Solution Manual for Probability and Statistical Inference 8 Edition 8th Eighth PDF

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Contents Preface v 1 Probability 1 1.1 Basic Concepts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 1.2 Properties of Probability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 1.3 Methods of Enumeration . . . . . . . . . . . . . . . . . ....

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Contents Preface

v

1 Probability 1.1 Basic Concepts . . . . . 1.2 Properties of Probability 1.3 Methods of Enumeration 1.4 Conditional Probability 1.5 Independent Events . . 1.6 Bayes’s Theorem . . . .

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1 1 2 3 4 6 7

2 Discrete Distributions 2.1 Random Variables of the Discrete Type . . . . 2.2 Mathematical Expectation . . . . . . . . . . . . 2.3 The Mean, Variance, and Standard Deviation . 2.4 Bernoulli Trials and the Binomial Distribution 2.5 The Moment-Generating Function . . . . . . . 2.6 The Poisson Distribution . . . . . . . . . . . .

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11 11 15 16 19 22 24

3 Continuous Distributions 3.1 Continuous-Type Data . . . . . . . . . . . . 3.2 Exploratory Data Analysis . . . . . . . . . . 3.3 Random Variables of the Continuous Type . 3.4 The Uniform and Exponential Distributions 3.5 The Gamma and Chi-Square Distributions . 3.6 The Normal Distribution . . . . . . . . . . . 3.7 Additional Models . . . . . . . . . . . . . .

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27 27 30 37 45 48 50 54

4 Bivariate Distributions 4.1 Bivariate Distributions . . . . . . . 4.2 The Correlation Coefficient . . . . 4.3 Conditional Distributions . . . . . 4.4 The Bivariate Normal Distribution

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57 57 59 61 66

5 Distributions of Functions of Random Variables 5.1 Distributions of Functions of a Random Variable . . . . . 5.2 Transformations of Two Random Variables . . . . . . . . 5.3 Several Independent Random Variables . . . . . . . . . . 5.4 The Moment-Generating Function Technique . . . . . . . 5.5 Random Functions Associated with Normal Distributions 5.6 The Central Limit Theorem . . . . . . . . . . . . . . . . . 5.7 Approximations for Discrete Distributions . . . . . . . . .

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69 69 71 74 77 79 82 84

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iii

Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.

iv 6 Estimation 6.1 Point Estimation . . . . . . . . . . . . . . . . . . . 6.2 Confidence Intervals for Means . . . . . . . . . . . 6.3 Confidence Intervals For Difference of Two Means 6.4 Confidence Intervals For Variances . . . . . . . . . 6.5 Confidence Intervals For Proportions . . . . . . . . 6.6 Sample Size . . . . . . . . . . . . . . . . . . . . . . 6.7 A Simple Regression Problem . . . . . . . . . . . . 6.8 More Regression . . . . . . . . . . . . . . . . . . .

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89 . 89 . 92 . 93 . 95 . 97 . 98 . 99 . 105

7 Tests of Statistical Hypotheses 7.1 Tests about Proportions . . . . . . . . . . . . 7.2 Tests about One Mean and One Variance . . 7.3 Tests of the Equality of Two Means . . . . . 7.4 Tests for Variances . . . . . . . . . . . . . . . 7.5 One-Factor Analysis of Variance . . . . . . . 7.6 Two-Factor Analysis of Variance . . . . . . . 7.7 Tests Concerning Regression and Correlation

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113 113 115 118 121 122 125 126

8 Nonparametric Methods 8.1 Chi-Square Goodness of Fit Tests . . . . . . . . . . . 8.2 Contingency Tables . . . . . . . . . . . . . . . . . . . 8.3 Order Statistics . . . . . . . . . . . . . . . . . . . . . 8.4 Distribution-Free Confidence Intervals for Percentiles 8.5 The Wilcoxon Tests . . . . . . . . . . . . . . . . . . 8.6 Run Test and Test for Randomness . . . . . . . . . . 8.7 Kolmogorov-Smirnov Goodness of Fit Test . . . . . . 8.8 Resampling . . . . . . . . . . . . . . . . . . . . . . .

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129 129 133 134 136 138 142 145 147

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9 Bayesian Methods 155 9.1 Subjective Probability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 155 9.2 Bayesian Estimation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 156 9.3 More Bayesian Concepts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 157 10 Some Theory 10.1 Sufficient Statistics . . . . . . . . . . . . 10.2 Power of a Statistical Test . . . . . . . . 10.3 Best Critical Regions . . . . . . . . . . . 10.4 Likelihood Ratio Tests . . . . . . . . . . 10.5 Chebyshev’s Inequality and Convergence 10.6 Limiting Moment-Generating Functions 10.7 Asymptotic Distributions of Maximum Likelihood Estimators . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . in Probability . . . . . . . . .

