαi = ¯yi. − ¯y.., for i = 1, 2, . . . , k,

ˆ βj = ¯y.j − ¯y.., for j = 1, 2, . . . , b.

Therefore,

R(α1, α2, . . . , αk, β1, β2, . . . , βb) = b

′

g −

T2

..

bk

=

Xk

i=1

T2

i.

b

+

Xb

j=1

T2

.j

k − 2

T2

..

bk

.

204 Chapter 13 One-Factor Experiments: General

To find R(β1, β2, . . . , βb | α1, α2, . . . , αk) we first find R(α1, α2, . . . , αk). Setting

βj = 0 in the model, we obtain the estimates (after applying the constraint

Pk

i=1

αi = 0)

ˆμ = ¯y.., and ˆαi = ¯yi. − ¯y.., for i = 1, 2, . . . , k.

The g vector is the same as in part (a) with the exception that T.1, T.2, . . . , T.b do

not appear. Thus one obtains

R(α1, α2, . . . , αk) =

Xk

i=1

T2

i.

b −

T2

..

bk

and thus

R(β1, β2, . . . , βb | α1, α2, . . . , αk) = R(α1, α2, . . . , αk, β1, β2, . . . , βb)

− R(α1, α2, . . . , αk) =

Xb

j=1

T2

.j

k −

T2

..

bk

= SSB.

13.48 Since

1 − β = P

F(3, 12) >

3.49

1 + (4)(1.5)

= P[F(12, 3) < 2.006] < 0.95.

Hence we do not have large enough samples. We then find, by trial and error, that

n = 16 is sufficient since

1 − β = P

F(3, 60) >

2.76

1 + (16)(1.5)

= P[F(60, 3) < 9.07] > 0.95.

13.49 We know φ2 = b

P4

i=1

2

i

4 2 = b

2 , when

P4

i=1

2

i

2 = 2.0.

If b = 10, φ = 2.24; v1 = 3 and v2 = 27 degrees of freedom.

If b = 9, φ = 2.12; v1 = 3 and v2 = 24 degrees of freedom.

If b = 8, φ = 2.00; v1 = 3 and v2 = 21 degrees of freedom.

From Table A.16 we see that b = 9 gives the desired result.

13.50 For the randomized complete block design we have

E(S2

1 ) = E

SSA

k − 1

= σ2 + b

Xk

i=1

α2

i

k − 1

.

Therefore,

λ =

v1[E(S2

1 )]

2σ2 −

v1

2

=

(k − 1)

σ2 + b

Pk

i=1

α2

i /(k − 1)

2σ2 −

k − 1

2

= b

Xk

i=1

α2

i

2σ2 ,

Solutions for Exercises in Chapter 13 205

and then

φ2 =

E(S2

1 ) − σ2

σ2 ·

v1

v1 + 1

=

[σ2 + b

Pk

i=1

α2

i /(k − 1)] − σ2

σ2 ·

k − 1

k

= b

Xk

i=1

α2

i

kσ2 .

13.51 (a) The model is yij = μ + αi + ǫij , where αi ∼ n(0, σ2

).

(b) Since s2 = 0.02056 and s2

1 = 0.01791, we have ˆσ2 = 0.02056 and s21

−s2

10 =

0.01791−0.02056

10 = −0.00027, which implies ˆσ2

= 0.

13.52 (a) The P-value of the test result is 0.7830. Hence, the variance component of pour

is not significantly different from 0.

(b) We have the ANOVA table as follows:

Source of Sum of Degrees of Mean Computed

Variation Squares Freedom Square f

Pours

Error

0.08906

1.02788

4

20

0.02227

0.05139

0.43

Total 1.11694 24

Since s21

−s2

5 = 0.02227−0.05139

5 < 0, we have ˆσ2

= 0.

13.53 (a) yij = μ + αi + ǫij , where αi ∼ n(x; 0, σ2

).

(b) Running an ANOVA analysis, we obtain the P-value as 0.0121. Hence, the loom

variance component is significantly different from 0 at level 0.05.

(c) The suspicion is supported by the data.

13.54 The hypotheses are

H0 : μ1 = μ2 = μ3 = μ4,

H1 : At least two of the μi’s are not equal.

α = 0.05.

Computation:

Source of Sum of Degrees of Mean Computed

Variation Squares Freedom Square f

Garlon levels

Error

3.7289

9.5213

3

12

1.2430

0.7934

1.57

Total 13.2502 15

P-value= 0.2487. Decision: Do not reject H0; there is insufficient evidence to claim

that the concentration levels of Garlon would impact the heights of shoots.

206 Chapter 13 One-Factor Experiments: General

13.55 Bartlett’s statistic is b = 0.8254. Conclusion: do not reject homogeneous variance

assumption.

13.56 The hypotheses are

H0 : τA = τB = τC = τD = τE = 0, ration effects are zero

H1 : At least one of the τi’s is not zero.

α = 0.01.

Computation:

Source of Sum of Degrees of Mean Computed

Variation Squares Freedom Square f

Lactation Periods

Cows

Rations

Error

245.8224

353.1224

89.2624

21.3392

4

4

4

12

61.4556

88.2806

22.3156

1.7783

12.55

Total 709.5464 24

P-value= 0.0003. Decision: Reject H0; different rations have an effect on the daily

milk production.

13.57 It can be shown that ¯yC = 76.16, ¯y1 = 81.20, ¯y2 = 81.54 and ¯y3 = 80.98. Since this is

a one-sided test, we find d0.01(3, 16) = 3.05 and

r

2s2

n

=

r

(2)(3.52575)

5

= 1.18756.

Hence

d1 =

81.20 − 76.16

1.18756

= 4.244, d2 =

81.54 − 76.16

1.18756

= 4.532, d3 =

80.98 − 76.16

1.18756

= 4.059,

which are all larger than the critical value. Hence, significantly higher yields are

obtained with the catalysts than with no catalyst.

13.58 (a) The hypotheses for the Bartlett’s test are

H0 : σ2

A = σ2

B = σ2

C = σ2

D ,

H1 : The variances are not all equal.

α = 0.05.

Critical region: We have n1 = n2 = n3 = n4 = 5, N = 20, and k = 4. Therefore,

we reject H0 when b < b4(0.05, 5) = 0.5850.

Computation: sA = 1.40819, sB = 2.16056, sC = 1.16276, sD = 0.76942 and

hence sp = 1.46586. From these, we can obtain that b = 0.7678.

Decision: Do not reject H0; there is no sufficient evidence to conclude that the

variances are not equal.

Solutions for Exercises in Chapter 13 207

(b) The hypotheses are

H0 : μA = μB = μC = μD,

H1 : At least two of the μi’s are not equal.

α = 0.05.

Computation:

Source of Sum of Degrees of Mean Computed

Variation Squares Freedom Square f

Laboratories

Error

85.9255

34.3800

3

16

28.6418

2.1488

13.33

Total 120.3055 19

P-value= 0.0001. Decision: Reject H0; the laboratory means are significantly

different.

(c) The normal probability plot is given as follows:

−2 −1 0 1 2

−2 −1 0 1 2

Theoretical Quantiles

Sample Quantiles

13.59 The hypotheses for the Bartlett’s test are

H0 : σ2

1 = σ2

2 = σ2

3 = σ2

4,

H1 : The variances are not all equal.

α = 0.01.

Critical region: We have n1 = n2 = n3 = 4, n4 = 9 N = 21, and k = 4. Therefore, we

reject H0 when

b < b4(0.01, 4, 4, 4, 9)

=

(4)(0.3475) + (4)(0.3475) + (4)(0.3475) + (9)(0.6892)

21

= 0.4939.

Computation: s2

1 = 0.41709, s2

2 = 0.93857, s2

3 = 0.25673, s2

4 = 1.72451 and hence

s2

p = 1.0962. Therefore,

b =

[(0.41709)3(0.93857)3(0.25763)3(1.72451)8]1/17

1.0962

= 0.79.

208 Chapter 13 One-Factor Experiments: General

Decision: Do not reject H0; the variances are not significantly different.

13.60 The hypotheses for the Cochran’s test are

H0 : σ2

A = σ2

B = σ2

C ,

H1 : The variances are not all equal.

α = 0.01.

Critical region: g > 0.6912.

Computation: s2

A = 29.5667, s2

B = 10.8889, s2

C = 3.2111, and hence

P

s2

i = 43.6667.

Now, g = 29.5667

43.6667 = 0.6771.

Decision: Do not reject H0; the variances are not significantly different.

13.61 The hypotheses for the Bartlett’s test are

H0 : σ2

1 = σ2

2 = σ2

3,

H1 : The variances are not all equal.

α = 0.05.

Critical region: reject H0 when

b < b4(0.05, 9, 8, 15) =

(9)(0.7686) + (8)(0.7387) + (15)(0.8632)

32

= 0.8055.

Computation: b = [(0.02832)8(0.16077)7(0.04310)14]1/29

0.067426 = 0.7822.

Decision: Reject H0; the variances are significantly different.

13.62 (a) The hypotheses are

H0 : α1 = α2 = α3 = α4 = 0,

H1 : At least one of the αi’s is not zero.

α = 0.05.

Computation:

Source of Sum of Degrees of Mean Computed

Variation Squares Freedom Square f

Diets

Blocks

Error

822.1360

17.1038

106.3597

3

5

15

274.0453

3.4208

7.0906

38.65

Total 945.5995 23

with P-value< 0.0001.

Decision: Reject H0; diets do have a significant effect on mean percent dry matter.

Solutions for Exercises in Chapter 13 209

(b) We know that ¯yC = 35.8483, ¯yF = 36.4217, ¯yT = 45.1833, ¯yA = 49.6250, and

r

2s2

n

=

r

(2)(7.0906)

6

= 1.5374.

Hence,

dAmmonia =

49.6250 − 35.8483

1.5374

= 8.961,

dUrea Feeding =

36.4217 − 35.8483

1.5374

= 0.3730,

dUrea Treated =

45.1833 − 35.8483

1.5374

= 6.0719.

Using the critical value d0.05(3, 15) = 2.61, we obtain that only “Urea Feeding” is

not significantly different from the control, at level 0.05.

(c) The normal probability plot for the residuals are given below.

−2 −1 0 1 2

−4 −2 0 2 4

Theoretical Quantiles

Sample Quantiles

13.63 The hypotheses are

H0 : α1 = α2 = α3 = 0,

H1 : At least one of the αi’s is not zero.

α = 0.05.

Computation:

Source of Sum of Degrees of Mean Computed

Variation Squares Freedom Square f

Diet

Error

0.32356

0.20808

2

12

0.16178

0.01734

9.33

Total 0.53164 14

210 Chapter 13 One-Factor Experiments: General

with P-value= 0.0036.

Decision: Reject H0; zinc is significantly different among the diets.

13.64 (a) The gasoline manufacturers would want to apply their results to more than one

model of car.

(b) Yes, there is a significant difference in the miles per gallon for the three brands

of gasoline.

(c) I would choose brand C for the best miles per gallon.

13.65 (a) The process would include more than one stamping machine and the results might

differ with different machines.

(b) The mean plot is shown below.

1

1

1

3.2 3.4 3.6 3.8 4.0 4.2 4.4

Material

Number of Gaskets

2

2

2

cork plastic rubber

(c) Material 1 appears to be the best.

(d) Yes, there is interaction. Materials 1 and 3 have better results with machine 1

but material 2 has better results with machine 2.

13.66 (a) The hypotheses are

H0 : α1 = α2 = α3 = 0,

H1 : At least one of the αi’s is not zero.

Computation:

Source of Sum of Degrees of Mean Computed

Variation Squares Freedom Square f

Paint Types

Error

227875.11

336361.83

2

15

113937.57

22424.12

5.08

Total 564236.94 17

with P-value= 0.0207.

Decision: Reject H0 at level 0.05; the average wearing quality differs significantly

for three paints.

(b) Using Tukey’s test, it turns out the following.

Solutions for Exercises in Chapter 13 211

¯y1. ¯y3. ¯y2.

197.83 419.50 450.50

Types 2 and 3 are not significantly different, while Type 1 is significantly different

from Type 2.

(c) We plot the residual plot and the normal probability plot for the residuals as

follows.

1.0 1.5 2.0 2.5 3.0

−200 −100 0 100 200

Type

Residual

−2 −1 0 1 2

−200 −100 0 100 200

Theoretical Quantiles

Sample Quantiles

It appears that the heterogeneity in variances may be violated, as is the normality

assumption.

(d) We do a log transformation of the data, i.e., y′ = log(y). The ANOVA result has

changed as follows.

Source of Sum of Degrees of Mean Computed

Variation Squares Freedom Square f

Paint Types

Error

2.6308

3.2516

2

15

1.3154

0.2168

6.07

Total 5.8824 17

with P-value= 0.0117.

Decision: Reject H0 at level 0.05; the average wearing quality differ significantly

for three paints. The residual and normal probability plots are shown here:

1.0 1.5 2.0 2.5 3.0

−0.8 −0.6 −0.4 −0.2 0.0 0.2 0.4 0.6

Type

Residual −

2

−

1

0

1

2

−0.8 −0.6 −0.4 −0.2 0.0 0.2 0.4 0.6

Theoretical Quantiles

Sample Quantiles

While the homogeneity of the variances seem to be a little better, the normality

assumption may still be invalid.

212 Chapter 13 One-Factor Experiments: General

13.67 (a) The hypotheses are

H0 : α1 = α2 = α3 = α4 = 0,

H1 : At least one of the αi’s is not zero.

Computation:

Source of Sum of Degrees of Mean Computed

Variation Squares Freedom Square f

Locations

Error

0.01594

0.00616

3

16

0.00531

0.00039

13.80

Total 0.02210 19

with P-value= 0.0001.

Decision: Reject H0; the mean ozone levels differ significantly across the locations.

(b) Using Tukey’s test, the results are as follows.

¯y4. ¯y1. ¯y3. ¯y2.

0.078 0.092 0.096 0.152

Location 2 appears to have much higher ozone measurements than other locations.

Chapter 14

Factorial Experiments (Two or More

Factors)

14.1 The hypotheses of the three parts are,

(a) for the main effects temperature,

H

′

0 : α1 = α2 = α3 = 0,

H

′

1 : At least one of the αi’s is not zero;

(b) for the main effects ovens,

H

′′

0 : β1 = β2 = β3 = β4 = 0,

H

′′

1 : At least one of the βi’s is not zero;

(c) and for the interactions,

H

′′′

0 : (αβ)11 = (αβ)12 = · · · = (αβ)34 = 0,

H

′′′

1 : At least one of the (αβ)ij ’s is not zero.

α = 0.05.

Critical regions: (a) f1 > 3.00; (b) f2 > 3.89; and (c) f3 > 3.49.

