Solution Manual for Probability & Statistical Inference 10th Edition by Hogg, Tanis & Zimmerman

SOLUTION MANUAL

SOLUTION MANUAL

SOLUTION MANUAL

2023

COMPLETE Elaborated Test Bank for

Probability and Statistical Inference 10th Edition Robert Hogg Elliot Tanis Dale Zimmerman 2023

Probability and Statistical Inference 10th Edition by Robert Hogg, Elliot Tanis & Dale Zimmerman ISBN-10 1319247229 ISBN-13 978-1319247225 Solution Manual - Probability and Statistical Inference 10th Edition by Robert Hogg, Elliot Tanis & Dale Zimmerman - ISBN-10 1319247229 | ISBN-13 978-1319247225

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This is the authentic Solution Manual that is designed for your specific textbook with 9 Chapters.

Chapter:1. Probability Chapter:2. Discrete Distributions Chapter:3. Continuous Distributions Chapter:4. Bivariate Distributions Chapter:5. Distributions of Functions of Random Variables Chapter:6. Point Estimation Chapter:7. Interval Estimation Chapter:8. Tests of Statistical Hypotheses Chapter:9. More Tests 9.1 Chi-Square Goodness-of-Fit Tests 9.2 Contingency Tables 9.3 One-Factor Analysis of Variance 9.4 Two-Way Analysis of Variance 9.5 General Factorial and 2k Factorial Designs 9.6 Tests Concerning Regression and Correlation 9.7 Statistical Quality Control

Probability and Statistical Inference 10th Edition Solution Manual

Contents Preface

v

1 Probability 1 1.1 Properties of Probability ................................................................................................................ 1 1.2 Methods of Enumeration ............................................................................................................... 2 1.3 Conditional Probability ................................................................................................................ 3 1.4 Independent Events ....................................................................................................................... 4 1.5 Bayes’ Theorem ............................................................................................................................. 5 2 Discrete Distributions 7 2.1 Random Variables of the Discrete Type ....................................................................................... 7 2.2 Mathematical Expectation ............................................................................................................. 9 2.3 Special Mathematical Expectations ............................................................................................. 11 2.4 The Binomial Distribution ..........................................................................................................14 2.5 The Hypergeometric Distribution .............................................................................................16 2.6 The Negative Binomial Distribution ........................................................................................ 17 2.7 The Poisson Distribution ...........................................................................................................18 3 Continuous Distributions 19 3.1 Random Variables of the Continuous Type................................................................................ 19 3.2 The Exponential, Gamma, and Chi-Square Distributions ...................................................... 26 3.3 The Normal Distribution ............................................................................................................28 3.4 Additional Models ........................................................................................................................ 30 4 Bivariate Distributions 33 4.1 Bivariate Distributions of the Discrete Type .......................................................................... 33 4.2 The Correlation Coefficient.......................................................................................................... 34 4.3 Conditional Distributions ...........................................................................................................36 4.4 Bivariate Distributions of the Continuous Type ..................................................................... 37 4.5 The Bivariate Normal Distribution............................................................................................41 5 Distributions of Functions of Random Variables 45 5.1 Functions of One Random Variable ............................................................................................ 45 5.2 Transformations of Two Random Variables ............................................................................... 47 5.3 Several Random Variables ........................................................................................................... 52 5.4 The Moment-Generating Function Technique ........................................................................... 53 5.5 Random Functions Associated with Normal Distributions.................................................... 55 5.6 The Central Limit Theorem .......................................................................................................58 5.7 Approximations for Discrete Distributions .................................................................................. 59

5.8 5.9 iv

Chebyshev’s Inequality and Convergence in Probability ........................................................... 61 Limiting Moment-Generating Functions ..................................................................................... 62 Contents

6 Point Estimation 63 6.1 Descriptive Statistics.................................................................................................................. 63 6.2 Exploratory Data Analysis ......................................................................................................... 65 6.3 Order Statistics........................................................................................................................... 70 6.4 Maximum Likelihood Estimation.............................................................................................. 73 6.5 A Simple Regression Problem .................................................................................................. 76 6.6 Asymptotic Distributions of Maximum Likelihood Estimators ................................................................................................................ 81 6.7 Sufficient Statistics....................................................................................................................... 81 6.8 Bayesian Estimation .................................................................................................................. 84 7 Interval Estimation 87 7.1 Confidence Intervals for Means.................................................................................................. 87 7.2 Confidence Intervals for the Difference of Two Means ............................................................ 88 7.3 Confidence Intervals For Proportions ......................................................................................... 90 7.4 Sample Size.................................................................................................................................. 91 7.5 Distribution-Free Confidence Intervals for Percentiles .............................................................. 92 7.6 More Regression .......................................................................................................................... 93 7.7 Resampling Methods ................................................................................................................... 99 8 Tests of Statistical Hypotheses 107 8.1 Tests About One Mean ............................................................................................................. 107 8.2 Tests of the Equality of Two Means ........................................................................................ 109 8.3 Tests for Variances .................................................................................................................... 111 8.4 Tests about Proportions ............................................................................................................ 113 8.5 Some Distribution-Free Tests.................................................................................................... 114 8.6 Power of a Statistical Test ...................................................................................................... 118 8.7 Best Critical Regions ................................................................................................................ 121 8.8 Likelihood Ratio Tests ............................................................................................................. 124 9 More Tests 127 9.1 Chi-Square Goodness-of-Fit Tests ............................................................................................ 127 9.2 Contingency Tables ................................................................................................................... 130 9.3 One-Factor Analysis of Variance ............................................................................................... 131 9.4 Two-Way Analysis of Variance.................................................................................................. 134 9.5 General Factorial and 2k Factorial Designs .............................................................................. 135 9.6 Tests Concerning Regression and Correlation .......................................................................... 136 9.7 Statistical Quality Control ....................................................................................................... 137

Preface

v

Preface This solutions manual provides answers for the even-numbered exercises in Probability and Statistical Inference, tenth edition, by Robert V. Hogg, Elliot A. Tanis, and Dale L. Zimmerman. Complete solutions are given for most of these exercises. You, the instructor, may decide how many of these solutions and answers you want to make available to your students. Note that the answers for the odd-numbered exercises are given in the textbook. Our hope is that this solutions manual will be helpful to each of you in your teaching.

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 for down load at http://www.math.hope.edu/tanis/. Short descriptions of these procedures are provided on the “Maple Card.” 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). You can download a slightly revised edition of this manual at http://www.math.hope.edu/tanis/MapleManual.pdf.

We also want to acknowledge the many suggestions/corrections that were made by our accuracy checker, Kyle Siegrist.

If you find an error or wish to make a suggestion, please send them to dale-zimmerman@uiowa.edu.

These errata will be posted on http://homepage.divms.uiowa.edu/∼dzimmer/.

E.A.T. D.L.Z.

Chapter 1 Probability

Chapter 1

Probability 1.1

Properties of Probability

1.1-2 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 A ∩ C = φ, P (A ∩ C) = 0. It follows that P (A ∩ B ∩ C′) = 0.17. Thus P (A′ ∩ B ∩ C′) = 0.06 and P (B′ ∩ C) = 0.09.

1.1-4 (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.

1.1-6 (a) P (A ∪ B) = 0.5 + 0.6 − 0.4 = 0.7; (b)

A = (A ∩ B′) ∪ (A ∩ B) P (A) = P (A ∩ B′) + P (A ∩ B) 0.5 = P (A ∩ B′) + 0.4 ′ P (A ∩ B ) = 0.1;

(c) P (A′ ∪ B′) = P [(A ∩ B)′] = 1 − P (A ∩ B) = 1 − 0.4 = 0.6.

1.1-8 Let A ={lab work done}, B ={referral to a specialist}, P (A) = 0.41, P (B) = 0.53, P ([A ∪ B]′) = 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.

1.1-10

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).

1.1-12 (a) 1/3; (b) 2/3; (c) 0; (d) 1/2.

1

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2

Section 1.2 Methods of Enumeration √ √ 1.1-14 P (A) = 2[r − r( 3/2)] = 1 − 3 .

2r 2 1.1-16 Note that the respective probabilities are p0, p1 = p0/4, p2 = p0/42, · · ·. ∞ Σ p0 = 1 k 4 k=0 p0 = 1 1 − 1/4 3 p0 = 4 15 1 1 − p0 − p1 = 1 − = .

16 16

1.2

Methods of Enumeration

1.2-2 (a) (4)(5)(2) = 40; (b) (2)(2)(2) = 8.

1.2-4 (a) 4

6

= 80;

3 (b) 4(26) = 256; (c)

(4 − 1 + 3)!

(4 − 1)!3!

= 20.

1.2-6 S ={DDD, DDFD, DFDD, FDDD, DDFFD, DFDFD, FDDFD, DFFDD, FDFDD, FFDDD, FFF, FFDF, FDFF, DFFF FFDDF, FDFDF, DFFDF, FDDFF, DFDFF, DDFFF } so there are 20 possibilities. Note that the winning player (2 choices) must win the last set and two of the previous sets, so the number of outcomes is 2 3 4 2 + + = 20.

2 2 2 1.2-8 3 · 3 · 212 = 36,864.

1.2-10

n−1 + n−1 r r−1

(n − 1)!

=

r!(n − 1 − r)!

(n − r)(n − 1)! + r(n − 1)!

=

1.2-12

0

r!(n − r)!

n n Σ (−1)r(1)n−r = = (1 − 1) = r

n

(−1)r

n

r=0

2n = (1 + 1)n =

n n Σ r=0

1.2-14 5 − 1 + 29 29 1.2-16 (a)

(n − 1)! (r − 1)!(n − r)!

+

19 3

=

r

102,486 =

351,325

n Σ r=0

33! = 40,920.

29!4!

52 − 19 6 52 9

(1)r(1)n−r =

r=0

= 0.2917;

n . r

=

n! n = . r!(n − r)! r

n . r

Chapter 1 Probability 19 3

3 10 2

7 1

3 0

5 1

2 0

6 2

7,695

9 (b) = = 0.00622.