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159 159 160 164 166 167 168

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11 Quality Improvement Through Statistical Methods 11.1 Time Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Statistical Quality Control . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.3 General Factorial and 2k Factorial Designs . . . . . . . . . . . . . . . . . . . . . . . . .

Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.

171 171 174 177

Preface This solutions manual provides answers for the even-numbered exercises in Probability and Statistical Inference, 8th edition, by Robert V. Hogg and Elliot A. Tanis. Complete solutions are given for most of these exercises. You, the instructor, may decide how many of these answers you want to make available to your students. Note that the answers for the odd-numbered exercises are given in the textbook. All of the figures in this manual were generated using Maple, a computer algebra system. Most of the figures were generated and many of the solutions, especially those involving data, were solved using procedures that were written by Zaven Karian from Denison University. We thank him for providing these. These procedures are available free of charge for your use. They are available on the CD-ROM in the textbook. Short descriptions of these procedures are provided on the “Maple Card” on the CD-ROM. Complete descriptions of these procedures are given in Probability and Statistics: Explorations with MAPLE, second edition, 1999, written by Zaven Karian and Elliot Tanis, published by Prentice Hall (ISBN 0-13-021536-8). Our hope is that this solutions manual will be helpful to each of you in your teaching. If you find an error or wish to make a suggestion, send these to Elliot Tanis at [email protected] and he will post corrections on his web page, http://www.math.hope.edu/tanis/. R.V.H. E.A.T.

v

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Chapter 1

Probability 1.1

Basic Concepts

1.1-2 (a) S = {bbb, gbb, bgb, bbg, bgg, gbg, ggb, ggg}; (b) S = {female, male};

(c) S = {000, 001, 002, 003, . . . , 999}.

1.1-4 (a)

Clutch size: Frequency:

(b)

h(x)

4 3

5 5

6 7

7 27

8 26

9 37

10 8

11 2

12 0

13 1

14 1

0.30 0.25 0.20 0.15 0.10 0.05 2

4

6

8

10

12

14

x

Figure 1.1–4: Clutch sizes for the common gallinule (c) 9.

1

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2

Chapter 1 1.1-6 (a)

No. Boxes: Frequency:

(b)

h(x) 0.20

4 10

5 19

6 13

7 8

8 13

9 7

10 9

11 5

12 2

13 4

14 4

15 2

16 2

19 1

24 1

0.18 0.16 0.14 0.12 0.10 0.08 0.06 0.04 0.02 2

4

6

8 10 12 14 16 18 20 22 24

x

Figure 1.1–6: Number of boxes of cereal

1.1-8 (a) f (1) =

2 3 3 2 , f (2) = , f (3) = , f (4) = . 10 10 10 10

1.1-10 This is an experiment. 1.1-12 (a) 50/204 = 0.245; 93/329 = 0.283; (b) 124/355 = 0.349; 21/58 = 0.362; (c) 174/559 = 0.311; 114/387 = 0.295; (d) Although James’ batting average is higher that Hrbek’s on both grass and artificial turf, Hrbek’s is higher over all. Note the different numbers of at bats on grass and artificial turf and how this affects the batting averages.

1.2

Properties of Probability

1.2-2 Sketch a figure and fill in the probabilities of each of the disjoint sets. Let A = {insure more than one car}, P (A) = 0.85. Let B = {insure a sports car}, P (B) = 0.23.

Let C = {insure exactly one car}, P (C) = 0.15.

It is also given that P (A ∩ B) = 0.17. Since P (A ∩ C) = 0, it follows that

P (A ∩ B ∩ C 0 ) = 0.17. Thus P (A0 ∩ B ∩ C 0 ) = 0.06 and P (A0 ∩ B 0 ∩ C) = 0.09.

1.2-4 (a) S = {HHHH, HHHT, HHTH, HTHH, THHH, HHTT, HTTH, TTHH, HTHT, THTH, THHT, HTTT, THTT, TTHT, TTTH, TTTT}; (b) (i) 5/16, (ii) 0, (iii) 11/16, (iv) 4/16, (v) 4/16, (vi) 9/16, (vii) 4/16. 1.2-6 (a) 1/6; (b) P (B) = 1 − P (B 0 ) = 1 − P (A) = 5/6; (c) P (A ∪ B) = P (S) = 1.