Computations: From the computer printout we have

Source of Sum of Degrees of Mean Computed

Variation Squares Freedom Square f

Temperatures

Ovens

Interaction

Error

5194.08

4963.12

3126.26

3833.50

2

3

6

12

2597.0400

1654.3733

521.0433

319.4583

8.13

5.18

1.63

Total 17, 116.96 23

213

214 Chapter 14 Factorial Experiments (Two or More Factors)

Decision: (a) Reject H′

0; (b) Reject H′′

0 ; (c) Do not reject H′′′

0 .

14.2 The hypotheses of the three parts are,

(a) for the main effects brands,

H

′

0 : α1 = α2 = α3 = 0,

H

′

1 : At least one of the αi’s is not zero;

(b) for the main effects times,

H

′′

0 : β1 = β2 = β3 = 0,

H

′′

1 : At least one of the βi’s is not zero;

(c) and for the interactions,

H

′′′

0 : (αβ)11 = (αβ)12 = · · · = (αβ)33 = 0,

H

′′′

1 : At least one of the (αβ)ij ’s is not zero.

α = 0.05.

Critical regions: (a) f1 > 3.35; (b) f2 > 3.35; and (c) f3 > 2.73.

Computations: From the computer printout we have

Source of Sum of Degrees of Mean Computed

Variation Squares Freedom Square f

Brands

Times

Interaction

Error

32.7517

227.2117

17.3217

254.7025

2

2

4

27

16.3758

113.6058

4.3304

9.4334

1.74

12.04

0.46

Total 531.9875 35

Decision: (a) Do not reject H′

0; (b) Reject H′′

0 ; (c) Do not reject H′′′

0 .

14.3 The hypotheses of the three parts are,

(a) for the main effects environments,

H

′

0 : α1 = α2 = 0, (no differences in the environment)

H

′

1 : At least one of the αi’s is not zero;

(b) for the main effects strains,

H

′′

0 : β1 = β2 = β3 = 0, (no differences in the strains)

H

′′

1 : At least one of the βi’s is not zero;

Solutions for Exercises in Chapter 14 215

(c) and for the interactions,

H

′′′

0 : (αβ)11 = (αβ)12 = · · · = (αβ)23 = 0, (environments and strains do not interact)

H

′′′

1 : At least one of the (αβ)ij ’s is not zero.

α = 0.01.

Critical regions: (a) f1 > 7.29; (b) f2 > 5.16; and (c) f3 > 5.16.

Computations: From the computer printout we have

Source of Sum of Degrees of Mean Computed

Variation Squares Freedom Square f

Environments

Strains

Interaction

Error

14, 875.521

18, 154.167

1, 235.167

42, 192.625

1

2

2

42

14, 875.521

9, 077.083

617.583

1004.586

14.81

9.04

0.61

Total 76, 457.479 47

Decision: (a) Reject H′

0; (b) Reject H′′

0 ; (c) Do not reject H′′′

0 . Interaction is not

significant, while both main effects, environment and strain, are all significant.

14.4 (a) The hypotheses of the three parts are,

H

′

0 : α1 = α2 = α3 = 0

H

′

1 : At least one of the αi’s is not zero;

H

′′

0 : β1 = β2 = β3 = 0,

H

′′

1 : At least one of the βi’s is not zero;

H

′′′

0 : (αβ)11 = (αβ)12 = · · · = (αβ)33 = 0,

H

′′′

1 : At least one of the (αβ)ij ’s is not zero.

α = 0.01.

Critical regions: for H′

0, f1 > 3.21; for H′′

0 , f2 > 3.21; and for H′′′

0 , f3 > 2.59.

Computations: From the computer printout we have

Source of Sum of Degrees of Mean Computed

Variation Squares Freedom Square f

Coating

Humidity

Interaction

Error

1, 535, 021.37

1, 020, 639.15

1, 089, 989.63

5, 028, 396.67

2

2

4

45

767, 510.69

510, 319.57

272, 497.41

111, 742.15

6.87

4.57

2.44

Total 76, 457.479 47

Decision: Reject H′

0; Reject H′′

0 ; Do not reject H′′′

0 . Coating and humidity do not

interact, while both main effects are all significant.

216 Chapter 14 Factorial Experiments (Two or More Factors)

(b) The three means for the humidity are ¯yL = 733.78, ¯yM = 406.39 and ¯yH = 638.39.

Using Duncan’s test, the means can be grouped as

¯yM ¯yL ¯yH

406.39 638.39 733.78

Therefore, corrosion damage is different for medium humidity than for low or high

humidity.

14.5 The hypotheses of the three parts are,

(a) for the main effects subjects,

H

′

0 : α1 = α2 = α3 = 0,

H

′

1 : At least one of the αi’s is not zero;

(b) for the main effects muscles,

H

′′

0 : β1 = β2 = β3 = β4 = β5 = 0,

H

′′

1 : At least one of the βi’s is not zero;

(c) and for the interactions,

H

′′′

0 : (αβ)11 = (αβ)12 = · · · = (αβ)35 = 0,

H

′′′

1 : At least one of the (αβ)ij ’s is not zero.

α = 0.01.

Critical regions: (a) f1 > 5.39; (b) f2 > 4.02; and (c) f3 > 3.17.

Computations: From the computer printout we have

Source of Sum of Degrees of Mean Computed

Variation Squares Freedom Square f

Subjects

Muscles

Interaction

Error

4, 814.74

7, 543.87

11, 362.20

2, 099.17

2

4

8

30

2, 407.37

1, 885.97

1, 420.28

69.97

34.40

26.95

20.30

Total 25, 819.98 44

Decision: (a) Reject H′

0; (b) Reject H′′

0 ; (c) Reject H′′′

0 .

14.6 The ANOVA table is shown as

Source of Sum of Degrees of Mean Computed

Variation Squares Freedom Square f P-value

Additive

Temperature

Interaction

Error

1.7578

0.8059

1.7934

1.8925

1

3

3

24

1.7578

0.2686

0.5978

0.0789

22.29

3.41

7.58

< 0.0001

0.0338

0.0010

Total 6.2497 32

Solutions for Exercises in Chapter 14 217

Decision: All main effects and interaction are significant.

An interaction plot is given here.

1

1

3.0 3.2 3.4 3.6 3.8 4.0

Additive

Adhesiveness

2

2

3

3

4

4

0 1

Temperature

1234

50

60

70

80

14.7 The ANOVA table is

Source of Sum of Degrees of Mean Computed

Variation Squares Freedom Square f P-value

Temperature

Catalyst

Interaction

Error

430.475

2, 466.650

326.150

264.500

3

4

12

20

143.492

616.663

27.179

13.225

10.85

46.63

2.06

0.0003

< 0.0001

0.0745

Total 3, 487.775 39

Decision: All main effects are significant and the interaction is significant at level

0.0745. Hence, if 0.05 significance level is used, interaction is not significant.

An interaction plot is given here.

1

1

1

1

1

40 45 50 55 60 65 70

Amount of Catalyst

Extraction Rate

2

2

2

2

2

3

3

3

3

3

4

4 4

4

4

0.5 0.6 0.7 0.8 0.9

Temperature

1234

50

60

70

80

Duncan’s tests, at level 0.05, for both main effects result in the following.

(a) For Temperature:

¯y50 ¯y80 ¯y70 ¯y60

52.200 59.000 59.800 60.300

218 Chapter 14 Factorial Experiments (Two or More Factors)

(b) For Amount of Catalyst:

¯y0.5 ¯y0.6 ¯y0.9 ¯y0.7 ¯y0.8

44.125 56.000 58.125 64.625 66.250

14.8 (a) The ANOVA table is

Source of Sum of Degrees of Mean Computed

Variation Squares Freedom Square f P-value

Nickel

pH

Nickel*pH

Error

31, 250.00

6, 606.33

670.33

8, 423.33

1

2

2

12

31, 250.00

3, 303.17

335.17

701.94

44.52

4.71

0.48

< 0.0001

0.0310

0.6316

Total 3, 487.775 39

Decision: Nickel contents and levels of pH do not interact to each other, while

both main effects of nickel contents and levels of pH are all significant, at level

higher than 0.0310.

(b) In comparing the means of the six treatment combinations, a nickel content level

of 18 and a pH level of 5 resulted in the largest values of thickness.

(c) The interaction plot is given here and it shows no apparent interactions.

1

1

1

80 100 120 140 160 180 200

pH

Thickness

2

2 2

5 5.5 6

Nickel

12

10

18

14.9 (a) The ANOVA table is

Source of Sum of Degrees of Mean Computed

Variation Squares Freedom Square f P-value

Tool

Speed

Tool*Speed

Error

675.00

12.00

192.00

72.67

1

1

1

8

675.00

12.00

192.00

9.08

74.31

1.32

21.14

< 0.0001

0.2836

0.0018

Total 951.67 11

Decision: The interaction effects are significant. Although the main effects of

speed showed insignificance, we might not make such a conclusion since its effects

might be masked by significant interaction.

Solutions for Exercises in Chapter 14 219

(b) In the graph shown, we claim that the cutting speed that results in the longest

life of the machine tool depends on the tool geometry, although the variability of

the life is greater with tool geometry at level 1.

1

1

5 10 15 20 25 30 35

Tool Geometry

Life

2

2

1 2

Speed

12

High

Low

(c) Since interaction effects are significant, we do the analysis of variance for separate

tool geometry.

(i) For tool geometry 1, an f-test for the cutting speed resulted in a P-value =

0.0405 with the mean life (standard deviation) of the machine tool at 33.33

(4.04) for high speed and 23.33 (4.16) for low speed. Hence, a high cutting

speed has longer life for tool geometry 1.

(ii) For tool geometry 2, an f-test for the cutting speed resulted in a P-value =

0.0031 with the mean life (standard deviation) of the machine tool at 10.33

(0.58) for high speed and 16.33 (1.53) for low speed. Hence, a low cutting

speed has longer life for tool geometry 2.

For the above detailed analysis, we note that the standard deviations for the mean

life are much higher at tool geometry 1.

(d) See part (b).

14.10 (a) yijk = μ + αi + βj + (αβ)ij + ǫijk, i = 1, 2, . . . , a; j = 1, 2, . . . , b; k = 1, 2, . . . , n.

(b) The ANOVA table is

Source of Sum of Degrees of Mean Computed

Variation Squares Freedom Square f P-value

Dose

Position

Error

117.9267

15.0633

13.0433

1

2

2

117.9267

7.5317

6.5217

18.08

1.15

0.0511

0.4641

Total 146.0333 5

(c) (n − 1) − (a − 1) − (b − 1) = 5 − 1 − 2 = 2.

(d) At level 0.05, Tukey’s result for the furnace position is shown here:

¯y2 ¯y1 ¯y3

19.850 21.350 23.700

220 Chapter 14 Factorial Experiments (Two or More Factors)

Although Tukey’s multiple comparisons resulted in insignificant differences among

the furnace position levels, based on a P-value of 0.0511 for the Dose and on the

plot shown we can see that Dose=2 results in higher resistivity.

1

1

1

16 18 20 22 24 26

Furnace Position

Resistivity

2

2

2

1 2 3

Dose

12

12

14.11 (a) The ANOVA table is

Source of Sum of Degrees of Mean Computed

Variation Squares Freedom Square f P-value

Method

Lab

Method*Lab

Error

0.000104

0.008058

0.000198

0.000222

1

6

6

14

0.000104

0.001343

0.000033

0.000016

6.57

84.70

2.08

0.0226

< 0.0001

0.1215

Total 0.00858243 27

(b) Since the P-value = 0.1215 for the interaction, the interaction is not significant.

Hence, the results on the main effects can be considered meaningful to the scientist.

(c) Both main effects, method of analysis and laboratory, are all significant.

(d) The interaction plot is show here.

1

1

1

1

1

1

1

0.10 0.11 0.12 0.13 0.14 0.15 0.16

Lab

Sulfur Percent

2

2

2

2

2

2

2

1 2 3 4 5 6 7

Method

12

12

(e) When the tests are done separately, i.e., we only use the data for Lab 1, or Lab

2 alone, the P-values for testing the differences of the methods at Lab 1 and 7

Solutions for Exercises in Chapter 14 221

are 0.8600 and 0.1557, respectively. In this case, usually the degrees of freedom

of errors are small. If we compare the mean differences of the method within the

overall ANOVA model, we obtain the P-values for testing the differences of the

methods at Lab 1 and 7 as 0.9010 and 0.0093, respectively. Hence, methods are

no difference in Lab 1 and are significantly different in Lab 7. Similar results may

be found in the interaction plot in (d).

14.12 (a) The ANOVA table is

Source of Sum of Degrees of Mean Computed

Variation Squares Freedom Square f P-value

Time

Copper

Time*Copper

Error

0.025622

0.008956

0.012756

0.005933

2

2

4

18

0.012811

0.004478

0.003189

0.000330

38.87

13.58

9.67

< 0.0001

0.0003

0.0002

Total 0.053267 26

(b) The P-value < 0.0001. There is a significant time effect.

(c) The P-value = 0.0003. There is a significant copper effect.

(d) The interaction effect is significant since the P-value = 0.0002. The interaction

plot is show here.

1

1

1

0.24 0.26 0.28 0.30 0.32 0.34 0.36

Copper Content

Algae Concentration

2

2

2

3

3

3

1 2 3

Time

123

5

12

18

The algae concentrations for the various copper contents are all clearly influenced

by the time effect shown in the graph.

14.13 (a) The interaction plot is show here. There seems no interaction effect.

1

1

1.90 1.95 2.00 2.05 2.10 2.15

Treatment

Magnesium Uptake

2

2

1 2

Time

12

12

222 Chapter 14 Factorial Experiments (Two or More Factors)

(b) The ANOVA table is

Source of Sum of Degrees of Mean Computed

Variation Squares Freedom Square f P-value

Treatment

Time

Treatment*Time

Error

0.060208

0.060208

0.000008

0.003067

1

1

1

8

0.060208

0.060208

0.000008

0.000383

157.07

157.07

0.02

< 0.0001

< 0.0001

0.8864

Total 0.123492 11

(c) The magnesium uptake are lower using treatment 2 than using treatment 1, no

matter what the times are. Also, time 2 has lower magnesium uptake than time

1. All the main effects are significant.

(d) Using the regression model and making “Treatment” as categorical, we have the

following fitted model:

ˆy = 2.4433 − 0.13667 Treatment − 0.13667 Time − 0.00333Treatment × Time.

(e) The P-value of the interaction for the above regression model is 0.8864 and hence

it is insignificant.

14.14 (a) A natural linear model with interaction would be

y = β0 + β1x1 + β2x2 + β12x1x2.

The fitted model would be

ˆy = 0.41772 − 0.06631x1 − 0.00866x2 + 0.00416x1x2,

with the P-values of the t-tests on each of the coefficients as 0.0092, 0.0379 and

0.0318 for x1, x2, and x1x2, respectively. They are all significant at a level larger

than 0.0379. Furthermore, R2

adj = 0.1788.