1,236,664 Σ 5 52 n 5 1.2-18 (a) P (A) = (1/2) = 1 − (1/2) ; n=1

(1/2)n = 1 − (1/2)10;

(b) P (B) = Σ1n0=1 (c) P (A ∪ B) = P (B) = 1 − (1/2)10; (d) P (A ∩ B) = P (A) = 1 − (1/2)5; (e) P (C) = P (B) − P (A) = (1/2)5 − (1/2)10; (f) P (B′) = 1 − P (B) = (1/2)10.

1.3

Conditional Probability 1041

1.3-2 (a)

1456 392 ; (b) 633 649 . (c) 823

;

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

13 12 1 · 1.3-4 (a) P (HH) = = ; 52 51 17 13 13 13 · (b) P (HC) = = ; 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 1.3-6 Let H ={died from heart disease}; P ={at least one parent had heart disease}.

P (H | P ′) =

N (H ∩ P ′) 110 = .

N (P ′) 648

1 3 2 1 = ; · 1.3-8 (a) 20 · 19 18 1140 3 2 (b)

17 1

20 3 9

(c) Σ

k=1

3 2

·

1 17

=

1 380

;

17 1 2k − 2 · = 35 = 0.4605; 20 20 − 2k 76 2k

4

Section 1.4 Independent Events (d) Draw second. The probability of winning is 1 − 0.4605 = 0.5395.

52 51 50 49 48 47 8,808,975 · · · · · = = 0.74141; 52 52 52 52 52 52 11,881,376 (b) P (A′) = 1 − P (A) = 0.25859.

1 1 1 1.3-12 (a) It doesn’t matter because P (B1) = , P (B5) = , P (B18) = ; 18 18 18 2 1 (b) P (B) = = on each draw.

18 9 1.3-14 (a) 5 · 4 · 3 = 60; (b) 5 · 5 · 5 = 125.

3 5 2 4 23 1.3-16 5 · 8 + 5 · 8 = 40 .

1.3-10 (a) P (A) =

1.4

Independent Events 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

1.4-2 (a)

1.4-4 Proof of (b): P (A′ ∩ B)

Proof of (c): P (A′ ∩ B′)

1.4-6

P [A ∩ (B ∩ C)]

= = = P [A ∩ (B ∪ C)] = = = = = P [A′ ∩ (B ∩ C′)]

= = = = = = = = =

= P (B)P (A′ |B) = P (B)[1 —P (A|B)] = P (B)[1 − P (A)] = P (B)P (A′). = = = = = =

P [(A ∪ B)′] 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 (A′)P (B′).

P [A ∩ B ∩ C] P (A)P (B)P (C) P (A)P (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 (A′ ∩ C′ ∩ B) P (B)[P (A′ ∩ C′) | B] P (B)[1 —P (A ∪ C |B)] P (B)[1 —P (A ∪ C)] P (B)P [(A ∪ C)′] P (B)P (A′ ∩ C′) P (B)P (A′)P (C′) P (A′)P (B)P (C′) P (A′)P (B ∩ C′).

Chapter 1 Probability

5

P [A′ ∩ B′ ∩ C′]

= P [(A ∪ B ∪ C)′] = 1 —P (A ∪ B ∪ C) = 1 — P (A) − P (B) − P (C) + P (A)P (B) + P (A)P (C)+ P (B)P (C) − PA)P (B)P (C) = [1 — P (A)][1 − P (B)][1 − P (C)] = P (A′)P (B′)P (C′).

1 2 3 1 4 3 5 2 3 2 · · + · · + · · = .

6 6 6 6 6 6 6 6 6 9 3 3 9 · = ; 1.4-10 (a) 4 4 16 1 3 3 2 9 (b) · + · = ; 4 4 4 4 16 2 1 2 4 10 · + · = . (c) 4 4 4 4 16 1.4-8

1.4-12 (a)

1

3

2 (b)

1 2

(c)

1 2 5!

(d)

1

2

;

2 3

3

3! 2!

1 2

2

; 2

1 ; 2 1 3 1 2

2

2

.

1.4-14 (a) 1 − (0.4)3 = 1 − 0.064 = 0.936; (b) 1 − (0.4)8 = 1 − 0.00065536 = 0.99934464. Σ ∞ 1 4 2k = 5 ; 1.4-16 (a) 5 5 9 k=0 1 4 3 1 4 3 2 1 1 3 (b) + · · + · · · · = .

5 5 4 3 5 4 3 2 1 5 1.4-18 (a) 7; (b) (1/2)7; (c) 63; (d) No! (1/2)63 = 1/9,223,372,036,854,775,808.

1.4-20 No. The equations that must hold are (1 − p1)(1 − p2) = p1(1 − p2) + p2(1 − p1) = p1p2.

There are no real solutions.

1.5

Bayes’ Theorem

1.5-2 (a)

(b)

= P (A ∩ G) + P (B ∩ G) = P (A)P (G |A) + P (B)P (G |B) = (0.40)(0.85) + (0.60)(0.75) = 0.79; P (A ∩ G) P (A | G) = P (G) (0.40)(0.85) = = 0.43.

0.79 P (G)

6

Section 1.5 Bayes’ Theorem 1.5-4 Let event B denote an accident and let A1 be the event that age of the driver is 16–25.

Then (0.1)(0.05) P (A1 | B) = (0.1)(0.05) + (0.55)(0.02) + (0.20)(0.03) + (0.15)(0.04) 50 50 = = = 0.179.

50 + 110 + 60 + 60 280 1.5-6 Let B be the event that the policyholder dies. Let A1, A2, A3 be the events that the deceased is standard, preferred and ultra-preferred, respectively. Then (0.60)(0.01) P (A1 | B) = (0.60)(0.01) + (0.30)(0.008) + (0.10)(0.007) 60 60 = = 0.659; = 60 + 24 + 7 91 24 = 0.264; P (A2 | B) = 91 7 = 0.077.

P (A3 | B) = 91 1.5-8 Let A be the event that the tablet is under warranty.

(0.40)(0.10) P (B1 | A) = (0.40)(0.10) + (0.30)(0.05) + (0.20)(0.03) + (0.10)(0.02) 40 40 = = 0.635; = 40 + 15 + 6 + 2 63 15 P (B2 | A) = = 0.238; 63 6 = 0.095; P (B3 | A) = 63 2 = 0.032.

P (B4 | A) = 63 1.5-10 (a) P (D+) = (0.02)(0.92) + (0.98)(0.05) = 0.0184 + 0.0490 = 0.0674; 0.0490 0.0184 (b) P (A− | D+) = = 0.727; P (A+ | D+) = = 0.273; 0.0674 0.0674 (0.98)(0.95) 9310 (c) P (A− | D−) = = = 0.998; (0.02)(0.08) + (0.98)(0.95) 16 + 9310 P (A+ | D−) = 0.002; (d) Yes, particularly those in part (b).

1.5-12 Let D = {defective roll}. Then P (I ∩ D) P (I | D) = P (D) P (I) · P (D | I) = P (I) · P (D | I) + P (II) · P (D | II) (0.60)(0.03) = (0.60)(0.03) + (0.40)(0.01) 0.018 0.018 = = = 0.818.

0.018 + 0.004 0.022

Chapter 2 Discrete Distributions

7

Chapter 2

Discrete Distributions 2.1

Random Variables of the Discrete Type

2.1-2 (a)

0.6, 0.3 , 0.1,

f(x) =

x = 1, x = 5, x = 10;

(b) f(x) 0.6 0.5 0.4 0.3 0.2 0.1

x 1

2

3

4

5

6

7

8

9

10

Figure 2.1–2: A probability histogram 9 Σ

2.1-4 (a)

log10

x=1

x+1 x

=

9 Σ

[log10(x + 1) − log10 x]

x=1

= log10 2 − log10 1 + log 10 3 − log10 2 + · · · + log 10 10 − log10 9 = log10 10 = 1; (b) F (x) =

0, log— n,≤ 10 1,

x < 1, n 1 9 ≤ x.

x < n,

n = 2, 3, . . . , 9,

8

Section 2.1 Random Variables of the Discrete Type 1 2.1-6 (a) f (x) = (b)

10

x = 0, 1, 2, · · · , 9;

,

N ({0})/150 = 11/150 = 0.073;

N ({5})/150 = 13/150 = 0.087;

N ({1})/150 = 14/150 = 0.093; N ({6})/150 = 22/150 = 0.147; N ({2})/150 = 13/150 = 0.087; N ({7})/150 = 16/150 = 0.107; N ({3})/150 = 12/150 = 0.080; N ({8})/150 = 18/150 = 0.120; N ({4})/150 = 16/150 = 0.107; N ({9})/150 = 15/150 = 0.100. (c)

f(x), h(x)

x 1

2

3

4

5

6

7

8

9

Figure 2.1–6: Michigan daily lottery digits

2.1-8 (a) f (x) =

6− |7 − x|

, x = 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12;

36 (b)

f(x) 0.16 0.14 0.12 0.10 0.08 0.06 0.04 0.02

x 1

2

3

4

5

6

7

8

9

10 11 12

Figure 2.1–8: Probability histogram for the sum of a pair of dice

Chapter 2 Discrete Distributions

9

2.1-10 (a) The space of W is S = {0, 1, 2, 3, 4, 5, 6, 7}.

1 1 1 P (W = 0) = P (X = 0, Y = 0) = · = , assuming independence.

2 4 8 1 1 1 P (W = 1) = P (X = 0, Y = 1) = · = , 2 4 8 1 1 1 P (W = 2) = P (X = 2, Y = 0) = · = , 2 4 8 1 1 1 P (W = 3) = P (X = 2, Y = 1) = · = , 2 4 8 1 1 1 P (W = 4) = P (X = 0, Y = 4) = · = , 2 4 8 1 1 1 P (W = 5) = P (X = 0, Y = 5) = · = , 2 4 8 1 1 1 P (W = 6) = P (X = 2, Y = 4) = · = , 2 4 8 1 1 1 P (W = 7) = P (X = 2, Y = 5) = · = .