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3

Probability 1.2-8 (a) P (A ∪ B) = 0.4 + 0.5 − 0.3 = 0.6; (b)

A P (A) 0.4 P (A ∩ B)

= (A ∩ B 0 ) ∪ (A ∩ B) = P (A ∩ B 0 ) + P (A ∩ B) = P (A ∩ B 0 ) + 0.3 = 0.1;

(c) P (A0 ∪ B 0 ) = P [(A ∩ B)0 ] = 1 − P (A ∩ B) = 1 − 0.3 = 0.7. 1.2-10 Let A ={lab work done}, B ={referral to a specialist}, P (A) = 0.41, P (B) = 0.53, P ([A ∪ B]0 ) = 0.21. P (A ∪ B) = P (A) + P (B) − P (A ∩ B) 0.79 = 0.41 + 0.53 − P (A ∩ B) P (A ∩ B) = 0.41 + 0.53 − 0.79 = 0.15.

1.2-12

A∪B∪C = A ∪ (B ∪ C) P (A ∪ B ∪ C) = P (A) + P (B ∪ C) − P [A ∩ (B ∪ C)] = P (A) + P (B) + P (C) − P (B ∩ C) − P [(A ∩ B) ∪ (A ∩ C)] = P (A) + P (B) + P (C) − P (B ∩ C) − P (A ∩ B) − P (A ∩ C) + P (A ∩ B ∩ C).

1.2-14 (a) 1/3; (b) 2/3; (c) 0; (d) 1/2. 1.2-16 (a) S = {(1, 2), (1, 3), (1, 4), (1, 5), (2, 3), (2, 4), (2, 5), (3, 4), (3, 5), (4, 5)}; (b) (i) 1/10; (ii) 5/10. √ √ 2[r − r( 3/2)] 3 =1− . 1.2-18 P (A) = 2r 2

1.2-20 Note that the respective probabilities are p0 , p1 = p0 /4, p2 = p0 /42 , . . .. ∞ X p0 = 1 4k k=0

p0 1 − 1/4 p0

=

1

=

3 4

1 − p 0 − p1 = 1 −

1.3

1 15 = . 16 16

Methods of Enumeration

1.3-2 (4)(3)(2) = 24. 1.3-4 (a) (4)(5)(2) = 40; (b) (2)(2)(2) = 8. µ ¶ 6 1.3-6 (a) 4 = 80; 3 (b) 4(26 ) = 256; (c) 1.3-8

9 P4

(4 − 1 + 3)! = 20. (4 − 1)!3! =

9! = 3024. 5!

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4

Chapter 1 1.3-10 S ={ HHH, HHCH, HCHH, CHHH, HHCCH, HCHCH, CHHCH, HCCHH, CHCHH, CCHHH, CCC, CCHC, CHCC, HCCC, CCHHC, CHCHC, HCCHC, CHHCC, HCHCC, HHCCC } so there are 20 possibilities. 1.3-12 3 · 3 · 212 = 36, 864. ¶ ¶ µ µ n−1 n−1 + 1.3-14 r−1 r

1.3-16

0

=

2n

=

(n − r)(n − 1)! + r(n − 1)! n! = = = r!(n − r)! r!(n − r)! µ ¶ n µ ¶ n X X n r n r n−r n (−1) (1 − 1) = . (−1) (1) = r r r=0 r=0 (1 + 1)n =

n µ ¶ X n

r

r=0

1.3-18

µ

(n − 1)! (n − 1)! + r!(n − 1 − r)! (r − 1)!(n − r)!

=

n n1 , n 2 , . . . , n s

¶

(1)r (1)n−r =

n µ ¶ X n r=0

r

.

¶µ ¶µ ¶ µ ¶ n − n 1 n − n 1 − n2 n − n1 − · · · − ns−1 ··· n2 n3 ns

=

µ

=

n! (n − n1 )! · n1 !(n − n1 )! n2 !(n − n1 − n2 )!

n n1

· =

µ ¶ n . r

(n − n1 − n2 − · · · − ns−1 )! (n − n1 − n2 )! ··· n3 !(n − n1 − n2 − n3 )! ns !0!

n! . n1 !n2 ! . . . ns !