(b) The new fitted model is

ˆy = 0.3368 − 0.15965x1 + 0.02684x2 + 0.00416x1x2 + 0.02333x2

1 − 0.00155x2

2,

with P-values of the t-tests on each of the coefficients as 0.0004, < 0.0001, 0.0003,

0.0156, and < 0.0001 for x1, x2. x1x2, x2

1, and x2

2, respectively. Furthermore,

R2

adj = 0.7700 which is much higher than that of the model in (a). Model in (b)

would be more appropriate.

14.15 The ANOVA table is given here.

Solutions for Exercises in Chapter 14 223

Source of Sum of Degrees of Mean Computed

Variation Squares Freedom Square f P-value

Main effect

A

B

C

Two-factor

Interaction

AB

AC

BC

Three-factor

Interaction

ABC

Error

2.24074

56.31815

17.65148

31.47148

31.20259

2156074

26.79852

148.04000

1

2

2

2

2

4

4

36

2.24074

28.15907

8.82574

15.73574

15.60130

5.39019

6.69963

4.11221

0.54

6.85

3.83

3.83

3.79

1.31

1.63

0.4652

0.0030

0.1316

0.0311

0.0320

0.2845

0.1881

Total 335.28370 53

(a) Based on the P-values, only AB and AC interactions are significant.

(b) The main effect B is significant. However, due to significant interactions mentioned

in (a), the insignificance of A and C cannot be counted.

(c) Look at the interaction plot of the mean responses versus C for different cases of

A.

1

1

1

13.5 14.0 14.5 15.0 15.5 16.0

C

y

2

2

2

1 2 3

A

12

12

Apparently, the mean responses at different levels of C varies in different patterns

for the different levels of A. Hence, although the overall test on factor C is

insignificant, it is misleading since the significance of the effect C is masked by

the significant interaction between A and C.

14.16 (a) When only A, B, C, and BC factors are in the model, the P-value for BC

interaction is 0.0806. Hence at level of 0.05, the interaction is insignificant.

(b) When the sum of squares of the BC term is pooled with the sum of squares of

the error, we increase the degrees of freedom of the error term. The P-values of

224 Chapter 14 Factorial Experiments (Two or More Factors)

the main effects of A, B, and C are 0.0275, 0.0224, and 0.0131, respectively. All

these are significant.

14.17 Letting A, B, and C designate coating, humidity, and stress, respectively, the ANOVA

table is given here.

Source of Sum of Degrees of Mean Computed

Variation Squares Freedom Square f P-value

Main effect

A

B

C

Two-factor

Interaction

AB

AC

BC

Three-factor

Interaction

ABC

Error

216, 384.1

19, 876, 891.0

427, 993, 946.4

31, 736, 625

699, 830.1

58, 623, 693.2

36, 034, 808.9

335, 213, 133.6

1

2

2

2

2

4

4

72

216, 384.1

9, 938, 445.5

213, 996, 973.2

15, 868, 312.9

349, 915.0

13, 655, 923.3

9, 008, 702.2

4, 655, 738.0

0.05

2.13

45.96

3.41

0.08

3.15

1.93

0.8299

0.1257

< 0.0001

0.0385

0.9277

0.0192

0.1138

Total 910, 395, 313.1 89

(a) The Coating and Humidity interaction, and the Humidity and Stress interaction

have the P-values of 0.0385 and 0.0192, respectively. Hence, they are all significant.

On the other hand, the Stress main effect is strongly significant as well.

However, both other main effects, Coating and Humidity, cannot be claimed as

insignificant, since they are all part of the two significant interactions.

(b) A Stress level of 20 consistently produces low fatigue. It appears to work best

with medium humidity and an uncoated surface.

14.18 The ANOVA table is given here.

Source of Sum of Degrees of Mean Computed

Variation Squares Freedom Square f P-value

A

B

C

AB

AC

BC

ABC

Error

1.90591

0.02210

38.93402

0.88632

0.53594

0.45435

0.42421

2.45460

3

1

1

3

3

1

3

32

0.63530

0.02212

38.93402

0.29544

0.17865

0.45435

0.14140

0.07671

8.28

0.29

507.57

3.85

2.33

5.92

1.84

0.0003

0.5951

< 0.0001

0.0185

0.0931

0.0207

0.1592

Total 45.61745 47

Solutions for Exercises in Chapter 14 225

(a) Two-way interactions of AB and BC are all significant and main effect of A and

C are all significant. The insignificance of the main effect B may not be valid due

to the significant BC interaction.

(b) Based on the P-values, Duncan’s tests and the interaction means, the most important

factor is C and using C = 2 is the most important way to increase percent

weight. Also, using factor A at level 1 is the best.

14.19 The ANOVA table shows:

Source of Sum of Degrees of Mean Computed

Variation Squares Freedom Square f P-value

A

B

C

AB

AC

BC

ABC

Error

0.16617

0.07825

0.01947

0.12845

0.06280

0.12644

0.14224

0.47323

2

2

2

4

4

4

8

81

0.08308

0.03913

0.00973

0.03211

0.01570

0.03161

0.01765

0.00584

14.22

6.70

1.67

5.50

2.69

5.41

3.02

< 0.0001

0.0020

0.1954

0.0006

0.0369

0.0007

0.0051

Total 1.19603 107

There is a significant three-way interaction by Temperature, Surface, and Hrc. A plot

of each Temperature is given to illustrate the interaction

1

1

1

0.55 0.60 0.65 0.70 0.75

Temperature=Low

Hrc

Gluing Power

2

2

2

3

3

3

20 40 60

Surface

123

123

1

1 1

0.45 0.50 0.55 0.60 0.65

Temperature=Medium

Hrc

Gluing Power 2

2

2

3

3

3

20 40 60

Surface

123

123

1

1

1

0.40 0.45 0.50 0.55 0.60

Temperature=High

Hrc

Gluing Power

2

2

2

3

3

3

20 40 60

Surface

123

123

226 Chapter 14 Factorial Experiments (Two or More Factors)

14.20 (a) yijk = μ + αi + βj + γk + (βγ)jk + ǫijk;

P

j

βj = 0,

P

k

γk = 0,

P

j

(βγ)jk = 0,

P

k

(βγ)jk = 0, and ǫijk ∼ n(x; 0, σ2).

(b) The P-value of the Method and Type of Gold interaction is 0.10587. Hence, the

interaction is at least marginally significant.

(c) The best method depends on the type of gold used.

The tests of the method effect for different type of gold yields the P-values as

0.9801 and 0.0099 for “Gold Foil” and “Goldent”, respectively. Hence, the methods

are significantly different for the “Goldent” type.

Here is an interaction plot.

1 1

1

600 650 700 750 800

Method

Hardness

2

2

2

1 2 3

Type

12

foil

goldent

It appears that when Type is “Goldent” and Method is 1, it yields the best

hardness.

14.21 (a) Yes, the P-values for Brand ∗ Type and Brand ∗ Temp are both < 0.0001.

(b) The main effect of Brand has a P-value < 0.0001. So, three brands averaged

across the other two factore are significantly different.

(c) Using brand Y , powdered detergent and hot water yields the highest percent

removal of dirt.

14.22 (a) Define A, B, and C as “Powder Temperature,” “Die Temperature,” and “Extrusion

Rate,” respectively. The ANOVA table shows:

Source of Sum of Degrees of Mean Computed

Variation Squares Freedom Square f P-value

A

B

C

AB

AC

Error

78.125

3570.125

2211.125

0.125

1.125

1.25

1

1

1

1

1

2

78.125

3570.125

2211.125

0.125

1.125

0.625

125.00

5712.20

3537.80

0.20

1.80

0.0079

0.0002

0.0003

0.6985

0.3118

Total 5861.875 7

Solutions for Exercises in Chapter 14 227

The ANOVA results only show that the main effects are all significant and no

two-way interactions are significant.

(b) The interaction plots are shown here.

1

1

100 110 120 130 140

Powder Temperature

Radius

2

2

150 190

DieTemp

12

220

250

1

1

110 120 130 140

Powder Temperature

Radius

2

2

150 190

Rate

12

12

24

(c) The interaction plots in part (b) are consistent with the findings in part (a) that

no two-way interactions present.

14.23 (a) The P-values of two-way interactions Time×Temperature, Time×Solvent,

Temperature × Solvent, and the P-value of the three-way interaction

Time×Temperature×Solvent are 0.1103, 0.1723, 0.8558, and 0.0140, respectively.

(b) The interaction plots for different levels of Solvent are given here.

1 1

1

91 92 93 94 95

Solvent=Ethanol

Time

Amount of Gel

2 2

2

4 8 16

Temp

12

80

120

1

1

1 91 92 93 94 95

Solvent=Toluene

Time

Amount of Gel

2

2

2

4 8 16

Temp

12

80

120

(c) A normal probability plot of the residuals is given and it appears that normality

assumption may not be valid.

−2 −1 0 1 2

−0.3 −0.2 −0.1 0.0 0.1 0.2 0.3

Normal Q−Q Plot

Theoretical Quantiles

Sample Quantiles

228 Chapter 14 Factorial Experiments (Two or More Factors)

14.24 (a) The two-way interaction plots are given here and they all show significant interactions.

1

1

1

0.50 0.55 0.60 0.65

Surface

Power

2

2

2

3 3

3

1 2 3

Temp

123

123

1

1

1

0.50 0.55 0.60 0.65

Hardness

Power

2 2

2

3

3

3

20 40 60

Temp

123

123

1

1

1

0.50 0.54 0.58 0.62

Hardness

Power

2

2

2

3

3

3

20 40 60

Surface

123

123

(b) The normal probability plot of the residuals is shown here and it appears that

normality is somewhat violated at the tails of the distribution.

−2 −1 0 1 2

−0.1 0.0 0.1 0.2

Normal Q−Q Plot

Theoretical Quantiles

Sample Quantiles

14.25 The ANOVA table is given.

Source of Sum of Degrees of Mean Computed

Variation Squares Freedom Square f P-value

Filters

Operators

Interaction

Error

4.63389

10.31778

1.65722

4.44000

2

3

6

24

2.31694

3.43926

0.27620

0.18500

8.39

12.45

1.49

0.0183

0.0055

0.2229

Total 21.04889 35

Solutions for Exercises in Chapter 14 229

Note the f values for the main effects are using the interaction term as the denominator.

(a) The hypotheses are

H0 : σ2

= 0,

H1 : σ2

6= 0.

Decision: Since P-value = 0.2229, the null hypothesis cannot be rejected. There

is no significant interaction variance component.

(b) The hypotheses are

H

′

0 : σ2

= 0. H

′′

0 : σ2

= 0.

H

′

1 : σ2

6= 0. H

′′

1 : σ2

6= 0.

Decisions: Based on the P-values of 0.0183, and 0.0055 for H′

0 and H′

1, respectively,

we reject both H′

0 and H′′

0 . Both σ2

and σ2

are significantly different from

zero.

(c) ˆσ2 = s2 = 0.185; ˆσ2

= 2.31691−0.185

12 = 0.17766, and ˆσ2

= 3.43926−0.185

9 = 0.35158.

14.26 ˆσ2

= (42.6289/2−14.8011/4

12 = 1.4678 Brand.

ˆσ2

= (299.3422/2−14.8011/4

12 = 12.1642 Time.

s2 = 0.9237.

14.27 The ANOVA table with expected mean squares is given here.

Source of Degrees of Mean (a) (b)

Variation Freedom Square Computed f Computed f

A

B

C

AB

AC

BC

ABC

Error

3

1

2

3

6

2

6

24

140

480

325

15

24

18

2

5

f1 = s2

1 /s2

5 = 5.83

f2 = s2

2 /s2

p1 = 78.82

f3 = s2

3 /s2

5 = 13.54

f4 = s2

4 /s2

p2 = 2.86

f5 = s2

5 /s2

p2 = 4.57

f6 = s2

6 /s2

p1 = 4.09

f7 = s2

7 /s2 = 0.40

f1 = s2

1 /s2

5 = 5.83

f2 = s2

2 /s2

6 = 26.67

f3 = s2

3 /s2

5 = 13.54

f4 = s2

4 /s2

7 = 7.50

f5 = s2

5 /s2

7 = 12.00

f6 = s2

6 /s2

7 = 9.00

f7 = s2

7 /s2 = 0.40

Total 47

In column (a) we have found the following main effects and interaction effects significant

using the pooled estimates: σ2

, σ2

, and σ2

.

s2

p1 = (12 + 120)/30 = 4.4 with 30 degrees of freedom.

s2

p2 = (12 + 120 + 36)/32 = 5.25 with 32 degrees of freedom.

s2

p3 = (12 + 120 + 36 + 45)/35 = 6.09 with 35 degrees of freedom.

In column (b) we have found the following main effect and interaction effect significant

when sums of squares of insignificant effects were not pooled: σ2

and σ2

.

230 Chapter 14 Factorial Experiments (Two or More Factors)

14.28

P4

i=1

γ2

k = 0.24 and φ =

q

(16)(0.24)

(3)(0.197) = 2.55. With α = 0.05, v1 = 2 and v2 = 39 we find

from A.16 that the power is approximately 0.975. Therefore, 2 observations for each

treatment combination are sufficient.

14.29 The power can be calculated as

1 − β = P

"

F(2, 6) > f0.05(2, 6)

σ2 + 3σ2

σ2 + 3σ2

+ 12σ2

#

= P

F(2, 6) >

(5.14)(0.2762)

2.3169

= P[F(2, 6) > 0.6127] = 0.57.

14.30 (a) A mixed model. Inspectors (αi in the model) are random effects. Inspection level

(βj in the model) is a fixed effect.

yijk = μ + αi + βj + (αβ)ij + ǫijk;

αi ∼ n(x; 0, σ2

), (αβ)ij ∼ n(x; 0, σ2

), ǫijk ∼ n(x; 0, σ2),

X

j

βj = 0.

(b) The hypotheses are

H0 : σ2

= 0. H0 : σ2

= 0. H0 : β1 = β2 = β3 = 0,

H1 : σ2

6= 0. H1 : σ2

6= 0 H1 : At least one βi’s is not 0.

f2,36 = 0.02 with P-value = 0.9806. There does not appear to be an effect due to

the inspector.

f4,36 = 0.04 with P-value = 0.9973. There does not appear to be an effect due to

the inspector by inspector level.

f2,4 = 54.77 with P-value = 0.0012. Mean inspection levels were significantly

different in determining failures per 1000 pieces.

14.31 (a) A mixed model.

(b) The ANOVA table is

Source of Sum of Degrees of Mean Computed

Variation Squares Freedom Square f P-value

Material

Brand

Material*Brand

Error

1.03488

0.60654

9, 70109

0.09820

2

2

4

9

0.51744

0.30327

0.17527

0.01091

47.42

1.73

16.06

< 0.0001

0.2875

0.0004

Total 2.44071 17

(c) No, the main effect of Brand is not significant. An interaction plot is given.