2 4 8 1 That is, f (w) = P (W = w) = , w ∈ S. 8 (b) f (w)

w 1

2

3

4

5

6

7

Figure 2.1–10: Probability histogram of sum of two special dice 2.1-12 Let x equal the number of orange balls and 144 − x the number of blue balls. Then x x − 1 144 − x 143 − x x 144 − x 144 − x x · + · = · + · 144 143 144 143 144 143 144 143 x2 − x + 144 · 143 − 144x − 143x + x2 = 2 · 144x − 2 · x2 x2 − 144x + 5,148 = 0 (x − 78)(x − 66) = 0 Thus there are 78 orange balls and 66 blue balls.

2.2

Mathematical Expectation 4

2.2-2 E(X) = (−1) E(X2) = (−1)

+ (0)

9 2 4

1

+ (1)

9 1 + (0)2 9

4 9

+ (1)2 9

= 0; 4

= 9

8

; 9

10

Section 2.3 Special Mathematical Expectations 8

E(3X2 − 2X + 4) = 3

− 2(0) + 4 =

20

9

1

9 Σ 6 2.2-4

1

f (x) =

=

x=0

2 c =

49 2

E(Payment) =

49 ∞ Σ

10

+c

1

.

3

+

1 2

+

1 3

+

1 4

+

1 5

+

1 6

; 1·

1

+2·

2

1 3

+3·

1

+4·

4

1

+5·

1

5

6

=

71

units.

490

2 ∞ Σ 1 = 6 π = 1, so this is a pdf. = 6 2 2 π2 x=1 x2 π2 6 x=1 π x ∞ ∞ Σ 6 6 Σ 1 = +∞ E(X) = x 2 2 = 2 πx π x

2.2-6 Note that

6

x=1

x=1

and it is well known that the sum of this harmonic series is not finite.

2.2-8 E(|X − c|) =

1Σ |x − c|, where S = {1, 2, 3, 5, 15, 25, 50}.

7 x∈S

When c = 5, E(|X − 5|) =

1 7

[(5 − 1) + (5 − 2) + (5 − 3) + (5 − 5) + (15 − 5) + (25 − 5) + (50 − 5)] .

If c is either increased or decreased by 1, this expectation is increased by 1/7. Thus c = 5, the median, minimizes this expectation while b = E(X) = 101/7 minimizes E[(X − b)2]. + (− 1) · 21 = −6 = −1 ; 36 36 36 6 15 21 −6 −1 + (−1) · = = (1) · ; 36 36 36 6 6 30 −6 −1 + (−1) · = = · .

(4) 36 36 36 6

2.2-10 (1) ·

15

2.2-12 (a) The average class size is (b)

0.4 ,

f(x) =

0.3, 0.3 ,

(16)(25) + (3)(100) + (1)(300) = 50; 20

x = 25, x = 100, x = 300;

(c) E(X) = 25(0.4) + 100(0.3) + 300(0.3) = 130.

Chapter 2 Discrete Distributions

2.3

11

Special Mathematical Expectations

2.3-2 (a)

µ

=

3 Σ x

3!

1 x 3 3−x x! (3 − x)! 4 4 x=1 2 k 1 Σ 2!

1 3 2−k = 3 4 k! (2 − k)! 4 4 k=0 2 3 3 1 1 = ; = 3 + 4 4 4 4 3Σ 3!

1 x 3 3−x = x(x – 1) x! (3 − x)!

4 4 x=2 2 3 1 3 1 = 2(3) +6 =

E[X(X − 1)]

E(X)

4 = σ2 =

3

4

4

=

4

;

E[X(X − 1)] + E(X) − µ2

= (2) µ

4

1 2= 2 1

6

= (2)

(b)

4

3

1

4 3

4 1

4

4

3

+

–

4 3

+

4

1

4

2

3

=3

4

1

3

4

4

E(X) 4 Σ x

4!

1 x 1 4−x x! (4 − x)! 2 2 x=1 3 1 Σ 3!

1 k 1 3−k = 4 2 k! (3 − k)! 2 2 k=0 3 1 = 2; = 4 1 1+ 2 2 2 =

E[X(X − 1)]

4 Σ

=

x(x − 1)

x=2

1 = 2(6)

σ2

= (12)

x! (4 − x)!

1

+ (6)(4)

4

1 = 12

2 1 2

2

x

1

4

2 1

= 48

4!

4 + 2

2 −

2

1

2 4

4−x

2 1

+ (12)

2

; 4 2 = 1.

2

2.3-4 E[(X − µ)/σ] = (1/σ)[E(X) − µ] = (1/σ)(µ − µ) = 0; E{[(X − µ)/σ]2} = (1/σ2)E[(X − µ)2] = (1/σ2)(σ2) = 1.

3 2 3 2.3-6 f (1) = , f(2) = , f (3) = , 8 8 8

2

4

;

12

Section 2.3 Special Mathematical Expectations 3

µ = 1·

+2·

2

+3·

3

= 2, 8 8 8 3 2 3 3 σ2 = 12 · + 22 · + 32 · − 22 = .

8 8 4 8 4 Σ 2x − 1 2.3-8 E(X) = x 16 · x=1

50

=

= 3.125;

16 4

2

E(X )

=

Σ

x2 ·

2x − 1 16

–

25 8

x=1

85 ; 8 85

= Var(X) = σ

8 √

=

55

2

55

=

= 0.8594;

64

= 0.9270.

8 2.3-10 µ = E(X) = (−5)(1/16) + (−1)(5/8) + (3)(5/16) = 0 so E[(X −µ)3] = E(X3) = (−5)3(1/16)+(−1)3(5/8)+(33)(5/16) = (−125−10+135)/16 = 0.

Thus the index of skewness likewise equals 0. The distribution is not symmetric, however.

364 2.3-12 (a) f (x) = (b)

365

1 ,

365

x = 1, 2, 3, . . . ;

1 = 365, 1 365 364 365 = 132,860, 2 1 365

µ =

σ2 =

σ

x−1

= 364.500;

(c) P (X > 400)

=

P (X < 300) =

364 400 = 0.3337, 365 364 299 1− = 0.5597.

365

2.3-14 P (X ≥ 100) = P (X > 99) = (0.99)99 = 0.3697.

2.3-16 (a) f (x) = (1/2)x−1, x = 2, 3, 4, . . .; (b)

M (t)

∞ Σ

=

E[etx] =

=

∞ Σ 2 (et/2)x

etx(1/2)x−1

x=2

x=2

2(et/2)2 =

1 − et/2

=

e2t 2 − et

,

t < ln 2;

Chapter 2 Discrete Distributions

(c)

13

4e2t − e3t (2 − et)2 µ = M ′(0) = 3; (2 − et)2(8e2t − 3e3t) − (4e2t − e3t)2 ∗ (2 − et)(−et) M ′(′ t) =

M′(t)

2

σ

=

= M ′′(0) − µ

2

(2 − et)4 = 11 − 9 = 2;

(d) (i) P (X ≤ 3) = 3/4; (ii) P (X ≥ 5) = 1/8; (iii) P (X = 3) = 1/4.

P (X > k + j) 2.3-18 P (X > k + j | X > k) = P (X > k) k+j q j = = q = P (X > j). k q 2.3-20 (a) Construct a square in the first quadrant with lengths of sides equal to one. Within this square construct squares with one vertex at the origin and side lengths equal to 1 − 1/2x; 1 2 (2x − 1)2 (2x−1 − 1)2 1 2 − = (b) f (x) = 1− 1− − 2x 22x−2 2x−1 22x x 2 x 2 (2 − 1) (2 − 1) 22(2x−1 − 1)2 (2x − 2)2 = − – = 22x 22x 22x 22x 2x x 2x 2 x 2 −2·2 +1−2 +2 ·2 −4 2x+1 − 3 = , x = 1, 2, 3, . . . ; = 22x ∞ 22x ∞ Σ Σ 2 · 2x 3 3 4/4 3/4 2x+1 = − = 2 − 1 = 1; (c) – = – 1 − 1/2 1 − 1/4 2x 2x x x=1 x=1 22x 2 2 4 Σ ∞ t x ∞ 2(2e ) − 3(etx) = 2ett − 3et; 2 − e 4 − et (d) M (t) = E(e ) = Σ etx · 2x+1 − 3 = 4x x tX 4 22x x=1 x=1 4et− 2e2t + 2e2t – 12et − 3e2t + 3e2t (e) M′(t) = (2 − et)2 (4 − et)2 4et

12et – (2 − et)2 (4 − et)2

=

µ = M ′(0) = 4 − (f)

12

8

; 9 3 (2 − et)2 · 4et − 4et − 4et · 2(2 − et)(−et)

′′

M (t)

=

=

−

(2 − et) 4 (4 − et)2 · 12et − 12et − 12et · 2(4 − et)(−et) (4 − et)4

σ2

=

M ′′(0) − µ2 = 4 + 8 — 1

108 + 72 81

−

8 3

2

=

8 .

3

14

Section 2.4 The Binomial Distribution

2.4

The Binomial Distribution

11 7 2.4-2 f (−1) = , f (1) = 18 ; 18 11 7 4 µ = (−1) + (1) =− ; 18 18 18 2 4 11 4 + 1+ σ2 = –1 + 18 18 18

2

7 18

=

77 81

.

The variance can also be computed by noting that X2 = 1 so 4 2 77 2 2 − = . σ =1−µ = 1− 18 81 2.4-4 (a) X is b(7, 0.15); (b)

(i) P (X ≥ 2)

= 1 − P (X ≤ 1) = 1 − 0.7166 = 0.2834;

(ii) P (X = 1)

= P (X ≤ 1) − P (X ≤ 0) = 0.7166 − 0.3206 = 0.3960;

(iii)

P (X ≤ 3)

= 0.9879.

15 − X is b(15, 0.25);

2.4-6 (a) X is b(15, 0.75);

(b) P (X ≥ 10) = P (15 − X ≤ 5) = 0.8516; (c) P (X ≤ 10) = P (15 − X ≥ 5) = 1 − P (15 − X ≤ 4) = 1 − 0.6865 = 0.3135; (d) P (X = 10) = P (X ≥ 10) − P (X ≥ 11) = P (15 − X ≤ 5) − P (15 − X ≤ 4) = 0.8516 − 0.6865 = 0.1651; √ (e) µ = (15)(0.75) = 11.25, σ2 = (15)(0.75)(0.25) = 2.8125; σ = 2.8125 = 1.67705.