¶ 52 − 19 102, 486 6 µ ¶ 1.3-20 (a) = = 0.2917; 52 351, 325 9 µ ¶µ ¶µ ¶µ ¶µ ¶µ ¶µ ¶ 19 10 7 3 5 2 6 7, 695 1 0 1 0 2 2 3 µ ¶ = 0.00622. (b) = 52 1, 236, 664 9 µ ¶ 45 1.3-22 = 886,163,135. 36 µ

1.4

19 3

¶µ

Conditional Probability

1.4-2 (a)

1041 ; 1456

(b)

392 ; 633

(c)

649 . 823

(d) The proportion of women who favor a gun law is greater than the proportion of men who favor a gun law.

Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.

5

Probability 1.4-4 (a) P (HH) = (b) P (HC) =

13 12 1 · = ; 52 51 17 13 13 13 · = ; 52 51 204

(c) P (Non-Ace Heart, Ace) + P (Ace of Hearts, Non-Heart Ace) =

12 4 1 3 51 1 · + · = = . 52 51 52 51 52 · 51 52

1.4-6 Let A = {3 or 4 kings}, B = {2, 3, or 4 kings}. P (A|B)

=

=

P (A ∩ B) N (A) = P (B) N (B) µ ¶µ ¶ µ ¶µ ¶ 4 48 4 48 + 3 10 9 ¶µ ¶ µ ¶µ ¶ µ ¶µ 4¶ µ = 0.170. 4 48 4 48 4 48 + + 2 11 3 10 4 9

1.4-8 Let H ={died from heart disease}; P ={at least one parent had heart disease}. P (H | P 0 ) =

110 N (H ∩ P 0 ) = . 0 N (P ) 648

3 2 1 1 · · = ; 20 19 18 1140 µ ¶µ ¶ 3 17 1 1 2 1 µ ¶ · (b) = ; 20 17 760 3 ¶ µ ¶µ 17 3 9 X 35 1 2 2k − 2 µ ¶ (c) = = 0.4605. · 20 20 − 2k 76 k=1 2k

1.4-10 (a)

(d) Draw second. The probability of winning in 1 − 0.4605 = 0.5395. µ ¶µ ¶ µ ¶µ ¶ 2 8 2 8 1 2 1 1 4 0 5 1.4-12 µ ¶ · + µ ¶ · = . 10 10 5 5 5 5 5 52 51 50 49 48 47 8, 808, 975 · · · · · = = 0.74141; 52 52 52 52 52 52 11, 881, 376 (b) P (A0 ) = 1 − P (A) = 0.25859.

1.4-14 (a) P (A) =

1.4-16 (a) It doesn’t matter because P (B1 ) = (b) P (B) = 1.4-18

1 1 1 , P (B5 ) = , P (B18 ) = ; 18 18 18

1 2 = on each draw. 18 9

23 3 5 2 4 · + · = . 5 8 5 8 40

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6

Chapter 1 1.4-20 (a) P (A1 ) = 30/100; (b) P (A3 ∩ B2 ) = 9/100;

(c) P (A2 ∪ B3 ) = 41/100 + 28/100 − 9/100 = 60/100;

(d) P (A1 | B2 ) = 11/41; (e) P (B1 | A3 ) = 13/29.

1.5

Independent Events

1.5-2 (a)

P (A ∩ B) = P (A)P (B) = (0.3)(0.6) = 0.18; P (A ∪ B) = P (A) + P (B) − P (A ∩ B) = 0.3 + 0.6 − 0.18 = 0.72.

(b) P (A|B) =

P (A ∩ B) 0 = = 0. P (B) 0.6

1.5-4 Proof of (b): P (A0 ∩ B)

Proof of (c): P (A0 ∩ B 0 )

1.5-6

= P (B)P (A0 |B) = P (B)[1 − P (A|B)] = P (B)[1 − P (A)] = P (B)P (A0 ). = P [(A ∪ B)0 ] = 1 − P (A ∪ B) = 1 − P (A) − P (B) + P (A ∩ B) = 1 − P (A) − P (B) + P (A)P (B) = [1 − P (A)][1 − P (B)] = P (A0 )P (B 0 ).

P [A ∩ (B ∩ C)]

= P [A ∩ B ∩ C] = P (A)P (B)P (C) = P (A)P (B ∩ C).

P [A ∩ (B ∪ C)]

= P [(A ∩ B) ∪ (A ∩ C)] = P (A ∩ B) + P (A ∩ C) − P (A ∩ B ∩ C) = P (A)P (B) + P (A)P (C) − P (A)P (B)P (C) = P (A)[P (B) + P (C) − P (B ∩ C)] = P (A)P (B ∪ C).

P...