Solutions for Exercises in Chapter 14 231

1

1

1

4.6 4.8 5.0 5.2 5.4 5.6

Brand

Year

2 2

3 2

3

3

A B C

Material

123

AB

C

Although brand A has highest means in general, it is not always significant,

especially for Material 2.

14.32 (a) Operators (αi) and time of day (βj) are random effects.

yijk = μ + αi + βj + (αβ)ij + ǫijk;

αi ∼ n(x; 0, σ2

), βj ∼ n(x; 0, σ2

), (αβ)ij ∼ n(x; 0, σ2

), ǫijk ∼ n(x; 0, σ2).

(b) σ2

= σ2

= 0 (both estimates of the variance components were negative).

(c) The yield does not appear to depend on operator or time.

14.33 (a) A mixed model. Power setting (αi in the model) is a fixed effect. Cereal type (βj

in the model) is a random effect.

yijk = μ + αi + βj + (αβ)ij + ǫijk;

X

i

αi = 0, βj ∼ n(x; 0, σ2

), (αβ)ij ∼ n(x; 0, σ2

), ǫijk ∼ n(x; 0, σ2).

(b) No. f2,4 = 1.37 and P-value = 0.3524.

(c) No. The estimate of σ2

is negative.

14.34 (a) The ANOVA table is given.

Source of Sum of Degrees of Mean Computed

Variation Squares Freedom Square f P-value

Sweetener

Flour

Interaction

Error

0.00871

0.00184

0.01015

0.01600

3

1

3

16

0.00290

0.00184

0.00338

0.00100

2.90

1.84

3.38

0.0670

0.1941

0.0442

Total 0.03670 23

Sweetener factor is close to be significant, while the P-value of the Flour shows

insignificance. However, the interaction effects appear to be significant.

232 Chapter 14 Factorial Experiments (Two or More Factors)

(b) Since the interaction is significant with a P-value = 0.0442, testing the effect

of sweetener on the specific gravity of the cake samples by flour type we get

P-value = 0.0077 for “All Purpose” flour and P-value = 0.6059 for “Cake” flour.

We also have the interaction plot which shows that sweetener at 100% concentration

yielded a specific gravity significantly lower than the other concentrations

for all-purpose flour. For cake flour, however, there were no big differences in the

effect of sweetener concentration.

1

1

1

1

0.80 0.82 0.84 0.86 0.88

Sweetener

Attribute

2

2

2

2

0 50 75 100

Flour

12

all−pupose

cake

14.35 (a) The ANOVA table is given.

Source of Sum of Degrees of Mean Computed

Variation Squares Freedom Square f P-value

Sauce

Fish

Sauce*Fish

Error

1, 031.3603

16, 505.8640

724.6107

3, 381.1480

1

2

2

24

1, 031.3603

8, 252.9320

362.3053

140.8812

7.32

58.58

2.57

0.0123

< 0.0001

0.0973

Total 21, 642.9830 29

Interaction effect is not significant.

(b) Both P-values of Sauce and Fish Type are all small enough to call significance.

14.36 (a) The ANOVA table is given here.

Source of Sum of Degrees of Mean Computed

Variation Squares Freedom Square f P-value

Plastic Type

Humidity

Interaction

Error

142.6533

143.7413

133.9400

50.5950

2

3

6

12

71.3267

47.9138

22.3233

4.2163

16.92

11.36

5.29

0.0003

0.0008

0.0070

Total 470.9296 23

The interaction is significant.

(b) The SS for AB with only Plastic Type A and B is 24.8900 with 3 degrees of

freedom. Hence f = 24.8900/3

4.2163 = 1.97 with P-value = 0.1727. Hence, there is no

significant interaction when only A and B are considered.

Solutions for Exercises in Chapter 14 233

(c) The SS for the single-degree-of-freedom contrast is 143.0868. Hence f = 33.94

with P-value < 0.0001. Therefore, the contrast is significant.

(d) The SS for Humidity when only C is considered in Plastic Type is 2.10042. So,

f = 0.50 with P-value = 0.4938. Hence, Humidity effect is insignificant when

Type C is used.

14.37 (a) The ANOVA table is given here.

Source of Sum of Degrees of Mean Computed

Variation Squares Freedom Square f P-value

Environment

Stress

Interaction

Error

0.8624

40.8140

0.0326

0.6785

1

2

2

12

0.8624

20.4020

0.0163

0.0565

15.25

360.94

0.29

0.0021

< 0.0001

0.7547

Total 42.3875 17

The interaction is insignificant.

(b) The mean fatigue life for the two main effects are all significant.

14.38 The ANOVA table is given here.

Source of Sum of Degrees of Mean Computed

Variation Squares Freedom Square f P-value

Sweetener

Flour

Interaction

Error

1.26893

1.77127

0.14647

2.55547

3

1

3

16

0.42298

1.77127

0.04882

0.15972

2.65

11.09

0.31

0.0843

0.0042

0.8209

Total 5.74213 23

The interaction effect is insignificant. The main effect of Sweetener is somewhat insignificant,

since the P-value = 0.0843. The main effect of Flour is strongly significant.

14.39 The ANOVA table is given here.

Source of Sum of Degrees of Mean Computed

Variation Squares Freedom Square f P-value

A

B

C

AB

AC

Error

1133.5926

26896.2593

40.1482

216.5185

1.6296

2.2963

2.5926

1844.0000

2

2

2

4

4

4

8

27

566.7963

13448.1296

20.0741

54.1296

0.4074

0.5741

0.3241

68.2963

8.30

196.91

0.29

0.79

0.01

0.01

0.00

0.0016

< 0.0001

0.7477

0.5403

0.9999

0.9999

1.0000

Total 30137.0370 53

234 Chapter 14 Factorial Experiments (Two or More Factors)

All the two-way and three-way interactions are insignificant. In the main effects, only

A and B are significant.

14.40 (a) Treating Solvent as a class variable and Temperature and Time as continuous

variable, only three terms in the ANOVA model show significance. They are (1)

Intercept; (2) Coefficient for Temperature and (3) Coefficient for Time.

(b) Due to the factor that none of the interactions are significant, we can claim that

the models for ethanol and toluene are equivalent apart from the intercept.

(c) The three-way interaction in Exercise 14.23 was significant. However, the general

patterns of the gel generated are pretty similar for the two Solvent levels.

14.41 The ANOVA table is displayed.

Source of Sum of Degrees of Mean Computed

Variation Squares Freedom Square f P-value

Surface

Pressure

Interaction

Error

2.22111

39.10778

112.62222

565.72000

2

2

4

9

1.11056

19.55389

28.15556

62.85778

0.02

0.31

0.45

0.9825

0.7402

0.7718

Total 719.67111 17

All effects are insignificant.

14.42 (a) This is a two-factor fixed-effects model with interaction.

yijk = μ + αi + βj+)αβ)ij + ǫijk,

X

i

αi = 0,

X

j

βj = 0,

X

i

(αβ)ij =

X

j

(αβ)ij = 0, ǫijk ∼ n(x; 0, σ)

(b) The ANOVA table is displayed.

Source of Sum of Degrees of Mean Computed

Variation Squares Freedom Square f P-value

Time

Temperature

Interaction

Error

0.16668

0.27151

0.03209

0.01370

3

2

6

12

0.05556

0.13575

0.00535

0.00114

48.67

118.91

4.68

< 0.0001

< 0.0001

0.0111

Total 0.48398 23

The interaction is insignificant, while two main effects are significant.

(c) It appears that using a temperature of −20◦C with drying time of 2 hours would

speed up the process and still yield a flavorful coffee. It might be useful to try

some additional runs at this combination.

Solutions for Exercises in Chapter 14 235

14.43 (a) Since it is more reasonable to assume the data come from Poisson distribution,

it would be dangerous to use standard analysis of variance because the normality

assumption would be violated. It would be better to transform the data to get at

least stable variance.

(b) The ANOVA table is displayed.

Source of Sum of Degrees of Mean Computed

Variation Squares Freedom Square f P-value

Teller

Time

Interaction

Error

40.45833

97.33333

8.66667

25.50000

3

2

6

12

13.48611

48.66667

1.44444

2.12500

6.35

22.90

0.68

0.0080

< 0.0001

0.6694

Total 171.95833 23

The interaction effect is insignificant. Two main effects, Teller and Time, are all

significant.

(c) The ANOVA table using a squared-root transformation on the response is given.

Source of Sum of Degrees of Mean Computed

Variation Squares Freedom Square f P-value

Teller

Time

Interaction

Error

0.05254

1.32190

0.11502

0.35876

3

2

6

12

0.17514

0.66095

0.01917

0.02990

5.86

22.11

0.64

0.0106

< 0.0001

0.6965

Total 2.32110 23

Same conclusions as in (b) can be reached. To check on whether the assumption

of the standard analysis of variance is violated, residual analysis may used to do

diagnostics.

Chapter 15

2k Factorial Experiments and

Fractions

15.1 Either using Table 15.5 (e.g., SSA = (−41+51−57−63+67+54−76+73)2

24 = 2.6667) or running

an analysis of variance, we can get the Sums of Squares for all the factorial effects.

SSA = 2.6667, SSB = 170.6667, SSC = 104.1667, SS(AB) = 1.500.

SS(AC) = 42.6667, SS(BC) = 0.0000, SS(ABC) = 1.5000.

15.2 A simplified ANOVA table is given.

Source of Degrees of Computed

Variation Freedom f P-value

A

B

AB

C

AC

BC

ABC

Error

1

1

1

1

1

1

1

8

1294.65

43.56

20.88

116.49

16.21

0.00

289.23

< 0.0001

0.0002

0.0018

< 0.0001

0.0038

0.9668

< 0.0001

Total 15

All the main and interaction effects are significant, other than BC effect. However,

due to the significance of the 3-way interaction, the insignificance of BC effect cannot

be counted. Interaction plots are given.

237

238 Chapter 15 2k Factorial Experiments and Fractions

1

1

5 10 15 20

C=−1

A

y

2

2

−1 1

B

12

−1

1

1

1

6 8 10 12 14 16 18

C=1

A

y

2

2

−1 1

B

12

−1

1

15.3 The AD and BC interaction plots are printed here. The AD plot varies with levels of

C since the ACD interaction is significant, or with levels of B since ABD interaction

is significant.

1

1

26.5 27.0 27.5 28.0 28.5 29.0

AD Interaction

D

y

2

2

−1 1

A

12

−1

1

1

1

26.0 27.0 28.0 29.0

BC Interaction

C

y

2

2

−1 1

B

12

−1

1

15.4 The ANOVA table is displayed.

Source of Degrees of Computed

Variation Freedom f P-value

A

B

AB

C

AC

BC

ABC

D

AD

BD

ABD

CD

ACD

BCD

ABCD

Error

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

16

57.85

7.52

6.94

127.86

7.08

10.96

1.26

44.72

4.85

4.85

1.14

6.52

1.72

1.20

0.87

< 0.0001

0.0145

0.0180

< 0.0001

0.0171

0.0044

0.2787

< 0.0001

0.0427

0.0427

0.3017

0.0213

0.2085

0.2900

0.3651

Total 31

Solutions for Exercises in Chapter 15 239

All main effects and two-way interactions are significant, while all higher order interactions

are insignificant.

15.5 The ANOVA table is displayed.

Source of Degrees of Computed

Variation Freedom f P-value

A

B

C

D

AB

AC

AD

BC

BD

CD

Error

1

1

1

1

1

1

1

1

1

1

5

9.98

0.20

6.54

0.02

1.83

0.20

0.57

19.03

1.83

0.02

0.0251

0.6707

0.0508

0.8863

0.2338

0.6707

0.4859

0.0073

0.2338

0.8863

Total 15

One two-factor interaction BC, which is the interaction of Blade Speed and Condition

of Nitrogen, is significant. As of the main effects, Mixing time (A) and Nitrogen

Condition (C) are significant. Since BC is significant, the insignificant main effect B,

the Blade Speed, cannot be declared insignificant. Interaction plots for BC at different

levels of A are given here.

1

1

15.9 16.1 16.3 16.5

Speed and Nitrogen Interaction (Time=1)

Nitrogen Condition

y

2

2

1 2

Speed

12

1 1

15.6 15.7 15.8 15.9 16.0

Speed and Nitrogen Interaction (Time=2)

Nitrogen Condition

y

2

2

1 2

Speed

12

12

15.6 (a) The three effects are given as

wA =

301 + 304 − 269 − 292

4

= 11, wB =

301 + 269 − 304 − 292

4

= −6.5,

wAB =

301 − 304 − 269 + 292

4

= 5.

There are no clear interpretation at this time.

(b) The ANOVA table is displayed.

240 Chapter 15 2k Factorial Experiments and Fractions

Source of Degrees of Computed

Variation Freedom f P-value

Concentration

Feed Rate

Interaction

Error

1

1

1

4

35.85

12.52

7.41

0.0039

0.0241

0.0529

Total 7

The interaction between the Feed Rate and Concentration is closed to be significant

at 0.0529 level. An interaction plot is given here.

1

1

135 140 145 150

Feed Rate

y

2

2

−1 1

Concentration

12

−1

1

The mean viscosity does not change much at high level of concentration, while it

changes a lot at low concentration.

(c) Both main effects are significant. Averaged across Feed Rate a high concentration

of reagent yields significantly higher viscosity, and averaged across concentration

a low level of Feed Rate yields a higher level of viscosity.

15.7 Both AD and BC interaction plots are shown in Exercise 15.3. Here is the interaction

plot of AB.

1

1

26.5 27.0 27.5 28.0 28.5

AB Interaction

A

y

2

2

−1 1

B

12

−1

1

For AD, at the high level of A, Factor D essentially has no effect, but at the low level

of A, D has a strong positive effect. For BC, at the low level of B, Factor C has a

strong negative effect, but at the high level of B, the negative effect of C is not as

pronounced. For AB, at the high level of B, A clearly has no effect. At the low level

of B, A has a strong negative effect.

15.8 The two interaction plots are displayed.

Solutions for Exercises in Chapter 15 241

1

1

26 27 28 29 30

AD Interaction (B=−1)

A

y

2

2

−1 1

D

12

−1

1

1

1

25.0 25.5 26.0 26.5 27.0 27.5

AD Interaction (B=1)

A

y

2

2

−1 1

D

12

−1

1

It can be argued that when B = 1 that there is essentially no interaction between A

and D. Clearly when B = −1, the presence of a high level of D produces a strong

negative effect of Factor A on the response.

15.9 (a) The parameter estimates for x1, x2 and x1x2 are given as follows.

Variable Degrees of Freedom Estimate f P-value

x1

x2

x1x2

1

1

1

5.50

−3.25

2.50

5.99

−3.54

2.72

0.0039

0.0241

0.0529

(b) The coefficients of b1, b2, and b12 are wA/2, wB/2, and wAB/2, respectively.

(c) The P-values are matched exactly.