2.4-8 (a) 1 − 0.014 = 0.99999999; (b) 0.994 = 0.960596.

2.4-10 (a) X is b(8, 0.90); (b)

P (X = 8)

=

P (8 − X = 0) = 0.4305;

(ii) P (X ≤ 6)

=

P (8 − X ≥ 2)

=

1 − P (8 − X ≤ 1) = 1 − 0.8131 = 0.1869;

=

P (8 − X ≤ 2) = 0.9619.

(i)

(iii) P (X ≥ 6) 2.4-12 (a)

125/216, f (x) =

(b)

µ σ2 σ

x = −1,

75/216,

x = 1,

15/216,

x = 2,

1/216,

x = 3;

125 75 15 1 17 + (1) · 216 + (2) · 216 + (3) · = − ; 216 216 216 269 17 2 – – = E(X2) − µ2 = = 1.2392; 216 216 = 1.11; = (−1)·

Chapter 2 Discrete Distributions

15

(c) See Figure 2.4-12. f(x)

x

–1

1

2

3

Figure 2.4–12: Losses in chuck-a-luck 2.4-14 Let X equal the number of winning tickets when n tickets are purchased. Then P (X ≥ 1) =

1 − P (X = 0)

= (a)

1 − (0.9)n

1–

9 10

n

.

= 0.50

(0.9)n = 0.50 n ln 0.9

= ln 0.5 ln 0.5 n = = 6.58 ln 0.9

so n = 7; (b)

1 − (0.9)n

= 0.95

(0.9)n = 0.05 ln 0.05 n = = 28.43 ln 0.9 so n = 29.

2.4-16 It is given that X is b(10, 0.10). We are to find M so that P (1000X ≤ M ) ≥ 0.99) or P (X ≤ M/1000) ≥ 0.99. From Appendix Table II, P (X ≤ 4) = 0.9984 > 0.99. Thus M/1000 = 4 or M = 4000 dollars.

2.4-18 Let X equal the number of infected individuals. X is b(5, 0.05). The expected number of tests is 1 P (X = 0) + 6 P (X > 0) = 1 (0.7738) + 6 (1 − 0.7738) = 2.131.

2.4-20 (a) (i) b(5, 0.7); (ii) µ = 3.5, σ2 = 1.05; (iii) 0.1607; (b) (i) geometric, p = 0.3; (ii) µ = 10/3, σ2 = 70/9; (iii) 0.51; (c) (i) Bernoulli, p = 0.55; (ii) µ = 0.55, σ2 = 0.2475; (iii) 0.55;

16

Section 2.5 The Hypergeometric Distribution (d) (ii) µ = 2.1, σ2 = 0.89; (iii) 0.7; (e) (i) discrete uniform on 1, 2, . . . , 10; (ii) 5.5, 8.25; (iii) 0.2.

2.5

The Hypergeometric Distribution 4 4

4 0

1

4 2.5-2 (a)

=

8 4 4

(b) 2

8

70

4 0

;

2 =

4

.

70

2.5-4 P (X ≥ 1) = 1 − P (X = 0) = 1 −

3 0

17 5 20 5

=1−

91

137

228 = 228

= 0.60.

2.5-6 We have N = N1 + N2. Thus n Σ

E[X(X − 1)] =

x(x − 1)f (x)

x=0 n Σ

=

x(x − 1)

x=2

N1! x! (N1 − x)! n

= N1(N1 − 1)

·

N 2! (n − x)! (N2 − n + x)!

N

n Σ (x − 2( )N! 1(N −12−)! x)! x=2

·

(n − x)! (NN22 !− n + x)!

N n In the summation, let k = x − 2, and in the denominator, note that N (N − 1) N − 2 .

N N! = = n n!(N − n)! n(n − 1) n−2 Thus N1 − 2 N2 n−2 N1(N1 − 1) Σ k n − 2 −k E[X(X − 1)] = N (N − 1) k=0 N−2 n(n − 1) n−2 N1(N1 − 1)(n)(n − 1) = .

N (N − 1) 5 2.5-8 (a)

5 70 5

65 0

·

1 25

1 =

; 302,575,350

.

Chapter 2 Discrete Distributions 5 5 (b)

65 0

70 5 5 4

(c)

(d)

65 1

65 1

(e)

·

705

302,575,350

=

302,575,350

24

;

7,800 =

302,575,350

1

·

;

325

25

65 5

24 =

1

·

25

70 5 5 0

24 25

70 5 5 4

·

17

;

8,259,888

25

=

302,575,350

.

2.5-10 (a) P (2, 1, 6, 10) means that 2 is in position 1 so 1 cannot be selected. Thus 1 1 8 P (2, 1, 6, 10) = 0 1 5 56 4 = = ; 10 210 15 6 i−1 1 n−i r−1

(b) P (i, r, k, n) =

2.6

1

k−r .

n k

The Negative Binomial Distribution

2.6-2

10 − 1

1 5 1 5

5−1

2

2

=

126

=

1024

63

.

512

2.6-4 Let “being missed” be a success and let X equal the number of trials until the first success.

Then p = 0.01.

P (X ≤ 50) = 1 − 0.9950 = 1 − 0.605 = 0.395.

2.6-6 (a)

R(t) =

ln(1 − p + pet), pet

R′(t) = R′′ (t) = (b)

R(t) =

1 − p + pet t=0

= p,

(1 − p + pet)(pet) − (pet)(pet) (1 − p + pet)2 n ln(1 − p + pet),

R′(t) = R′′ (t) = n

npet 1 − p + pet t=0

= p(1 − p); t=0

= np,

(1 − p + pet)(pet) − (pet)(pet) (1 − p + pet)2

= np(1 − p); t=0

18

The Poisson Distribution (c)

(d)

ln p + t − ln[1 − (1 − p)et], (1 − p)et 1 − p = 1, ′ 1 + R (t) = 1 − (1 − p)et t=0 = 1 + p p 1−p R′′(t) = (−1){1 − (1 − p)et}2{−(1 − p)et} ; t=0 = p2 R(t) = r [ln p + t − ln{1 − (1 − p)et}] , 1 r R′(t) = r = , p 1 − (1 − p)et t=0 r(1 − p) = .

R′′(t) = r (−1){1 − (1 − p)et}−2{−(1 − p)et} t=0 p2 R(t) =

2.6-8 (0.7)(0.7)(0.3) = 0.147.

2.6-10 (a) Let X equal the number of boxes that must be purchased. Then 10 E(X) = Σ x=1

(b)

100 · E(X) 365

2.7

1 = 29.29; (11 − x)/10 ≈ 8 years.

The Poisson Distribution

2.7-2 λ = µ = σ2 = 3 so P (X = 2) = 0.423 − 0.199 = 0.224.

2.7-4

2.7-6

λ1e−λ = λ2e−λ 1!

2! e−λλ(λ − 6) = 0 λ = 6 Thus P (X = 4) = 0.285 − 0.151 = 0.134.

3

t

M (t) =

eλ(e −1),

M ′(t) =

eλ(e −1)λet,

M ′′(t) =

eλ(e t−1)λ2e2t + eλ(e t−1)λet,

t

t

t

t

t

M ′′′3(t) = eλ(e −1)λ3e3t3 + eλ(e2 −1)λ2 · 2e2t + eλ(e −1) λ2e2t + eλ(e −1)λet, E(X ) = M ′′′(0) = λ + 3λ + λ, √ γ = [λ3 + 3λ2 + λ − λ3 − 3λ(λ + λ2) + 3λ2 · λ]/λ3/2 = 1/ λ.

Thus the index of skewness is positive. As λ increases, the index of skewness deceases to zero.

2.7-8 np = 1000(0.005) = 5, (a) P (X ≤ 1) ≈ 0.040; (b) P (X = 4, 5, 6) = P (X ≤ 6) − P (X ≤ 3) ≈ 0.762 − 0.265 = 0.497. √ 2.7-10 σ = 9 = 3, P (3 < X < 15) = P (X ≤ 14) − P (X ≤ 3) = 0.959 − 0.021 = 0.938.

2.7-12 Since E(X) = 0.2, the expected loss is (0.2)($100,000) = $2,000.

Chapter 3 Continuous Distributions

19

Chapter 3

Continuous Distributions 3.1

Random Variables of the Continuous Type ∫1

c · (x − x2) dx =

c

⇒ c = 6; ∫ 6 0.6 0.6 6(x − x2) dx = 3x2 − 2x3 = 0.432; (b) P (0.3 < X < 0.6) = 0.3 0.3 ∫ π 0.5 0 2 6 (x − x ) d x = 1/2 ⇒ π 0.50 = 1/2. (c) 0

3.1 –2 (a)

0

3.1–4 X is U (4, 5); (a) µ = 9/2; (b) σ2 = 1/12; (c) 0.5. ∫ 200 ∫n 1 1 [n 3.1 –6 E(profit) = – 5(x − n)] [x − 0.5(n − x)] dx + dx 200 n 200 0 n 5x2 200 1 x2 (n − x)2 1 + + = 6nx − 200 2 4 2 n 200 0 1 = 3.25n2 + 1200n 100000− 200 – 1 derivative = [−6.5n + 1200] = 0 200 1200 n = ≈ 185.