15.10 The effects are given here.

A B C D AB AC AD BC

−0.2625 −0.0375 0.2125 0.0125 −0.1125 0.0375 −0.0625 0.3625

BD CD ABC ABD ACD BCD ABCD

0.1125 0.0125 −0.1125 0.0375 −0.0625 0.1125 −0.0625

The normal probability plot of the effects is displayed.

−1 0 1

−0.2 0.0 0.1 0.2 0.3

Normal Q−Q Plot

Theoretical Quantiles

Sample Quantiles

A

AB ABC

AD ACD ABCD

B

D CD

AC ABD

BD BCD

C

BC

(a) It appears that all three- and four-factor interactions are not significant.

(b) From the plot, it appears that A and BC are significant and C is somewhat

significant.

242 Chapter 15 2k Factorial Experiments and Fractions

15.11 (a) The effects are given here and it appears that B, C, and AC are all important.

A B C AB AC BC ABC

−0.875 5.875 9.625 −3.375 −9.625 0.125 −1.125

(b) The ANOVA table is given.

Source of Degrees of Computed

Variation Freedom f P-value

A

B

AB

C

AC

BC

ABC

Error

1

1

1

1

1

1

1

8

0.11

4.79

12.86

1.58

12.86

0.00

0.18

0.7528

0.0600

0.0071

0.2440

0.0071

0.9640

0.6861

Total 15

1

1

30 35 40 45 50

AC Interaction

A

y

2

2

−1 1

C

12

−1

1

(c) Yes, they do agree.

(d) For the low level of Cutting Angle, C, Cutting Speed, A, has a positive effect on

the life of a machine tool. When the Cutting Angle is large, Cutting Speed has a

negative effect.

15.12 A is not orthogonal to BC, B is not orthogonal to AC, and C is not orthogonal to AB.

If we assume that interactions are negligible, we may use this experiment to estimate

the main effects. Using the data, the effects can be obtained as A: 1.5; B: −6.5;

C: 2.5. Hence Factor B, Tool Geometry, seems more significant than the other two

factors.

15.13 Here is the block arrangement.

Block Block Block

1 2 1 2 1 2

(1)

c

ab

abc

a

b

ac

bc

(1)

c

ab

abc

a

b

ac

bc

(1)

c

ab

abc

a

b

ac

bc

Replicate 1 Replicate 2 Replicate 3

AB Confounded AB Confounded AB Confounded

Solutions for Exercises in Chapter 15 243

Analysis of Variance

Source of Variation Degrees of Freedom

Blocks

A

B

C

AC

BC

ABC

Error

5

1

1

1

1

1

1

12

Total 23

15.14 (a) ABC is confounded with blocks in the first replication and ABCD is confounded

with blocks in second replication.

(b) Computing the sums of squares by the contrast method yields the following

ANOVA table.

Source of Degrees of Mean Computed

Variation Freedom Square f P-value

Blocks

A

B

C

D

AB

AC

BC

AD

BD

CD

ABC

ABD

ACD

BCD

ABCD

Error

3

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

13

2.32

2.00

0.50

4.50

8.00

0.50

0.32

0.50

0.72

0.32

0.18

1.16

0.32

0.02

0.18

0.53

0.60

3.34

0.83

7.51

13.36

0.83

0.53

0.83

1.20

0.53

0.30

1.93

0.53

0.03

0.30

0.88

0.0907

0.3775

0.0168

0.0029

0.3775

0.4778

0.3775

0.2928

0.4778

0.5928

0.1882

0.4778

0.8578

0.5928

0.3659

Total 31

Only the main effects C and D are significant.

15.15 L1 = γ1 + γ2 + γ3 and L2 = γ1 + γ2 + γ4. For treatment combination (1) we find

L1 (mod 2) = 0. For treatment combination a we find L1 (mod 2) = 1 and L2 (mod 2) =

1. After evaluating L1 and L2 for all 16 treatment combinations we obtain the following

blocking scheme:

244 Chapter 15 2k Factorial Experiments and Fractions

Block 1 Block 2 Block 3 Block 4

(1)

ab

acd

bcd

c

abc

ad

bd

d

ac

bc

abd

a

b

cd

abcd

L1 = 0 L1 = 1 L1 = 0 L1 = 1

L2 = 0 L2 = 0 L2 = 1 L2 = 1

Since (ABC)(ABD) = A2B2CD = CD (mod 2), then CD is the other effect confounded.

15.16 (a) L1 = γ1+γ2+γ4 +γ5, L2 = γ1 +γ5. We find that the following treatment combinations

are in the principal block (L1 = 0, L2 = 0): (1), c, ae, bd, ace, abde, abcde.

The other blocks are constructed by multiplying the treatment combinations in

the principal block modulo 2 by a, b, and ab, respectively, to give the following

blocking arrangement:

Block 1 Block 2 Block 3 Block 4

(1)

c

ae

bd

ace

bcd

abde

abcde

a

ac

e

abd

ce

abcd

bde

bcde

b

bc

abe

d

abce

cd

ade

acde

ab

abc

bce

ad

bce

acd

de

cde

(b) (ABDE)(AE) = BD (mod 2). Therefore BD is also confounded with days.

(c) Yates’ technique gives the following sums of squares for the main effects:

SSA = 21.9453, SSB = 40.2753, SSC = 2.4753,

SSD = 7.7028, SSE = 1.0878.

15.17 L1 = γ1 + γ2 + γ3, L2 = γ1 + γ2.

Block Block Block

1 2 1 2 1 2

abc

a

b

c

ab

ac

bc

(1)

abc

a

b

c

ab

ac

bc

(1)

(1)

c

ab

abc

a

b

ac

bc

Rep 1 Rep 2 Rep 3

ABC Confounded ABC Confounded AB Confounded

Solutions for Exercises in Chapter 15 245

For treatment combination (1) we find L1 (mod 2) = 0 and L2 (mod 2) = 0. For

treatment combination a we find L1 (mod 2) = 1 and L2 (mod 2) = 1. Replicate 1 and

Replicate 2 have L1 = 0 in one block and L1 = 1 in the other. Replicate 3 has L2 = 0

in one block and L2 = 1 in the other.

Analysis of Variance

Source of Variation Degrees of Freedom

Blocks

A

B

C

AB

AC

BC

ABC

Error

5

1

1

1

1′

1

1

1′

11

Total 23

Relative information on ABC = 1

3 and relative information on AB = 2

3 .

15.18 (a) The ANOVA table is shown here.

Source of Degrees of Mean Computed

Variation Freedom Square f P-value

Operators

A

B

C

D

Error

1

1

1

1

1

10

0.1225

4.4100

3.6100

9.9225

2.2500

2.8423

0.04

1.55

1.27

3.49

0.79

0.2413

0.2861

0.0912

0.3945

Total 15

None of the main effects is significant at 0.05 level.

(b) ABC is confounded with operators since all treatments with positive signs in the

ABC contrast are in one block and those with negative signs are in the other

block.

15.19 (a) One possible design would be:

Machine

1

2

3

4

(1)

a

c

d

ab

b

abc

abd

ce

ace

e

cde

abce

bce

abe

abcde

acd

cd

ad

ac

bde

abde

bcde

be

ade

de

acde

ae

bcd

abcd

bd

bc

(b) ABD, CDE, and ABCE.

246 Chapter 15 2k Factorial Experiments and Fractions

15.20 (a) ˆy = 43.9 + 1.625x1 − 8.625x2 + 0.375x3 + 9.125x1x2 + 0.625x1x3 + 0.875x2x3.

(b) The Lack-of-fit test results in a P-value of 0.0493. There are possible quadratic

terms missing in the model.

15.21 (a) The P-values of the regression coefficients are:

Parameter Intercept x1 x2 x3 x1x2 x1x3 x2x3 x1x2x3

P-value < 0.0001 0.5054 0.0772 0.0570 0.0125 0.0205 0.7984 0.6161

and s2 = 0.57487 with 4 degrees of freedom. So x2, x3, x1x2 and x1x3 are important

in the model.

(b) t = ¯yf−¯yC √s2(1/nf+1/nC)

= 52.075−49.275 √(0.57487)(1/8+1/4)

= 6.0306. Hence the P-value = 0.0038 for

testing quadratic curvature. It is significant.

(c) Need one additional design point different from the original ones.

15.22 (a) No.

(b) It could be as follows.

Machine

1

2

3

4

(1)

a

b

d

ad

e

abe

ade

bc

abc

c

bcd

abce

bce

ace

abcde

acd

cd

abcd

ac

abd

bd

ad

ab

cde

acde

bcde

ce

bde

abde

de

be

ADE, BCD and ABCE are confounded with blocks.

(c) Partial confounding.

15.23 To estimate the quadratic terms, it might be good to add points in the middle of the

edges. Hence (−1, 0), (0,−1), (1, 0), and (0, 1) might be added.

15.24 The alias for each effect is obtained by multiplying each effect by the defining contrast

and reducing the exponents modulo 2.

A ≡CDE, AB ≡BCDE, BD≡ABCE, B ≡ABCDE, AC ≡DE,

BE≡ABCD, C ≡ADE, AD≡CE, ABC≡BDE, D ≡ACE,

AE≡CD, ABD≡BCE, E ≡ACD, BC ≡ABDE, ABE≡BCD,

15.25 (a) With BCD as the defining contrast, we have L = γ2+γ3+γ4. The 1

2 fraction corresponding

to L = 0 (mod 2 is the principal block: {(1), a, bc, abc, bd, abd, cd, acd}.

(b) To obtain 2 blocks for the 1

2 fraction the interaction ABC is confounded using

L = γ1 + γ2 + γ3:

Solutions for Exercises in Chapter 15 247

Block 1 Block 2

(1)

bc

abd

acd

a

abc

bd

cd

(c) Using BCD as the defining contrast we have the following aliases:

A≡ABCD, AB≡ACD, B≡CD, AC≡ABD,

C≡BD, AD≡ABC, D≡BC.

Since AD and ABC are confounded with blocks there are only 2 degrees of freedom

for error from the unconfounded interactions.

Analysis of Variance

Source of Variation Degrees of Freedom

Blocks

A

B

C

D

Error

1

1

1

1

1

2

Total 7

15.26 With ABCD and BDEF as defining contrasts, we have

L1 = γ1 + γ2 + γ3 + γ4, L2 = γ2 + γ4 + γ5 + γ6.

The following treatment combinations give L1 = 0, L2 = 0 (mod 2) and thereby suffice

as the 1

4 fraction:

{(1), ac, bd, abcd, abe, bce, ade, abf, bcf, adf, cdf, ef, acef, bdef, abcdef}.

The third defining contrast is given by

(ABCD)(BDEF) = AB2CD2EF = ACEF (mod 2).

The effects that are aliased with the six main effects are:

A≡BCD ≡ABDEF≡CEF, B≡ACD ≡DEF≡ABCEF,

C≡ABD ≡BCDEF≡AEF, D≡ABC ≡BEF ≡ACDEF,

E≡ABCDE≡BDF ≡ACF, F ≡ABCDF≡BDE≡ACE.

248 Chapter 15 2k Factorial Experiments and Fractions

15.27 (a) With ABCE and ABDF, and hence (ABCE)(ABDF) = CDEF as the defining

contrasts, we have

L1 = γ1 + γ2 + γ3 + γ5, L2 = γ1 + γ2 + γ4 + γ6.

The principal block, for which L1 = 0, and L2 = 0, is as follows:

{(1), ab, acd, bcd, ce, abce, ade, bde, acf, bcf, df, abdf, aef, bef, cdef, abcdef}.

(b) The aliases for each effect are obtained by multiplying each effect by the three

defining contrasts and reducing the exponents modulo 2.

A ≡BCE ≡BDF ≡ACDEF, B ≡ACE ≡ADF ≡BCDEF,

C ≡ABE ≡ABCDF≡DEF, D ≡ABCDE≡ABF ≡CEF,

E ≡ABC ≡ABDEF≡CDF, F ≡ABCEF ≡ABD ≡CDE,

AB ≡CE ≡DF ≡ABCDEF, AC ≡BE ≡BCDF≡ADEF,

AD ≡BCDE≡BF ≡ACEF, AE ≡BC ≡BDEF≡ACDF,

AF ≡BCEF ≡BD ≡ACDE, CD ≡ABDE ≡ABCF ≡EF,

DE ≡ABCD≡ABEF ≡CF, BCD≡ADE ≡ACF ≡BEF,

DCF≡AEF ≡ACD ≡BDE, .

Since E and F do not interact and all three-factor and higher interactions are

negligible, we obtain the following ANOVA table:

Source of Variation Degrees of Freedom

A

B

C

D

E

F

AB

AC

AD

BC

BD

CD

Error

1

1

1

1

1

1

1

1

1

1

1

1

3

Total 15

15.28 The ANOVA table is shown here and the error term is computed by pooling all the

interaction effects. Factor E is the only significant effect, at level 0.05, although the

decision on factor G is marginal.

Solutions for Exercises in Chapter 15 249

Source of Degrees of Mean Computed

Variation Freedom Square f P-value

A

B

C

D

E

F

G

Error

1

1

1

1

1

1

1

8

1.44

4.00

9.00

5.76

16.00

3.24

12.96

2.97

0.48

1.35

3.03

1.94

5.39

1.09

4.36

0.5060

0.2793

0.1199

0.2012

0.0488

0.3268

0.0701

Total 15

15.29 All two-factor interactions are aliased with each other. So, assuming that two-factor

as well as higher order interactions are negligible, a test on the main effects is given in

the ANOVA table.

Source of Degrees of Mean Computed

Variation Freedom Square f P-value

A

B

C

D

Error

1

1

1

1

3

6.125

0.605

4.805

0.245

1.053

5.81

0.57

4.56

0.23

0.0949

0.5036

0.1223

0.6626

Total 7

Apparently no main effects is significant at level 0.05. Comparatively factors A and C

are more significant than the other two. Note that the degrees of freedom on the error

term is only 3, the test is not very powerful.

15.30 Two-factor interactions are aliased with each other. There are total 7 two-factor interactions

that can be estimated. Among those 7, we picked the three, which are AC,

AF, and BD, that have largest SS values and pool the other 2-way interactions to the

error term. An ANOVA can be obtained.

250 Chapter 15 2k Factorial Experiments and Fractions

Source of Degrees of Mean Computed

Variation Freedom Square f P-value

A

B

C

D

E

F

AC

AF

BD

Error

1

1

1

1

1

1

1

1

1

6

81.54

166.54

5.64

4.41

40.20

1678.54

978.75

625.00

429.53

219.18

0.37

0.76

0.03

0.02

0.18

7.66

4.47

2.85

1.96

0.5643

0.4169

0.8778

0.8918

0.6834

0.0325

0.0790

0.1423

0.2111

Total 15

Main effect F, the location of detection, appears to be the only significant effect. The

AC interaction, which is aliased with BE, has a P-value closed to 0.05.

15.31 To get all main effects and two-way interactions in the model, this is a saturated design,

with no degrees of freedom left for error. Hence, we first get all SS of these effects and

pick the 2-way interactions with large SS values, which are AD, AE, BD and BE.

An ANOVA table is obtained.

Source of Degrees of Mean Computed

Variation Freedom Square f P-value

A

B

C

D

E

AD

AE

BD

BE

Error

1

1

1

1

1

1

1

1

1

6

388, 129.00

277, 202.25

4, 692.25

9, 702.25

1, 806.25

1, 406.25

462.25

1, 156.25

961.00

108.25

3, 585.49

2, 560.76

43.35

89.63

16.69

12.99

4.27

10.68

8.88

< 0.0001

< 0.0001

0.0006

< 0.0001

0.0065

0.0113

0.0843

0.0171

0.0247

Total 15

All main effects, plus AD, BD and BE two-way interactions, are significant at 0.05

level.

15.32 Consider a 24 design with letters A, B, C, and D, with design points

{(1), a, b, c, d, ab, ac, ad, bc, bd, cd, abc, abd, acd, bcd, abcd}

. Using E = ABCD, we have the following design:

{e, a, b, c, d, abe, ace, ade, bce, bde, cde, abc, abd, acd, bcd, abcde}.

Solutions for Exercises in Chapter 15 251

15.33 Begin with a 23 with design points

{(1), a, b, c, ab, ac, bc, abc}.

Now, use the generator D = AB, E = AC, and F = BC. We have the following

result:

{def, af, be, cd, abd, ace, bcf, abcdef}.

15.34 We can use the D = AB, E = −AC and F = BC as generators and obtain the result:

{df, aef, b, cde, abde, ac, bcef, abcdf}.

15.35 Here are all the aliases

A≡BD≡CE≡CDF≡BEF ≡ ≡ABCF ≡ADEF ≡ABCDE;

B≡AD≡CF ≡CDE≡AEF ≡ ≡ABCE ≡BDEF ≡ABCDF;

C≡AE≡BF ≡BDE≡ADF ≡ ≡CDEF ≡ABCD ≡ABCEF;

D≡AB≡EF ≡BCE≡ACF ≡ ≡BCDF ≡ACDE ≡ABDEF;

E≡AC ≡DF≡ABF ≡BCD≡ ≡ABDE ≡BCEF ≡ACDEF;

F ≡BC≡DE≡ACD≡ABE≡ ≡ACEF ≡ABDF ≡BCDEF.

15.36 (a) The defining relation is ABC = −I.

(b) A = −BC, B = −AC, and C = −AB.

(c) The mean squares for A, B, and C are 1.50, 0.34, and 5.07, respectively. So,

factor C, the amount of grain refiner, appears to be most important.

(d) Low level of C.

(e) All at the “low” level.

(f) A hazard here is that the interactions may play significant roles. The following

are two interaction plots.

1

1

0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0

AB Interaction

A

y

2

2

−1 1

B

12

−1

1

1

1

0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0

BC Interaction

B

y

2

2

−1 1

C

12

−1

1

252 Chapter 15 2k Factorial Experiments and Fractions

15.37 When the variables are centered and scaled, the fitted model is

ˆy = 12.7519 + 4.7194x1 + 0.8656x2 − 1.4156x3.

The lack-of-fit test results in an f-value of 81.58 with P-value < 0.0001. Hence, higherorder

terms are needed in the model.

15.38 The ANOVA table for the regression model looks like the following.

Coefficients Degrees of Freedom

Intercept

β1

β2

β3

β4

β5

Two-factor interactions

Lack of fit

Pure error

1

1

1

1

1

1

10

16

32

Total 63

15.39 The defining contrasts are

AFG, CEFG, ACDF, BEG, BDFG, CDG, BCDE, ABCDEFG, DEF, ADEG.

15.40 Begin with the basic line for N = 24; permute as described in Section 15.12 until 18

columns are formed.

15.41 The fitted model is

ˆy =190, 056.67 + 181, 343.33x1 + 40, 395.00x2 + 16, 133.67x3 + 45, 593.67x4

− 29, 412.33x5 + 8, 405.00x6.

The t-tests are given as

Variable t P-value

Intercept

x1

x2

x3

x4

x5

x6

4.48

4.27

0.95

0.38

1.07

−0.69

0.20

0.0065

0.0079

0.3852

0.7196

0.3321

0.5194

0.8509

Only x1 and x2 are significant.

Solutions for Exercises in Chapter 15 253

15.42 An ANOVA table is obtained.

Source of Degrees of Mean Computed

Variation Freedom Square f P-value

Polymer 1

Polymer 2

Polymer 1*Polymer 2

Error

1

1

1

4

172.98

180.50

1.62

0.17

1048.36

1093.94

9.82

< 0.0001

< 0.0001

0.0351

Total 7

All main effects and interactions are significant.

15.43 An ANOVA table is obtained.

Source of Degrees of Mean Computed

Variation Freedom Square f P-value

Mode

Type

Mode*Type

Error

1

1

1

12

2, 054.36

4, 805.96

482.90

27.67

74.25

173.71

17.45

< 0.0001

< 0.0001

0.0013

Total 15

All main effects and interactions are significant.

15.44 Two factors at two levels each can be used with three replications of the experiment,

giving 12 observations. The requirement that there must be tests on main effects

and the interactions suggests that partial confounding be used The following design is

indicated:

Block Block Block

1 2 1 2 1 2

(1)

ab

a

b

a

ab

(1)

b

(1)

a

ab

a

Rep 1 Rep 2 Rep 3

15.45 Using the contrast method and compute sums of squares, we have

Source of Variation d.f. MS f

A

B

C

D

E

F

Error

1

1

1

1

1

1

8

0.0248

0.0322

0.0234

0.0676

0.0028

0.0006

0.0201

1.24

1.61

1.17

3.37

0.14

0.03

254 Chapter 15 2k Factorial Experiments and Fractions

15.46 With the defining contrasts ABCD, CDEFG, and BDF, we have

L1 = γ1 + γ2 + γ3 + γ4,

L2 = γ3 + γ4 + γ5 + γ6 + γ7.

L3 = γ2 + γ4 + γ6.

The principal block and the remaining 7 blocks are given by

Block 1 Block 2 Block 3 Block 4

(1), eg

abcd, bdg

adf, bcf

cdef, abcdeg

bde, adefg

bcefg, cdfg

acg, abef

ace, abfg

a, aeg

bcd, abdg

df, abcf

acdef, bcdeg

abde, defg

abcefg, acdfg

cg, bef

ce, bfg

b, beg

acd, dg

abdf, cf

bcdef, acdeg

de, abdefg

cefg, bcdfg

abcg, aef

abce, afg

c, ceg

abd, bcdg

acdf, bf

def, abdeg

bcde, acdefg

befg, dfg

ag, abcef

ae, abcfg

Block 5 Block 6 Block 7 Block 8

d, deg

abc, bg

af, bcdf

cef, abceg

be, aefg

bcdefg, cfg

acdg, abdef

acde, abdfg

e, g

abcde, bdeg

adef, bcef

cdf, abcdg

bd, adfg

bcfg, cdefg

aceg, abf

ac, abefg

f, efg

abcdf, bdfg

ad, bc

cde, abcdefg

bdef, adeg

bceg, cdg

acfg, abe

acef, abg

ab, abeg

cd, adg

bdf, acf

abcdef, cdeg

ade, bdefg

acefg, abcdfg

bcg, ef

bce, fg

The two-way interactions AB ≡ CD, AC ≡ BD, AD ≡ BC, BD ≡ F, BF ≡ D and

DF ≡ B.

15.47 A design (where L1 = L2 = L3 = L4 = 0 (mod 2) are used) is:

{(1), abcg, abdh, abef, acdf, aceh, adeg, afgh,

bcde, bcfh, bdfg, cdgh, cefg, defh, degh, abcdefgh}

15.48 In the four defining contrasts, BCDE, ACDF, ABCG, and ABDH, the length of

interactions are all 4. Hence, it must be a resolution IV design.

15.49 Assuming three factors the design is a 23 design with 4 center runs.

15.50 (a) Consider a 23−1

III design with ABC ≡ I as defining contrast. Then the design

points are

Solutions for Exercises in Chapter 15 255

x1 x2 x3

−1

1

−1

1

0

0

−1

−1

1

1

0

0

−1

1

1

−1

0

0

For the noncentral design points, ¯x1 = ¯x2 = ¯x3 = 0 and ¯x2

1 = ¯x2

2 = ¯x2

3 = 1. Hence

E(¯yf −¯y0) = β0+β1¯x1+β2¯x2+β3¯x3+β11¯x2

1 +β22¯x2

2 +β33¯x2

3 −β0 = β11+β22+β33.

(b) It is learned that the test for curvature that involves ¯yf − ¯y0 actually is testing

the hypothesis β11 + β22 + β33 = 0.

Chapter 16

Nonparametric Statistics

16.1 The hypotheses

H0 : ˜μ = 20 minutes

H1 : ˜μ > 20 minutes.

α = 0.05.

Test statistic: binomial variable X with p = 1/2.

Computations: Subtracting 20 from each observation and discarding the zeroes. We

obtain the signs

− + + − + + − + + +

for which n = 10 and x = 7. Therefore, the P-value is

P = P(X ≥ 7 | p = 1/2) =

X10

x=7

b(x; 10, 1/2)

= 1 −

X6

x=0

b(x; 10, 1/2) = 1 − 0.8281 = 0.1719 > 0.05.

Decision: Do not reject H0.

16.2 The hypotheses

H0 : ˜μ = 12

H1 : ˜μ 6= 12.

α = 0.02.

Test statistic: binomial variable X with p = 1/2.

Computations: Replacing each value above and below 12 by the symbol “+” and “−”,

respectively, and discarding the two values which equal to 12. We obtain the sequence

− + − − + + + + − + + − + + − +

257

258 Chapter 16 Nonparametric Statistics

for which n = 16, x = 10 and n/2 = 8. Therefore, the P-value is

P = 2P(X ≥ 10 | p = 1/2) = 2

X16

x=10

b(x; 16, 1/2)

= 2(1 −

X9

x=0

b(x; 16, 1/2)) = 2(1 − 0.7728) = 0.4544 > 0.02.

Decision: Do not reject H0.

16.3 The hypotheses

H0 : ˜μ = 2.5

H1 : ˜μ 6= 2.5.

α = 0.05.

Test statistic: binomial variable X with p = 1/2.

Computations: Replacing each value above and below 2.5 by the symbol “+” and “−”,

respectively. We obtain the sequence

− − − − − − − + + − + − − − − −

for which n = 16, x = 3. Therefore, μ = np = (16)(0.5) = 8 and σ =

p

(16)(0.5)(0.5) =

2. Hence z = (3.5 − 8)/2 = −2.25, and then

P = 2P(X ≤ 3 | p = 1/2) ≈ 2P(Z < −2.25) = (2)(0.0122) = 0.0244 < 0.05.

Decision: Reject H0.

16.4 The hypotheses

H0 : ˜μ1 = ˜μ2

H1 : ˜μ1 < ˜μ2.

α = 0.05.

Test statistic: binomial variable X with p = 1/2.

Computations: After replacing each positive difference by a “+” symbol and negative

difference by a “−” symbol, respectively, and discarding the two zero differences, we

have n = 10 and x = 2. Therefore, the P-value is

P = P(X ≤ 2 | p = 1/2) =

X2

x=0

b(x; 10, 1/2) = 0.0547 > 0.05.

Decision: Do not reject H0.

Solutions for Exercises in Chapter 16 259

16.5 The hypotheses

H0 : ˜μ1 − ˜μ2 = 4.5

H1 : ˜μ1 − ˜μ2 < 4.5.

α = 0.05.

Test statistic: binomial variable X with p = 1/2.

Computations: We have n = 10 and x = 4 plus signs. Therefore, the P-value is

P = P(X ≤ 4 | p = 1/2) =

X4

x=0

b(x; 10, 1/2) = 0.3770 > 0.05.

Decision: Do not reject H0.

16.6 The hypotheses

H0 : ˜μA = ˜μB

H1 : ˜μA 6= ˜μB.

α = 0.05.

Test statistic: binomial variable X with p = 1/2.

Computations: We have n = 14 and x = 12. Therefore, μ = np = (14)(1/2) = 7 and

σ =

p

(14)(1/2)(1/2) = 1.8708. Hence, z = (11.5 − 7)/1.8708 = 2.41, and then

P = 2P(X ≥ 12 | p = 1/2) = 2P(Z > 2.41) = (2)(0.0080) = 0.0160 < 0.05.

Decision: Reject H0.

16.7 The hypotheses

H0 : ˜μ2 − ˜μ1 = 8

H1 : ˜μ2 − ˜μ1 < 8.

α = 0.05.

Test statistic: binomial variable X with p = 1/2.

Computations: We have n = 13 and x = 4. Therefore, μ = np = (13)(1/2) = 6.5 and

σ =

p

(13)(1/2)(1/2) = 1.803. Hence, z = (4.5 − 6.5)/1.803 = −1.11, and then

P = P(X ≥ 4 | p = 1/2) = P(Z < −1.11) = 0.1335 > 0.05.

Decision: Do not reject H0.

16.8 The hypotheses

H0 : ˜μ = 20

H1 : ˜μ > 20.

α = 0.05.

Critical region: w≤11 for n = 10.

Computations:

260 Chapter 16 Nonparametric Statistics

di −3 12 5 −5 8 5 −8 15 6 4

Rank 1 9 4 4 7.5 4 7.5 10 6 2

Therefore, w=12.5.

Decision: Do not reject H0.

16.9 The hypotheses

H0 : ˜μ = 12

H1 : ˜μ 6= 12.

α = 0.02.

Critical region: w≤20 for n = 15.

Computations:

di −3 1 −2 −1 6 4 1 2 −1 3 −3 1 2 −1 2

Rank 12 3.5 8.5 3.5 15 14 3.5 8.5 3.5 12 12 3.5 8.5 3.5 8.5

Now, w=43 and w+ = 77, so that w = 43.

Decision: Do not reject H0.

16.10 The hypotheses

H0 : ˜μ1 − ˜μ2 = 0

H1 : ˜μ1 − ˜μ2 < 0.

α = 0.02.

Critiral region: w+ ≤ 1 for n = 5.

Computations:

Pair 1 2 3 4 5

di −5 −2 1 −4 2

Rank 5 2.5 1 4 2.5

Therefore, w+ = 3.5.

Decision: Do not reject H0.

16.11 The hypotheses

H0 : ˜μ1 − ˜μ2 = 4.5

H1 : ˜μ1 − ˜μ2 < 4.5.

α = 0.05.

Critiral region: w+ ≤ 11.