6.5 ∫c 3.1–8 (a) (i) x3/4 dx = 1 0

c4/16 = 1 c = 2; ∫x (ii) F (x) = f (t) dt −∞

∫x

t3/4 dt

= 0

= x4/16,

F (x) =

0,

−∞ < x < 0,

x4/16,

0 ≤ x < 2,

1,

2 ≤ x < ∞;

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20

Section 3.1 Random Variables of the Continuous Type

(iii) F(x)

f(x) 2.0

1.0

1.5

0.75

1.0

0.5

0.5

0.25

0.4

0.8

1.2

1.6

x

x

2.0

0.4

0.8

1.2

1.6

2.0

Figure 3.1–8: (a) Continuous distribution pdf and cdf ∫2 (iv)

µ =

(x)(x3/4) dx =

0

∫2 E(X2)

=

(x2)(x3/4) dx =

8

; 5 8

0

8 σ

2

γ

2

8

−

−c

;

8

; 3 75 5 ∫ 2 x3x3/4 dx − 3 · (8/5) · (8/75) − (8/5)3

= =

=

0

(8/75)3/2

∫c (b) (i)

3

(3/16)x2 dx

=

= −

√ √ 3 8 75 70

= − 1.04978;

1

c3/8 = 1

(ii) F (x)

c = 2; ∫x f (t) dt

=

−∞

∫x =

−2

(3/16)t2 dt x

t3 16 −2 x3 1 + , 16 2

= =

F (x) =

0, x3

−∞ < x < −2,

1 + , 16 2

1,

−2 ≤ x < 2, 2 ≤ x < ∞;

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Chapter 3 Continuous Distributions

21

(iii)

–2

f(x) 1.0

F(x) 1.0

0.8

0.8

0.6

0.6

0.4

0.4

0.2

0.2 x

–1

1

2

–2

–1

1

2

x

Figure 3.1–8: (b) Continuous distribution pdf and cdf ∫2 (iv)

µ =

−2 ∫ 2

(x)(3/16)(x2) dx = 0; (x2)(3/16)(x2) dx =

12

; 5 ∫ 2 x3(3/16)x2 dx − 3 · (0) · (12/5) − 03

σ2 =

−2

−2

γ = ∫ 1 c 0

(c) (i)

(12/5)3/2

= 0;

√ dx = 1 x 2c = 1 c = 1/2.

The pdf in part (c) is unbounded. ∫x f(t) dt (ii) F (x) = −∞

∫

x

1 √ dt 0 2 t √ h√ ix x, = t0 =

=

−∞ < x < 0,

0, √ x,

F (x) = 1,

0 ≤ x < 1, 1 ≤ x < ∞;

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22

Section 3.1 Random Variables of the Continuous Type (iii)

f(x) 2.0

F(x)

1.5

1.0

0.5

–0.2

0.2

0.4

0.6

0.8

x

x 1.2

1

0.2

0.4

0.6

0.8

1.0

Figure 3.1–8: (c) Continuous distribution pdf and cdf ∫ 1 (iv)

µ =

0

∫1 E(X2)

=

σ2 = γ = ∫ ∞ 3.1–10 (a) 1

c

0

1

√ (x)(1/2)/

xdx =

√ (x2)(1/2)/ xdx =

−

1

2

=

1

; 3 1 5

;

4

; 5 45 3 ∫ 1 x3/(2√ x ) dx − 3 · (1/3) · (4/45) − (1/3)3 0

(4/45)3/2

=

√ 2 45 21

= 0.63888.

dx = 1

x2 −c x 1

∞

= 1

c = 1; ∫ ∞ x (b) E(X) = dx = [ln x]∞, which is unbounded.

1 2 x 1

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Chapter 3 Continuous Distributions 3.1 –12 (a)

F (x) =

23

0,

−∞ < x < −1,

(x3 + 1)/2,

−1 ≤ x < 1,

1,

1 ≤ x < ∞;

f(x)

F(x)

1.4 1.2 1.0 0.8 0.6 0.4 0.2 x –1.0

–0.6

–0.2

0.2

0.6

1.0

x –1.0 –0.75–0.50–0.25

0.25 0.5 0.75 1.0

Figure 3.1–12: (a) f (x) = (3/2)x2 and F (x) = (x3 + 1)/2

(b) F (x) =

–1.0

–0.6

0,

−∞ < x < −1,

(x + 1)/2,

−1 ≤ x < 1,

1,

1 ≤ x < ∞;

f(x) 1.0

F(x) 1.0

0.8

0.8

0.6

0.6

0.4

0.4

0.2

0.2

–0.2

x 0.2

0.6

1.0

–1.0

–0.6

–0.2

x 0.2

0.6

1.0

Figure 3.1–12: (b) f (x) = 1/2 and F (x) = (x + 1)/2

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24

Section 3.1 Random Variables of the Continuous Type (c)

−∞ < x < −1,

0, 2

−1 ≤ x < 0,

1 − (1 − x)2/2,

0 ≤ x < 1,

1,

1 ≤ x < ∞;

( x + 1) /2, F (x) =

f(x) 1.0

F(x) 1.0

0.8

0.8

0.6

0.6

0.4

0.4

0.2

0.2 x

–1.0

–0.6

–0.2

0.2

0.6

1.0

–1.0

–0.6

–0.2

x 0.2

0.6

1.0

Figure 3.1–12: (c) f (x) and F (x) for Exercise 3.1-12(c) 3.1 –14 (a) σ2 = ∫

1

(b) σ2 =

2

−1 x (3/2)x ∫1

2

dx = 3/5 = 0.6;

2

x (1/2) dx = 1/3 = 0.333; −1 ∫ 0 ∫ 1 2 2 (c) σ = x (1 − x) dx = 1/6 = 0.1667. x (x + 1) dx + 2

−1

0

3.1 –16 F (x) = (x + 1)2/4,

−1 < x < 1.

(a) F (π0.64) = (π0.64 + 1)2/4 π0.64 + 1

=

0.64 √ = 2.56

π0.64 = 0.6; (b) (π0.25 + 1) /4 = 0.25 √ π0.25 + 1 = 1.00 2

π0.25 = 0; (c) (π0.81 + 1)2/4 = 0.81 √ π0.81 + 1 = 3.24 π0.81 = 0.8.

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Chapter 3 Continuous Distributions 3.1 –18 (b)

−∞ < x ≤ 0,

0, x ,

0 < x ≤ 1,

,

1 < x ≤ 2,

2 2 x − 1 1,

2 ≤ x < 3, 3 ≤ x < ∞;

2 1

F (x) =

25

2

f(x) 1.0

F(x) 1.0

0.8

0.8

0.6

0.6

0.4

0.4

0.2

0.2 x 0.5

1.0

1.5

2.0

2.5

3.0

x 0.5

1.0

1.5

2.0

2.5

3.0

Figure 3.1–18: f (x) and F (x) for Exercise 3.1-18(a) and (b) (c)

q1 = 0.25 2 q1 = 0.5;

(d) 1 ≤ m ≤ 2, not unique; q3 1 (e) − = 0.75 2 2 q3 5 = 2 4 5 . q3 = 2 ∫ 1 ∫ ∞ c dx = 1 3.1 –20 (a) xdx + 0 x3 1 h c i∞ 1 x2 – = 1 2x2 20 1 1c + = 1 2 2 ∫ c = 1; 4 ∫ 1 (b) E(X) = x2 dx + ∞ 1 dx = ; 0 1 x2 3 (c) the variance does not exist;

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26

Section 3.2 The Exponential, Gamma, and Chi-Square Distributions ∫ 1

∫ 2

1

3 xdx + (d) P (1/2 ≤ X ≤ 2) = 3 dx = 4 .

1/2 1 x (e) Since the median of this distribution is 1, π.82 must be greater than 1. Thus solve ∫ π.82 π 0.82 − 0.5 = (1/x3) dx = −(1/2x2) .82 = [1 − (1/π2 )]/2 1

⇒ 1 − (1/π.82 ) =

3.2

1

0.64 ⇒ 1/π.82

2

2

.82

= 0.36 ⇒ π.82 = 1/(0.6) = 5/3.

The Exponential, Gamma, and Chi-Square Distributions

3.2 –2 (a) f (x) =

2

0 ≤ x < ∞;

e−2x/3,

3

(b) P (X > 2) =

∫ ∞ 2

h i∞ 2 −2x/3 e dx = −e−2x/3 = e−4/3.

3 2

3.2–4 Let F (x) = P (X ≤ x). Then P (X > x + y | X > x) =

P (X > y)

1 − F (x + y) = 1 − F (y).

1 − F (x) That is, with g(x) = 1 − F (x), g(x + y) = g(x)g(y). This functional equation implies that 1 − F (x) = g(x) = acx = e(cx) ln a = ebx where b = c ln a. That is, F (x) = 1—ebx. Since F ( ∞) = 1, b must be negative, say b = − λ with λ > 0. Thus F (x) = 1—e−λx, 0 ≤x, the distribution function of an exponential distribution. ∫ ∞ 3 3.2 –6 (a) P (X > 40) = e−3x/100 dx 40 100 ∞

−e −3x/100 40 = e−1.2;

=

(b) Flaws occur randomly so we are observing a Poisson process.

3.2 –8 Either use integration by parts or

P (X ≤ x) α−1 Σ (λx)ke−λx . = 1− k!

F (x) =

k=0

Thus, with λ = 1/θ = 1/4 and α = 2, P (X < 5) =

1 − e−5/4 −

5 4

e−5/4

= 0.35536.

3.2 –10 The moment generating function of X is M (t) = (1 − θt)−α, t < 1/θ. Thus

M ′(t)

=

αθ(1 − θt)−α−1

M ′′(t)

=

α(α + 1)θ2(1 − θt)−α−2.

The mean and variance are

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Chapter 3 Continuous Distributions µ = σ2

27

M ′(0) = αθ

= M ′′(0) − (αθ)2 = α(α + 1)θ2 − (αθ)2 = αθ2.

3.2–12 (a) W has a gamma distribution with α = 7, θ = 1/16. (b) Using Table III in the Appendix, P (W ≤ 0.5) =

1−

6 Σ 8ke−8 k=0

k!

= 1 − 0.313 = 0.687, because here λw = (16)(0.5) = 8.

3.2–14 a = 5.226, b = 21.03.

3.2–16 Note that λ = 5/10 = 1/2 is the mean number of arrivals per minute. Thus θ = 2 and the pdf of the waiting time before the eighth toll is 1 x8−1e−x/2 f (x) = Γ(8)28 1 16/2−1 −x/2 = x e , 0 < x < ∞, 16 16/2 Γ 2 2 the pdf of a chi-square distribution with r = 16 degrees of freedom. Using Table IV, P (X > 26.30) = 0.05.