Computations:

Solutions for Exercises in Chapter 16 261

Woman 1 2 3 4 5 6 7 8 9 10

di −1.5 5.4 3.6 6.9 5.5 2.7 2.3 3.4 5.9 0.7

di − d0 −6.0 0.9 −0.9 2.4 1.0 −1.8 −2.2 −1.1 1.4 −3.8

Rank 10 1.5 1.5 8 3 6 7 4 5 9

Therefore, w+ = 17.5.

Decision: Do not reject H0.

16.12 The hypotheses

H0 : ˜μA − ˜μB = 0

H1 : ˜μA − ˜μB > 0.

α = 0.01.

Critiral region: z > 2.575.

Computations:

Day 1 2 3 4 5 6 7 8 9 10

di 2 6 3 5 8 −3 8 1 6 −3

Rank 4 15.5 7.5 13 19.5 7.5 19.5 1.5 15.5 7.5

Day 11 12 13 14 15 16 17 18 19 20

di 4 6 6 2 −4 3 7 1 −2 4

Rank 11 15.5 15.5 4 11 7.5 18 1.5 4 11

Now w = 180, n = 20, μW+ = (20)(21)/4 = 105, and σW+ =

p

(20)(21)(41)/24 =

26.786. Therefore, z = (180 − 105)/26.786 = 2.80

Decision: Reject H0; on average, Pharmacy A fills more prescriptions than Pharmacy

B.

16.13 The hypotheses

H0 : ˜μ1 − ˜μ2 = 8

H1 : ˜μ1 − ˜μ2 < 8.

α = 0.05.

Critiral region: z < −1.645.

Computations:

di 6 9 3 5 8 9 4 10

di − d0 −2 1 −5 −3 0 1 −4 2

Rank 4.5 1.5 10.5 7.5 − 1.5 9 4.5

di 8 2 6 3 1 6 8 11

di − d0 0 −6 −2 −5 −7 −2 0 3

Rank − 12 4.5 10.5 13 4.5 − 7.5

262 Chapter 16 Nonparametric Statistics

Discarding zero differences, we have w+ = 15, n = 13, μW+ = (13)(14)/4 = 45, 5, and

σW+ =

p

(13)(14)(27)/24 = 15.309. Therefore, z = (15 − 45.5)/14.309 = −2.13

Decision: Reject H0; the average increase is less than 8 points.

16.14 The hypotheses

H0 : ˜μA − ˜μB = 0

H1 : ˜μA − ˜μB 6= 0.

α = 0.05.

Critiral region: w ≤ 21 for n = 14.

Computations:

di 0.09 0.08 0.12 0.06 0.13 −0.06 0.12

Rank 7 5.5 10 2.5 12 2.5 10

di 0.11 0.12 −0.04 0.08 0.15 0.07 0.14

Rank 8 10 1 5.5 14 4 13

Hence, w+ = 101.5, w− = 3.5, so w = 3.5.

Decision: Reject H0; the different instruments lead to different results.

16.15 The hypotheses

H0 : ˜μB = ˜μA

H1 : ˜μB < ˜μA.

α = 0.05.

Critiral region: n1 = 3, n2 = 6 so u1 ≤ 2.

Computations:

Original data 1 7 8 9 10 11 12 13 14

Rank 1 2∗ 3∗ 4 5∗ 6 7 8 9

Now w1 = 10 and hence u1 = 10 − (3)(4)/2 = 4

Decision: Do not reject H0; the claim that the tar content of brand B cigarettes is

lower than that of brand A is not statistically supported.

16.16 The hypotheses

H0 : ˜μ1 = ˜μ2

H1 : ˜μ1 < ˜μ2.

α = 0.05.

Critiral region: u1 ≤ 2.

Computations:

Solutions for Exercises in Chapter 16 263

Original data 0.5 0.9 1.4 1.9 2.1 2.8 3.1 4.6 5.3

Rank 1∗ 2 3 4∗ 5 6∗ 7∗ 8 9

Now w1 = 18 and hence u1 = 18 − (4)(5)/2 = 8

Decision: Do not reject H0.

16.17 The hypotheses

H0 : ˜μA = ˜μB

H1 : ˜μA > ˜μB.

α = 0.01.

Critiral region: u2 ≤ 14.

Computations:

Original data 3.8 4.0 4.2 4.3 4.5 4.5 4.6 4.8 4.9

Rank 1∗ 2∗ 3∗ 4∗ 5.5∗ 5.5∗ 7 8∗ 9∗

Original Data 5.0 5.1 5.2 5.3 5.5 5.6 5.8 6.2 6.3

Rank 10 11 12 13 14 15 16 17 18

Now w2 = 50 and hence u2 = 50 − (9)(10)/2 = 5

Decision: Reject H0; calculator A operates longer.

16.18 The hypotheses

H0 : ˜μ1 = ˜μ2

H1 : ˜μ1 6= ˜μ2.

α = 0.01.

Critiral region: u ≤ 27.

Computations:

Original data 8.7 9.3 9.5 9.6 9.8 9.8 9.8 9.9 9.9 10.0

Rank 1∗ 2 3∗ 4 6∗ 6∗ 6∗ 8.5∗ 8.5 10

Original Data 10.1 10.4 10.5 10.7 10.8 10.9 11.0 11.2 11.5 11.8

Rank 11∗ 12 13∗ 14 15∗ 16 17∗ 18∗ 19 20

Here “∗” is for process 2. Now w1 = 111.5 for process 1 and w2 = 98.5 for process 2.

Therefore, u1 = 111.5 − (10)(11)/2 = 56.5 and u2 = 98.5 − (10)(11)/2 = 43.5, so that

u = 43.5.

Decision: Do not reject H0.

16.19 The hypotheses

H0 : ˜μ1 = ˜μ2

H1 : ˜μ1 6= ˜μ2.

264 Chapter 16 Nonparametric Statistics

α = 0.05.

Critiral region: u ≤ 5.

Computations:

Original data 64 67 69 75 78 79 80 82 87 88 91 93

Rank 1 2 3∗ 4 5∗ 6 7∗ 8 9∗ 10 11∗ 12

Now w1 = 35 and w2 = 43. Therefore, u1 = 35−(5)(6)/2 = 20 and u2 = 43−(7)(8)/2 =

15, so that u = 15.

Decision: Do not reject H0.

16.20 The hypotheses

H0 : ˜μ1 = ˜μ2

H1 : ˜μ1 6= ˜μ2.

α = 0.05.

Critiral region: Z < −1.96 or z > 1.96.

Computations:

Observation 12.7 13.2 13.6 13.6 14.1 14.1 14.5 14.8 15.0 15.0 15.4

Rank 1∗ 2 3.5∗ 3.5 5.5∗ 5.5 7 8 9.5∗ 9.5 11.5∗

Observation 15.4 15.6 15.9 15.9 16.3 16.3 16.3 16.3 16.5 16.8 17.2

Rank 11.5 13∗ 14.5∗ 14.5 17.5∗ 17.5∗ 17.5 17.5 20 21∗ 22

Observation 17.4 17.7 17.7 18.1 18.1 18.3 18.6 18.6 18.6 19.1 20.0

Rank 23 24.5 24.5∗ 26.5∗ 26.5 28 30 30∗ 30 32 33

Now w1 = 181.5 and u1 = 181.5 − (12)(13)/2 = 103.5. Then with μU1 = (21)(12)/2 =

126 and σU1 =

p

(21)(12)(34)/12 = 26.721, we find z = (103.5−126)/26.721 = −0.84.

Decision: Do not reject H0.

16.21 The hypotheses

H0 : Operating times for all three calculators are equal.

H1 : Operating times are not all equal.

α = 0.01.

Critiral region: h > χ2

0.01 = 9.210 with v = 2 degrees of freedom.

Computations:

Solutions for Exercises in Chapter 16 265

Ranks for Calculators

A B C

4 8.5 15

12 7 18

1 13 10

2 11 16

6 8.5 14

r1 = 25 5 17

3 r3 = 90

r2 = 56

Now h = 12

(18)(19)

h

252

5 + 562

7 + 902

6

i

− (3)(19) = 10.47.

Decision: Reject H0; the operating times for all three calculators are not equal.

16.22 Kruskal-Wallis test (Chi-square approximation)

h =

12

(32)(33)

210.52

9

+

1892

8

+

128.52

15

− (3)(33) = 20.21.

χ2

0.05 = 5.991 with 2 degrees of freedom. So, we reject H0 and claim that the mean

sorptions are not the same for all three solvents.

16.23 The hypotheses

H0 : Sample is random.

H1 : Sample is not random.

α = 0.1.

Test statistics: V , the total number of runs.

Computations: for the given sequence we obtain n1 = 5, n2 = 10, and v = 7. Therefore,

from Table A.18, the P-value is

P = 2P(V ≤ 7 when H0 is true) = (2)(0.455) = 0.910 > 0.1

Decision: Do not reject H0; the sample is random.

16.24 The hypotheses

H0 : Fluctuations are random.

H1 : Fluctuations are not random.

α = 0.05.

Test statistics: V , the total number of runs.

Computations: for the given sequence we find ˜x = 0.021. Replacing each measurement

266 Chapter 16 Nonparametric Statistics

by the symbol “+” if it falls above 0.021 and by the symbol “−” if it falls below 0.021

and omitting the two measurements that equal 0.021, we obtain the sequence

− − − − − + + + + +

for which n1 = 5, n2 = 5, and v = 2. Therefore, the P-value is

P = 2P(V ≤ 2 when H0 is true) = (2)(0.008) = 0.016 < 0.05

Decision: Reject H0; the fluctuations are not random.

16.25 The hypotheses

H0 : μA = μB

H1 : μA > μB.

α = 0.01.

Test statistics: V , the total number of runs.

Computations: from Exercise 16.17 we can write the sequence

B B B B B B A B B A A B A A A A A A

for which n1 = 9, n2 = 9, and v = 6. Therefore, the P-value is

P = P(V ≤ 6 when H0 is true) = 0.044 > 0.01

Decision: Do not reject H0.

16.26 The hypotheses

H0 : Defectives occur at random.

H1 : Defectives do not occur at random.

α = 0.05.

Critical region: z < −1.96 or z > 1.96.

Computations: n1 = 11, n2 = 17, and v = 13. Therefore,

μV =

(2)(11)(17)

28

+ 1 = 14.357,

σ2

V =

(2)(11)(17)[(2)(11)(17)− 11 − 17]

(282)(27)

= 6.113,

and hence σV = 2.472. Finally,

z = (13 − 14.357)/2.472 = −0.55.

Decision: Do not reject H0.

Solutions for Exercises in Chapter 16 267

16.27 The hypotheses

H0 : Sample is random.

H1 : Sample is not random.

α = 0.05.

Critical region: z < −1.96 or z > 1.96.

Computations: we find ¯x = 2.15. Assigning “+” and “−” signs for observations above

and below the median, respectively, we obtain n1 = 15, n2 = 15, and v = 19. Hence,

μV =

(2)(15)(15)

30

+ 1 = 16,

σ2

V =

(2)(15)(15)[(2)(15)(15)− 15 − 15]

(302)(29)

= 7.241,

which yields σV = 2.691. Therefore,

z = (19 − 16)/2.691 = 1.11.

Decision: Do not reject H0.

16.28 1 − γ = 0.95, 1 − α = 0.85. From Table A.20, n = 30.

16.29 n = 24, 1 − α = 0.90. From Table A.20, 1 − γ = 0.70.

16.30 1 − γ = 0.99, 1 − α = 0.80. From Table A.21, n = 21.

16.31 n = 135, 1 − α = 0.95. From Table A.21, 1 − γ = 0.995.

16.32 (a) Using the computations, we have

Student Test Exam di

L.S.A. 4 4 0

W.P.B. 10 2 8

R.W.K. 7 8 −1

J.R.L. 2 3 −1

J.K.L. 5 6.5 −1.5

D.L.P. 9 6.5 2.5

B.L.P. 3 10 −7

D.W.M. 1 1 0

M.N.M. 8 9 −1

R.H.S. 6 5 −1

rS = 1 −

(6)(125.5)

(10)(100 − 1)

= 0.24.

268 Chapter 16 Nonparametric Statistics

(b) The hypotheses

H0 : ρ = 0

H1 : ρ > 0

α = 0.025.

Critical region: rS > 0.648.

Decision: Do not reject H0.

16.33 (a) Using the following

Ranks Ranks

x y d x y d

1 6 −5 14 12 2

2 1 1 15 2 13

3 16 −13 16 6 10

4 9.5 −5.5 17 13.5 3.5

5 18.5 −13.5 18 13.5 4.5

6 23 −17 19 16 3

7 8 −1 20 23 −3

8 3 5 21 23 −2

9 9.5 −0.5 22 23 −1

10 16 −6 23 18.5 4.5

11 4 7 24 23 1

12 20 −8 25 6 19

13 11 2

we obtain rS = 1 − (6)(1586.5)

(25)(625−1) = 0.39.

(b) The hypotheses

H0 : ρ = 0

H1 : ρ 6= 0

α = 0.05.

Critical region: rS < −0.400 or rs > 0.400.

Decision: Do not reject H0.

16.34 The numbers come up as follows

Ranks Ranks Ranks

x y d x y d x y d

3 7 −4 4 6 −2 7 3 4

6 4.5 1.5 8 2 6 5 4.5 0.5

2 8 −6 1 9 −8 9 1 8

Solutions for Exercises in Chapter 16 269

X

d2 = 238.5, rS = 1 −

(6)(238.5)

(9)(80)

= −0.99.

16.35 (a) We have the following table:

Weight Chest Size di Weight Chest Size di Weight Chest Size di

3 6 −3 1 1 0 8 8 0

9 9 0 4 2 2 7 3 4

2 4 −2 6 7 −1 5 5 0

rS = 1 −

(6)(34)

(9)(80)

= 0.72.

(b) The hypotheses

H0 : ρ = 0

H1 : ρ > 0

α = 0.025.

Critical region: rS > 0.683.

Decision: Reject H0 and claim ρ > 0.

16.36 The hypotheses

H0 : ρ = 0

H1 : ρ 6= 0

α = 0.05.

Critical region: rS < −0.683 or rS > 0.683.

Computations:

Manufacture A B C D E F G H I

Panel rating 6 9 2 8 5 1 7 4 3

Price rank 5 1 9 8 6 7 2 4 3

di 1 8 −7 0 −1 −6 5 0 0

Therefore, rS = 1 − (6)(176)

(9)(80) = −0.47.

Decision: Do not reject H0.

16.37 (a)

P

d2 = 24, rS = 1 − (6)(24)

(8)(63) = 0.71.

(b) The hypotheses

H0 : ρ = 0

H1 : ρ > 0

270 Chapter 16 Nonparametric Statistics

α = 0.05.

Critical region: rS > 0.643.

Computations: rS = 0.71.

Decision: Reject H0, ρ > 0.

16.38 (a)

P

d2 = 1828, rS = 1 − (6)(1828)

(30)(899) = 0.59.

(b) The hypotheses

H0 : ρ = 0

H1 : ρ 6= 0

α = 0.05.