3.2 –18 Note that X = 80 + Y where Y has a gamma distribution with shape parameter 2 and

scale parameter 50. (a) E(X) = 80 + E(Y ) = 80 + 2 · 50 = 180 and Var(X) = Var(Y ) = 2 · 502 = 5000; (b) The same calculus argument shows that the median of the gamma distribution in general is (α − 1)θ. So the median is X is 80 + (1)50 = 130; (c) P (X < 200) = P (Y < 120) = 0.6916 using tables or computer software or calculus: ∫ 200 200 x − 80 x − 80 −(x−80)/50 −(x−80)/50dx = −(x−80)/50 e − e − e 50 502 80 80 = = ∫ 3 3.2 –20

E[v(T )]

=

= =

100(23−t − 1)e−t/5/5dt

0

∫ 3

−120 −120/50 e – e−120/50 + 1 50 17 −12/5 e = 0.6916 = 69.16%.

1− 5

∫ 3

– 20e−t/5dt + 100

e(3−t) ln 2e−t/5/5dt ∫ 3 0.6 3 ln − e−t ln 2e−t/5/5dt −100(1 − e ) + 100e 0

0

2

0

=

−(ln 2+0.2)t –e ln 2 + 0.2

−100(1 − e−0.6) + 100e3 ln

3

= $121.734.

0

2

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28

Section 3.3 The Normal Distribution M ′(t)

3.2 –22

r

=

M ′′(t)

=

M ′′′(t)

=

,

(1 − 2t)(r/2)+1 r(r + 2) , (1 − 2t)(r/2)+2 r(r + 2)(r + 4) (1 − 2t)(r/2)+3

⇒ E(X3) = M ′′′(0) = r(r + 2)(r + 4).

Thus E[(X − µ)3] = E(X3) − 3µσ2 − µ3 = r(r + 2)(r + 4) − 3(r)(2r) − r3 = 8r, r

so 8r γ= = (2r)3/2

8 , r

which decreases as r increases.

3.3

The Normal Distribution

3.3 –2 (a) 0.2823: (b) 0.4931; (c) 0.2474; (d) 0.0536;

(e) 0.1038; (f) 0.3174; (g) 0.0456; (h) 0.0026.

3.3 –4 (a) 1.282; (b)−1.645; (c) −1.66; (d) −1.82.

2

3.3 –6 M (t) = e166t+400t /2, − ∞ < t < ∞, so

(a) µ = 166; (b) σ2 = 400; (c) P (170 < X < 200) = P (0.2 < Z < 1.7) = 0.3761; (d) P (148 ≤ X ≤ 172) = P (−0.9 ≤ Z ≤ 0.3) = 0.4338.

3.3 –8 We must solve f ′′(x) = 0. We have

ln f (x) = f ′(x) f (x) f (x)f ′′(x) − [f ′(x)]2 [f(x)]2 f ′′(x) (x − µ)2 σ4

x−µ

√ − ln( 2π σ) − (x − µ)2/2σ2,

=

−2(x − µ) 2σ2

=

−1 σ2

=

f (x)

−1 σ2

+

1 = =

σ2 ±σ

or

f ′(x)

= 0

f (x) x = µ ± σ.

Here is an alternative solution: 1 − 2 φ(z) = √ e z /2 2π φ′(z) = −zφ(z) φ′′(z) =

−φ(z) + z2φ(z)

φ′′(z) = 0 ⇒ −1 + z2 = 0 ⇒ z = ±1, the inflection points. Now let

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Chapter 3 Continuous Distributions

f(x)

=

f ′(x)

=

f ′′(x)

=

f ′′(x)

√

1

exp −

σ 2π 1 ′ x−µ φ σ2 σ 1 ′′ x − µ φ σ3 σ 0 ⇒ −1 +

=

29 (x − µ)2 =

2σ2

x−µ σ

σ

2

=0 ⇒

x−µ

1 φ

σ

x−µ σ

= ±1.

Thus the points of inflection are at x = µ ± σ.

3.3–10

G(y)

P (Y ≤ y) = P (aX + b ≤ y) y−b = P X≤ if a > 0 a ∫ (y−b)/a 1 e −(x−µ) 2/2σ 2 dx √ = σ 2π −∞ =

Let w = ax + b so dw = a dx. Then ∫ y 1 2 2 2 √ e−(w−b−aµ) /2a σ dw G(y) = −∞ aσ 2π which is the distribution function of the normal distribution N (b + aµ, a2σ2). The case when a < 0 can be handled similarly.

3.3–12 X is N (500, 2500); so [(X − 500)/50]2 is χ2(1) and " # 2 P 2.706 ≤ X − 500 ≤ 5.024 = 0.975 − 0.900 = 0.075.

50 3.3 –14

G(x) =

P (X ≤ x)

= P (eY ≤ x) = P (Y ≤ ln x) ∫ ln x

1 2 √ e−(y−10) /2dy = Φ(ln x − 10) 2π −∞ 1 −(ln x− 2 1 , 0 < x < ∞. g(x) = G′(x) = 10) /2 x √ e 2π P (10,000 < X < 20,000) = P (ln 10,000 < Y < ln 20,000) =

= Φ(ln 20,000 − 10) − Φ(ln 10,000 − 10) = 0.461557 − 0.214863 = 0.246694.

3.3

–16 (a) N (0, 1); (b) N (−1, 1); (c) N (2, 1). Note the slopes of these graphs at 0.

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30

Section 3.4 Additional Models

3.4

Additional Models

3.4 –2 With b = ln 1.1,

G(w)

h a w ln 1.1 a i 1 − exp − + e ln 1.1 ln 1.1

=

G(64) − G(63) = 0.01 1 a = 0.00002646 = P (W ≤ 71| 70 < W )

37792.19477 P (70 < W ≤ 71)

=

using Maple

P (70 < W ) = 0.0217.

3.4 –4

λ(w) = H(w) =

aebw + c ∫w (aebt + c) dt 0

a bw e − 1 + cw b i h a G(w) = 1 − exp − ebw − 1 − cw , b a bw — 1) − cw – (e , bw b g(w) = (ae + c)e =

3.4–6

(a) 1/4 − 1/8 = 1/8;

(b) 1/4 − 1/4 = 0;

(c) 3/4 − 1/4 = 1/2;

(d) 1 − 1/2 = 1/2;

(e) 3/4 − 3/4 = 0;

(f) 1 − 3/4 = 1/4.

0<w<∞

0 < w < ∞.

3.4 –8 There is a discrete point of probability at x = 0, P (X = 0) = 1/3, and F ′(x) = (2/3)e−x

for 0 < x. Thus µ = E(X)

∫∞

x(2/3)e−xdx

=

(0)(1/3) +

=

(2/3) [−xe−x − e−x]∞

=

(0)2(1/3) +

0

E(X2)

0

= 2/3,

∫∞ x2(2/3)e−xdx 0

=

(2/3)[−x2e−x − 2xe−x − 2e−x]0∞

= 4/3,

so σ2 = Var(X) = 4/3 − (2/3)2 = 8/9.

3.4–10

T =

( X,

X ≤ 4,

4,

4 < X;

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Chapter 3 Continuous Distributions ∫ 4 E(T ) =

1

x

31 e−x/5 dx +

∫∞

5

0

4

1 5

4

4

−xe −x/5 − 5e −x/5 0 + 4 −e −x/5

=

e−x/5 dx ∞ 4

= 5 − 4e−4/5 − 5e−4/5 + 4e−4/5 = 5 − 5e−4/5 ≈ 2.753.

3.4–12 (a)

t = ln x x = et

(b)

dx = et dt t dx g(t) = f (et) = ete−e , dt t = α + β ln w dt

−∞ < t < ∞;

β =

dw

w eα+β ln we−e

h(w) =

β w

β α

0 < w < ∞.

βwβ−1e αe−w e ,

= ∫1

α+β ln w

6(1 − x)5 dx = 0.0198;

3.4 –14 (a) (0.03) 2/30

∫ 1 x6(1 − x)5 dx = 0.0042857;

(b) E(X) = (0.97)(0) + 0.03 0

The expected payment is E(X) · [$ 30,000] = $ 128.57. ∫1 ∫2 1 −x/10 1 −x/10 3.4–16 m e dx + (m/2) e dx = 200 0 10 1 10 m[1 − e−1/10] + 0.5m[e−1/10 − e−2/10] = 200 m − 0.5me−1/10 − 0.5me−2/10

=

200

m =

200 1 − 0.5e−1/10 − 0.5e−2/10

= 1447. ∫∞ 3.4–18 P (X > x) = x

3

t 4

4

4

e− (t/4) dt = e −(x/4) ; P (X > 5)

−369/256 e−625/256 = e . = −1 P (X > e ∫ 5 | X > 4) = P (X > 4) h 60 i60 2 2x 3.4–20 (a) 40 502 e − (x/50) dx = −e −(x/50) 2 40 = e −16/25 − e−36/25; h i∞ = e−64/25. (b) P (X > 80) = −e−(x/50)2 80

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Chapter 4 Bivariate Distributions

33

Chapter 4

Bivariate Distributions 4.1

Bivariate Distributions of the Discrete Type

4.1 –2

16

4

1 • 16

1 • 16

1 • 16

1 • 16

4 16

3

1 • 16

1 • 16

1 • 16

1 • 16

4 16

2

1 • 16

1 • 16

1 • 16

1 • 16

4 16

1

1 • 16

1 • 16

1 • 16

1 • 16

4

1

2

3

4

4 16

4 16

4 16

4 16

(e) Independent, because fX (x)fY (y) = f (x, y).

4.1–4

12 11 10 9 8 7 6 5 4 3 2 1 1

0

1 • 25

0 1 5

1

2 1 5

3

4

5

1 5

6 1 5

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8 1 5

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34

Section 4.2 The Correlation Coefficient (c) Not independent, because fX (x)fY (y) f (x, y) and also because the support is not rectangular.

25!

4.1–6 (0.30)7(0.40)8(0.20)6(0.10)4 = 0.00405.

7!8!6!4!