Critical region: rS < −0.364 or rS > 0.364.

Computations: rS = 0.59.

Decision: Reject H0, ρ 6= 0.

16.39 (a) The hypotheses

H0 : μA = μB

H1 : μA 6= μB

Test statistic: binomial variable X with p = 1/2.

Computations: n = 9, omitting the identical pair, so x = 3 and P-value is

P = P(X ≤ 3) = 0.2539.

Decision: Do not reject H0.

(b) w+ = 15.5, n = 9.

Decision: Do not reject H0.

16.40 The hypotheses:

H0 : μ1 = μ2 = μ3 = μ4.

H1 : At least two of the means are not equal.

α = 0.05.

Critical region: h > χ2

0.05 = 7.815 with 3 degrees of freedom.

Computaions:

Ranks for the Laboratories

A B C D

7 18 2 12

15.5 20 3 10.5

13.5 19 4 13.5

8 9 1 15.5

6 10.5 5 17

r1 = 50 r2 = 76.5 r3 = 15 r4 = 68.5

Solutions for Exercises in Chapter 16 271

Now

h =

12

(20)(21)

502 + 76.52 + 152 + 68.52

5

− (3)(21) = 12.83.

Decision: Reject H0.

16.41 The hypotheses:

H0 : μ29 = μ54 = μ84.

H1 : At least two of the means are not equal.

Kruskal-Wallis test (Chi-squared approximation)

h =

12

(12)(13)

62

3

+

382

5

+

342

4

− (3)(13) = 6.37,

with 2 degrees of freedom. χ2

0.05 = 5.991.

Decision: reject H0. Mean nitrogen loss is different for different levels of dietary protein.

Chapter 17

Statistical Quality Control

17.1 Let Y = X1 + X2 + · · · + Xn. The moment generating function of a Poisson random

variable is given by MX(t) = eμ(et−1). By Theorem 7.10,

MY (t) = eμ1(et−1) · eμ2(et−1) · · · eμn(et−1) = e(μ1+μ2+···+μn)(et−1),

which we recognize as the moment generating function of a Poisson random variable

with mean and variance given by

Pn

i=1

μi.

17.2 The charts are shown as follows.

2.385 0 5 10 15 20

2.390

2.395

2.400

2.405

2.410

2.415

2.420

UCL

LCL

Sample

X−bar

00 5 10 15 20

0.003

0.006

0.009

0.012

0.015 UCL

LCL

Sample

R

Although none of the points in R-chart is outside of the limits, there are many values

fall outside control limits in the ¯X -chart.

17.3 There are 10 values, out of 20, fall outside the specification ranges. So, 50% of the

units produced by this process will not confirm the specifications.

17.4 ¯¯X = 2.4037 and ˆσ = ¯R

d2

= 0.006935

2.326 = 0.00298.

17.5 Combining all 35 data values, we have

¯¯

x = 1508.491, ¯R = 11.057,

273

274 Chapter 17 Statistical Quality Control

so for ¯X -chart, LCL = 1508.491 − (0.577)(11.057) = 1502.111, and UCL = 1514.871;

and for R-chart, LCL = (11.057)(0) = 0, and UCL = (11.057)(2.114) = 23.374. Both

charts are given below.

1485 0 10 20 30

1490

1495

1500

1505

1510

1515

1520

1525

LCL

UCL

Sample

X

0 10 20 30

5

10

15

20

25

UCL

LCL = 0

Sample

Range

The process appears to be out of control.

17.6

β = P(Z < 3 − 1.5√5) − P(Z < −3 − 1.5√5)

= P(Z < −0.35) − P(Z < −6.35) ≈ 0.3632.

So,

E(S) = 1/(1 − 0.3632) = 1.57, and σS =

p

β(1 − β)2 = 0.896.

17.7 From Example 17.2, it is known than

LCL = 62.2740, and UCL = 62.3771,

for the ¯X -chart and

LCL = 0, and UCL = 0.0754,

for the S-chart. The charts are given below.

0 10 20 30

62.26

62.28

62.30

62.32

62.34

62.36

62.38

62.40

62.42

LCL

UCL

Sample number

X

0 10 20 30

0.01

0.03

0.05

0.07

0.09

UCL

LCL

Sample number

S

Solutions for Exercises in Chapter 17 275

The process appears to be out of control.

17.8 Based on the data, we obtain ˆp = 0.049, LCL = 0.049 − 3

q

(0.049)(0.951)

50 = −0.043, and

LCL = 0.049 + 3

q

(0.049)(0.951)

50 = 0.1406. Based on the chart shown below, it appears

that the process is in control.

00 5 10 15 20

0.03

0.06

0.09

0.12

0.15

UCL

LCL

Sample

p

17.9 The chart is given below.

00 5 10 15 20 25 30

0.03

0.06

0.09

0.12

0.15

UCL

LCL

Sample

p

Although there are a few points closed to the upper limit, the process appears to be

in control as well.

17.10 We use the Poisson distribution. The estimate of the parameter λ is ˆλ = 2.4. So, the

control limits are LCL = 2.4 − 3√2.4 = −2.25 and UCL = 2.4 + 3√2.4 = 7.048. The

control chart is shown below.

276 Chapter 17 Statistical Quality Control

00 5 10 15 20

1

2

3

4

5

6

7

8

LCL

UCL

Sample

Number of Defect

The process appears in control.

Chapter 18

Bayesian Statistics

18.1 For p = 0.1, b(2; 2, 0.1) =

2

2

(0.1)2 = 0.01.

For p = 0.2, b(2; 2, 0.2) =

2

2

(0.2)2 = 0.04. Denote by

A : number of defectives in our sample is 2;

B1 : proportion of defective is p = 0.1;

B2 : proportion of defective is p = 0.2.

Then

P(B1|A) =

(0.6)(0.01)

(0.6)(0.01) + (0.4)(0.04)

= 0.27,

and then by subtraction P(B2|A) = 1−0.27 = 0.73. Therefore, the posterior distribution

of p after observing A is

p 0.1 0.2

π(p|x = 2) 0.27 0.73

for which we get p∗ = (0.1)(0.27) + (0.2)(0.73) = 0.173.

18.2 (a) For p = 0.05, b(2; 9, 0.05) =

9

2

(0.05)2(0.95)7 = 0.0629.

For p = 0.10, b(2; 9, 0.10) =

9

2

(0.10)2(0.90)7 = 0.1722.

For p = 0.15, b(2; 9, 0.15) =

9

2

(0.15)2(0.85)7 = 0.2597.

Denote the following events:

A : 2 drinks overflow;

B1 : proportion of drinks overflowing is p = 0.05;

B2 : proportion of drinks overflowing is p = 0.10;

B3 : proportion of drinks overflowing is p = 0.15.

Then

P(B1|A) =

(0.3)(0.0629)

(0.3)(0.0629) + (0.5)(0.1722) + (0.2)(0.2597)

= 0.12,

P(B2|A) =

(0.5)(0.1722)

(0.3)(0.0629) + (0.5)(0.1722) + (0.2)(0.2597)

= 0.55,

277

278 Chapter 18 Bayesian Statistics

and P(B3|A) = 1 − 0.12 − 0.55 = 0.33. Hence the posterior distribution is

p 0.05 0.10 0.15

π(p|x = 2) 0.12 0.55 0.33

(b) p∗ = (0.05)(0.12) + (0.10)(0.55) + (0.15)(0.33) = 0.111.

18.3 (a) Let X = the number of drinks that overflow. Then

f(x|p) = b(x; 4, p) =

4

x

px(1 − p)4−x, for x = 0, 1, 2, 3, 4.

Since

f(1, p) = f(1|p)π(p) = 10

4

1

p(1 − p)3 = 40p(1 − p)3, for 0.05 < p < 0.15,

then

g(1) = 40

Z 0.15

0.05

p(1 − p)3 dp = −2(1 − p)4 (4p + 1)|0.15

0.05 = 0.2844,

and

π(p|x = 1) = 40p(1 − p)3/0.2844.

(b) The Bayes estimator

p∗ =

40

0.2844

Z 0.15

0.05

p2(1 − p)3 dp

=

40

(0.2844)(60)

p3 (20 − 45p + 36p2 − 10p3)

0

.

1

5

0.05 = 0.106.

18.4 Denote by

A : 12 condominiums sold are units;

B1 : proportion of two-bedroom condominiums sold 0.60;

B2 : proportion of two-bedroom condominiums sold 0.70.

For p = 0.6, b(12; 15, 0.6) = 0.0634 and for p = 0.7, b(12; 15, 0.7) = 0.1701. The prior

distribution is given by

p 0.6 0.7

π(p) 1/3 2/3

So, P(B1|A) = (1/3)(0.0634)

(1/3)(0.0634)+(2/3)(0.1701) = 0.157 and P(B2|A) = 1 − 0.157 = 0.843.

Therefore, the posterior distribution is

Solutions for Exercises in Chapter 18 279

p 0.6 0.7

π(p|x = 12) 0.157 0.843

(b) The Bayes estimator is p∗ = (0.6)(0.157) + (0.7)(0.843) = 0.614.

18.5 n = 10, ¯x = 9, σ = 0.8, μ0 = 8, σ0 = 0.2, and z0.025 = 1.96. So,

μ1 =

(10)(9)(0.04) + (8)(0.64)

(10)(0.04) + 0.64

= 8.3846, σ1 =

s

(0.04)(0.64)

(10)(0.04) + 0.64

= 0.1569.

To calculate Bayes interval, we use 8.3846 ± (1.96)(0.1569) = 8.3846 ± 0.3075 which

yields (8.0771, 8.6921). Hence, the probability that the population mean is between

8.0771 and 8.6921 is 95%.

18.6 n = 30, ¯x = 24.90, s = 2.10, μ0 = 30 and σ0 = 1.75.

(a) μ∗ = n¯x 2

0+μ0 2

n 2

0+ 2 = 2419.988

96.285 = 25.1336.

(b) σ∗ =

q

2

0 2

n 2

0+ 2 =

q

13.5056

96.285 = 0.3745, and z0.025 = 1.96. Hence, the 95% Bayes

interval is calculated by 25.13±(1.96)(0.3745) which yields $23.40 < μ < $25.86.

(c) P(24 < μ < 26) = P

24−25.13

0.3745 < Z < 26−25.13

0.3745

= P(−3.02 < Z < 2.32) =

0.9898 − 0.0013 = 0.9885.

18.7 (a) P(71.8 < μ < 73.4) = P

71.8−72 √5.76

< Z < 73.4−72 √5.76

= P(−0.08 < Z < 0.58) =

0.2509.

(b) n = 100, ¯x = 70, s2 = 64, μ0 = 72 and σ2

0 = 5.76. Hence,

μ1 =

(100)(70)(5.76) + (72)(64)

(100)(5.76) + 64

= 70.2,

σ1 =

s

(5.76)(64)

(100)(5.76) + 64

= 0.759.

Hence, the 95% Bayes interval can be calculated as 70.2 ± (1.96)(0.759) which

yields 68.71 < μ < 71.69.

(c) P(71.8 < μ < 73.4) = P

71.8−70.2

0.759 < Z < 73.4−70.2

0.759

= P(2.11 < Z < 4.22) =

0.0174.

18.8 Multiplying the likelihood function

f(x1, x2, . . . , xn|μ) =

1

(2π)25/210025 exp

"

−

1

2

X25

i=1

xi − μ

100

2

#

by the prior π(μ) = 1

60 for 770 < μ < 830, we obtain

f(x1, x2, . . . , xn, μ) =

1

(60)(2π)25/210025 exp

"

−

1

2

X25

i=1

xi − μ

100

2

#

= Ke−1

2 (μ−780

20 )2

,

280 Chapter 18 Bayesian Statistics

where K is a function of the sample values. Since the marginal distribution

g(x1, x2, . . . , xn) = √2π(20)K

1

√2π20

Z 830

770

e−1

2 ( μ−780

100 )2

dμ

= √2π(13.706)K.

Hence, the posterior distribution

π(μ|x1, x2, . . . , xn) =

f(x1, x2, . . . , xn)

g(x1, x2, . . . , xn)

=

1

√2π(13.706)

e−1

2 ( μ−780

20 )2

,

for 770 < μ < 830.

18.9 Multiplying the likelihood function and the prior distribution together, we get the joint

density function of θ as

f(t1, t2, . . . , tn, θ) = 2θn exp

"

−θ

Xn

i=1

ti + 2

!#

, for θ > 0.

Then the marginal distribution of (T1, T2, . . . , Tn) is

g(t1, t2, . . . , tn) = 2

Z

∞

0

θn exp

"

−θ

Xn

i=1

ti + 2

!#

dθ

=

2(n + 1)

Pn

i=1

ti + 2

n+1

Z

∞

0

θn exp

−θ

Pn

i=1

ti + 2

(n + 1)

Pn

i=1

ti + 2

−(n+1) dθ

=

2(n + 1)

Pn

i=1

ti + 2

n+1 ,

since the integrand in the last term constitutes a gamma density function with parameters

α = n + 1 and β = 1/

Pn

i=1

ti + 2

. Hence, the posterior distribution of θ

is

π(θ|t1, . . . , tn) =

f(t1, . . . , tn, θ)

g(t1, . . . , tn)

=

Pn

i=1

ti + 2

n+1

(n + 1)

θn exp

"

−θ

Xn

i=1

ti + 2

!#

,

for θ > 0, which is a gamma distribution with parameters α = n + 1 and β =

1/

Pn

i=1

ti + 2

.

Solutions for Exercises in Chapter 18 281

18.10 Assume that p(xi|λ) = e− xi

xi! , xi = 0, 1, . . . , for i = 1, 2, . . . , n and π(λ) = 1

24 λ2e− /2,

for λ > 0. The posterior distribution of λ is calculated as

π(λ|x1, . . . , xn) =

e−(n+1/2) λ

n P

i

=

1

xi+2

R

∞

0 e−(n+1/2) λ

n P

i

=

1

xi+2

dλ

=

(n + 1/2)n¯x+3

(n¯x + 3)

λ

n P

i

=

1

xi+2

e−(n+1/2) ,

which is a gamma distribution with parameters α = n¯x + 3 and β = (n + 1/2)−1,

with mean n¯x+3

n+1/2 . Hence, plug the data in we obtain the Bayes estimator of λ, under

squared-error loss, is λ∗ = 57+3

10+1/2 = 5.7143.

18.11 The likelihood function of p is

x−1

4

p5(1−p)x−5 and the prior distribution is π(p) = 1.

Hence the posterior distribution of p is

π(p|x) =

p5(1 − p)x−5

R 1

0 p5(1 − p)x−5 dp

=

(x + 2)

(6)(x − 4)

p5(1 − p)x−5,

which is a Beta distribution with parameters α = 6 and β = x − 4. Hence the Bayes

estimator, under the squared-error loss, is p∗ = 6

x+2 .