4.1 –8 (a) x = 0, 1, . . . , 6, y = max(0, x − 2), . . . , min(x, 4);

1/15, 2/15, 3/15,

fX (x) =

x = 0, 6, x = 1, 5, x = 2, 3, 4;

fY (y) = 3/15,

y = 0, 1, 2, 3, 4;

(b) x = 0, 1, 2, 3, 4, y = max(0, x − 2), . . . , x;

fX (x) =

1/12, 2/12, 3/12,

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

fY (y) =

3/12, 2/12, 1/12,

y = 0, 1, 2, y = 3, y = 4;

(c) x = 0, 1, 2, 3, 4, y = max(2 − x, x − 2), . . . , min(2 + x, 6 − x);

fX (x) =

1/13, 3/13, 5/13,

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

fY (y) =

1/13, 3/13, 5/13,

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

4.1 –10 (a) f (x, y) =

7! x!y!(7 − x − y)!

(b) X is b(7, 0.78),

4.2

(0.78)x(0.01)y(0.21)7−x−y,

0 ≤ x + y ≤ 7;

x = 0, 1, . . . , 7.

The Correlation Coefficient

4.2 –2 (c)

µX

= 0.5(0) + 0.5(1) = 0.5,

µY

= 0.2(0) + 0.6(1) + 0.2(2) = 1,

σ2X

=

(0 − 0.5)2(0.5) + (1 − 0.5)2(0.5) = 0.25,

σ2Y

=

(0 − 1)2(0.2) + (1 − 1)2(0.6) + (2 − 1)2(0.2) = 0.4,

Cov(X, Y ) =

(0)(0)(0.2) + (1)(2)(0.2) + (0)(1)(0.3) +

(1)(1)(0.3) − (0.5)(1) = 0.2, 0.2 √ = ρ = √ 0.4; √ 0.25 0.4 ! √ √ 0.4 (d) y = 1 + 0.4 √ (x − 0.5) = 0.6 + 0.8x.

0.25 4.2–4 Note that X is b(3, 1/6), Y is b(3, 1/2) so

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Chapter 4 Bivariate Distributions

35

(a) E(X) = 3(1/6) = 1/2; (b) E(Y ) = 3(1/2) = 3/2; (c) Var(X) = 3(1/6)(5/6) = 5/12; (d) Var(Y ) = 3(1/2)(1/2) = 3/4; 0 + (1)f (1, 1) + 2f (1, 2) + 2f (2, 1) − (1/2)(3/2)

(e) Cov(X, Y ) =

= (1)(1/6) + 2(1/8) + 2(1/24) − 3/4 = −1/4; −1/4 −1 (f) ρ = r =√ .

5 ·3 5 12 4 4.2–6 (b)

6

1

2

• 16

2 6

1

• 16

• 16

3 6

0

•1

•1

• 16

0

1

2

3 6

2 6

1 6

6

6

1

(c) Cov(X, Y ) = (1)(1)

− 6

σ2

(d)

X

=

2

2 3 5

=

1

3

− 6

4

= 9

−5

; 18

2 4 2 2 = = σ2Y, + − 3 9 6 6 −5/18 1

√ = − ; 2 (5/9)(5/9) 2 1s 2 (e) y = − 5/9 3 2 5/9 x − 3 1 y = 1 − x.

2 √ 3/2 = 0.886; 4.2 –8 (a) µ = 3, µ = 2, σ2 = 8/3, σ2 = 2, Cov(X, Y ) = 2, ρ = X Y Y 19X 2 29 227 ,σ ,µ = , Cov(X, Y ) = 181 , ρ = 181 = 0.7974; 227 2 = (b) µ = = ,σ X 12 Y 12 X 144 Y 144 144 227 (c) µX = 2, µY = 2, σ2 = 14/13, σ2 = 14/13, Cov(X, Y ) = 0, ρ = 0. ρ

=

X

Y

4.2 –10 (a) fX (1) = 0.15, fX (2) = 0.25, fX (3) = 0.45, fX (4) = 0.15;

fY (1) = 0.35, fY (2) = 0.65; µX = 2.60; µY = 1.65; σ2 = 0.8400; σ2 = 0.2275; X

(b)

Y

Cov(X, Y ) = −0.0900; ρ = −0.2059;

(c) E(C) = $34.70.

4.2 –12 Note that h(v)

=

E{[(X − µX ) + v(Y − µY )]2}

= E[(X − µX )2 + 2E[(X − µX )(Y − µY ) + E[(Y − µY )2] = σ2X+ 2Cov(X, Y )v + σ2 vY2 ≥ 0.

Thus the discriminant of this quadratic must be less than or equal to 0. So we have

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36

Section 4.3 Conditional Distributions [Cov(X, Y )]2 − σ2 σ2 X

4.3

≤ 0

Y

ρ2 ≤ 1 −1 ≤ ρ ≤ 1.

Conditional Distributions

4.3–2 2

1 4

3 4

g(x | 2)

1

3 4

1 4

g(x | 1)

1

2

h(y | 1)

3 − 2|x − y| , 4 x = 1, 2, for y = 1 or 2;

equivalently, g(x | y) =

h(y | 2)

3 − 2|x − y| , 4 y = 1, 2, for x = 1 or 2;

equivalently, h(y | x) = 2

1 4

3 4

1

3 4

1 4

1

2

µX|1 = 5/4, µX|2 = 7/4, µY |1 = 5/4, µY |2 = 7/4; 2 = σ2 = σ2 = 3/16. σ2 = σ X|1

X|2

Y |1

Y |2

4.3 –4 (a) X is b(400, 0.75);

(b) E(X) = 300, Var(X) = 75; (c) b(300, 2/3); (d) E(Y | X = 300) = 200, Var(Y | X = 300) = 200/3.

4.3–6 (a) P (X = 500) = 0.40, P (Y = 500) = 0.35, P (Y = 500 | X = 500) = 0.50, P (Y = 100 | X = 500) = 0.25; (b) µX = 485, µY = 510, σ2 = 118,275, σ2 = 130,900; X

Y

(c) µX|Y =100 = 2400/7, µY |X=500 = 525; (d) Cov(X, Y ) = 49650; (e) ρ = 0.399.

4.3 –8 (a) X and Y have a trinomial distribution with n = 30, pX = 1/6, pY = 1/6;

(b) The conditional pmf of X, given Y = y, is b n − y,

pX 1 − pY

= b(30 − y, 1/5).

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Chapter 4 Bivariate Distributions

37

(c) Since E(X) = 5 and Var(X) = 25/6, E(X2) = Var(X) + [E(X)]2 = 25/6 + 25 = 175/6.

Similarly, E(Y ) = 5, Var(Y ) = 25/6, E(Y 2) = 175/6. The correlation coefficient is s (1/6)(1/6) −= 1/5 ρ=− (5/6)(5/6) √ E(XY ) = −1/5 (25/6)(25/6) + (5)(5) = 145/6.

so Thus

E(X2 − 4XY + 3Y 2) =

175

−4

145

6 4.3 –10 (a) f (x, y) = 1/[10(10 − x)],

(b) fY

y (y) = Σ x=0

1

175

6

=

6

120

= 20.

6

x = 0, 1, · · · , 9, y = x, x + 1, · · · , 9;

,

y = 0, 1, . . . , 9;

10(10 − x)

(c) E(Y |x) = (x + 9)/2,

4.4

+3

x = 0, 1, . . . , 9.

Bivariate Distributions of the Continuous Type ∫ 1

4.4 –2 (a) fX (x)

=

(x + y) dy 0

1 y2 1 = x + 1 , 0 ≤ x ≤ 1; 2 0 2 ∫ 1 1 0 ≤ y ≤ 1; (x + y) dx = y + , fY (y) = 2 0 1 1 x+ y+ f (x, y) = x + y = fX (x)fY (y); 2 2 1 1 3 1 ∫ 1 7 x + x2 (b) (i) µX = x x + 1 dx = = ; 2 12 4 0 3 0 ∫ 1 7 (ii) µY = y y + 1 dy = ; 2 12 0 1 ∫ 1 2 1 4 1 1 5 x x+ x + x3 = , (iii) E(X2) = dx = 6 12 2 4 0 0 5 7 2 = 11 ; 2 2 2 σX = E(X ) − µ X = − 12 144 12 11 2 (iv) Similarly, σ = .

Y 144 =

xy +

∫ 1∫ 1 3

1

4.4–4 (a)

P 0≤X ≤ 2

2

= =

01

x2 2

∫ 23

dy dx

(1 − x 2) dx =

11

;

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2

16

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38

Section 4.4 Bivariate Distributions of the Continuous Type (b)

P

∫ 1 ∫ √y

1 ≤Y ≤1

=

2 1

2

3√ y dy = 1 − 1 2 2 ∫ 1 ∫ √y 3 dx dy = 2

P X ≥ 1, Y ≥ 1 2

2

0

∫1

= (c)

3 dx dy

2

1

∫2 1 = = 5

(d) X and Y are dependent.

;

1 2

√ 1 dy y− 2

3 2

1 2

3/2

1 2

–

8

1

3/2

;

2

4.4–6 (a) The variances are X2

σ

= E(X ) − 0 = 2

2

∫ 1

1

10

−1

x2 · 3 x2 dx = −1 2

and σY2 = E(Y 2) − 02

3x5

=

∫ 1 y2(1 − |y|) dy =

−1

y2(1 − y) dy

2

y (1 + y) dy + −1

1

=

−

5

∫ 1

∫ 0

=

3

0

1

1

−

1

1

; 3 4 6 ∫ 1∫ 1 3 2 ∫0∫1 3 2 x (1 − y) dxdy + (b) P (−X ≤ Y ) = x (1 + y) dxdy 0 −y 2 −1 −y 2 1 1 1 1 1 1 1 1 1 + − − + = . = − + − 2 4 8 10 2 4 8 10 2 ∫0 ∫1 15 15 x· x· 4.4–8 µX = E(X) = x2(1 − x2) dx + x2 dx 7 7 0 −1 15 x4

x6

0

3

15 x4

4

+

=

1

+ 7 4 0 − − 7 4 6 1 15 −1 1 15 1 5 = + + = , 7 4 6 7 4 14 ∫ 0 ∫ 1 15 15 2 2 2 x · x (1 − x ) dx + x2 · = x2 dx 7 7 −1 0 1 15 x5 7 0 15 x 5 x + 7 5 0 = − − 1 7 5 7 15 1 1 15 1 27 = − + = , 7 5 7 7 5 49 =

E(X2)

2

σX

=

27 − 49

5 2 14

= 83 ; 196

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Chapter 4 Bivariate Distributions

39

∫1 µY = E(Y )

y·

= 0

10 y3

=

7

3

10 7 +

(y + y4) dy y6 1 6 0

10 1 1 5 + = 7 3 6 7 ∫ 1 10 y2 · (y + y4) dy E(Y 2) = 7 0 =

10 y4

=

σ2

7

7

=

10 1 1 + 7 4 7

=

55

−

98

Y

1

+ y 4 7 0

5 7

=

2

=

55 98 5

.

98

4.4–10 The area of the space is ∫ 6 ∫ 14−2t2

∫ 6 dt1dt2 =

2

1

(13 − 2t2) dt2 = 20;

2

Thus P (T1 + T2 > 10)

∫ 4 ∫ 14−2t2 1 dt1dt2 2 10−t2 20 ∫ 44 t −2 = dt2 20 2 4 (4 — t )2 1 2 = − = .

40 10 2

=

4.4–12 E[a1u1(X1, X2) + a2u2(X1, X2)] Σ Σ [a1u1(x1, x2) + a2u2(x1, x2)]f (x1, x2) = (x1,x2) ∈R

=

Σ Σ Σ Σ a1(x1, x2) ∈R u1(x1, x2)f (x1, x2) + a2(x1, x2) ∈R u2(x1, x2)f (x1, x2)

=

a1E[u1(X1, X2)] + a2E[u2(X1, X2)]. ∫ 1

4.4–14 (a)

fX (x) = (y) = fY

x

∫ y 0

8xy dy = 4x(1 − x2), 8xy dx = 4y3,

0 ≤ x ≤ 1,

0 ≤ y ≤ 1;

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40

Section 4.4 Bivariate Distributions of the Continuous Type (b)

µX

∫ 1

=

x4x(1− x2) dx =

0

∫ 1 = 0

∫ 1 σX

=

,

15

4

∗ µY

8

y 4y3 dy = 5,

0

11

∫ 1 (x − 8/15) 4x(1 − x ) dx = 225 , 2 Y — ∗ σ = 0 2 (y 4/5)2 4y3 dy = 75 , ∫ 1∫ 1 0 x (x − 8/15)(y − 4/5)8xy dy dx = Cov(X, Y ) = √ 2 66 4/225 ; = ρ = √ (11/225)(2/75) 33 2

20 (c) y = 4.4

33

+

4x 11

2

2

4 , 225

.

1 2 1 2 –16 From Example 4.4–3, µ = , µY = , and E(Y ) = ; X 3 3 2 ∫ 1 E(X2) =

1 1 1 2 , σ2 = − X 6 6 3

2x2(1 − x) dx =

0

=

1 18

, σ2 =

∫ 1∫ 1 – Cov(X, Y ) =

0

x

so

2xy dy dx

ρ= √

1 3

0

1 =

x/8,

0 ≤ x ≤ 2,

1/8 dy = 1/4,

2 < x < 4,

1/8 dy

=

2 −

4

2

9

−

2 2 3

=

1

;

18

1 =

36

,

1/36 1 √ = .

2 1/18 1/18

∫x

4.4–18 (b)

2 3

Y

1

∫x fX (x) = ∫

x−2 4 x−2

1/8 dy

=

(6 − x)/8,

4 ≤ x ≤ 6;

∫ y+2 (c) fY (y) =

1/8 dx = 1/4,

0 ≤ y ≤ 4;

y

(d)

h(y | x) = (e) g(x | y) = 1/2,

1/x,

0 ≤ y ≤ x,

0 ≤ x ≤ 2,

1/2,

x − 2 < y < x,

2 < x < 4,

1/(6 − x), x − 2 ≤ y ≤ 4,

4 ≤ x ≤ 6;

y ≤ x ≤ y + 2,

0 ≤ y ≤ 4;

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Chapter 4 Bivariate Distributions

41

∫x

(f)

∫x

(g) E(X | y) =

dy =

x

0

E(Y | x) =

x

1

y

2

∫ y+2 x·

y

1 2

x

y2

1 ·2

4

dy = y ∫ 4 dy = x−2 6 − x x−2 y

0 ≤ x ≤ 2,

,

dx =

= x − 1, 4 = x + 2,

x−2

y2

2(6 − x) x−2

4 ≤ x < 6;

2

x2 y+2 = y + 1, 4 y

y

2 < x < 4,

0 ≤ y ≤ 4.

y

4

4

3

3

2

2

1

1 x 1

2

3

4

5

x

6

1

2

Figure 4.4–18: (h) y = E(Y | x) 4.4–20

(x + 1)y

x+1

dy =

0

Thus

E(Y )

=

E[E(Y | X)] =

∫ 2 x+1 1

x+2

1

x + 1 x+2 y x+2

E(Y | X) = and

4

5

6

(i) x = E(X | y)

∫ 1

E(Y | x) =

3

= 0

x+1 x+2

,

1 < x < 2.

X +1 , X +2 ∫2

dx =

1−

1

dx

x+2

1

= [x − ln(x + 2)]2 1 = 2 − ln 4 − (1 − ln 3) = 1 − ln(4/3) = 0.7123.

4.5

The Bivariate Normal Distribution

4.5–2

q(x, y)

= =

2 [y − µY − ρ(σ + (x − 2µX )2 Y /σX )(x − µX )] σ2 (1 − 2 ρ) Y σX 2 1 (y − µY ) 2ρ(x − µX )(y − µY ) − 1 − ρ2 σX σY σ2Y2 ρ (x− µX )2 (x − µ )2 X 2 + + (1 − ρ ) σ2X " σ X2 1 x−µ 2 x− µ y−µ X

=

1−ρ

2

X

σ

#

y−µ

2

X

– 2ρ

X

σ

σY

Y

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Y

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42

Section 4.5 The Bivariate Normal Distribution 42.5

–4 (a) E(Y | X = 76) = 80 + 13 10 " # 5 2 (b) Var(Y | X = 76) = 169 1 − = 144; 13 (c) P (Y ≤ 86 | X = 76) = P Z ≤

5 13

(76 − 70) = 83;

86 − 83 = Φ(0.25) = 0.5987.

12

4.5–6 (a) P (19.9 < Y < 26.9) = Φ(1.2) − Φ(−0.8) = 0.8849 − 0.2119 = 0.6730; Note that using an electronic device, you may get 0.88493 − 0.21186 = 0.67307; (b) E(Y | x) = 22.7 + 0.78(3.5/4.2)(x − 22.7) = 0.65x + 7.945; (c) Var(Y | x) = 12.25(1 − 0.782) = 4.7971; (d) P (19.9 < Y < 26.9 | X = 23) = Φ(1.8286) − Φ(−1.3674) = 0.9663 − 0.0857 = 0.8806; (e) P (19.9 < Y < 26.9 | X = 25) = Φ(1.2350) − Φ(−1.9610) = 0.8914 − 0.0249 = 0.8665; (f) 0.16 0.12 0.08 0.04 20 x

18

25

26 22 y

30

14

Figure 4.5–6: Conditional pdfs of Y , given x = 21, 23, 25

4.5–8 (a) P (13.6 < Y < 17.8) = Φ(0.20) − Φ(−0.85) = 0.3816; (b) E(Y | x) = 17 + 0(4/3)(x − 8) = 17; (c) Var(Y | x) = 16(1 − 02) = 16; (d) P (13.6 < Y < 17.8 | X = 9.1) = 0.3816.

4.5–10 (a) P (2.80 ≤ Y ≤ 5.35) = Φ(1.50) − Φ(0) = 0.4332; (b) E(Y | X = 82.3) = 2.80 + (−0.57)

1.7

(82.3 − 72.30) = 1.877; 10.5

Var(Y | X = 82.3) = 2.89[1 − (−0.57)2] = 1.9510; P (2.76 ≤ Y ≤ 5.34 | X = 82.3) =

Φ(2.479) − Φ(0.632) = 0.9934 − 0.7363 = 0.2571.

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Chapter 4 Bivariate Distributions

43

∫ ∞

∫ ∞ 2 2 1 1 −(x2+y 2)/2 2 2 4.5–12 fX (x) = (x +y )/2 (x +y −2)/2 dy − − e dy + e xye −∞ 2π −∞ 2π ∫∞ 1 −y2/2 −2x /2 1 √2π e dy + 0 = √2π e − 2 −∞ = √1 e x /2, −∞ < x < ∞.

2π Note that the first integrand is the product of two N (0, 1) pdfs and the integral of a pdf is equal to 1. The second integral is an odd function so it is equal to 0.

1 − 2 e y /2 , Similarly, fY (y) = √ −∞ < y < ∞.

2∞π ∞ ∫ ∫ ∞ It also follows that f (x, y) dx dy = ∫ f (x) dx = 1. −∞ −∞ −∞ X

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Chapter 5 Distributions of Functions of Random Variables

45

Chapter 5

Distributions of Functions of Random Variables 5.1

Functions of One Random Variable

5.1–2 The linear transformation is one-to-one with inverse transformation X = (Y − b)/a, so the pdf of Y is 1 g(y) = f [(y − b)/a], ac1 + b < y < ac2 + b. |a| 5.1–4 Here x =

√

y,

dx = dy

1 √ 2 y

and 0 < x < ∞ maps onto 0 < y < ∞. Thus

g(y) =

√ 1 . 1 −y/2 = e , y e−y/2 . √ .2 y.

2

0 < y < ∞.

5.1 –6 (a)

0, F (x) =

x < 0, ≤

0

2

2t dt = x , 1,

0 x < 1, 1 ≤ x;

√ (b) Let y = x2; so x = √y. Let Y be U (0, 1); then X = Y has the given x-distribution; (c) Repeat the procedure outlined in part (b) ten times.

5.1–8 The cdf of X is 1 F (x) =

1 + e −x

,

−∞ < x < ∞.

By Theorem 5.1-2, 1

Y = F (X) =

1 + e−X is U (0, 1).

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46

Section 5.1 Functions of One Random Variable 5.1–10

x dx dy f (x)

= = =

y 10/7 5 10 y 3/7 1 7 5 5 x − e , 0<x<∞ 2

10 7

g(y) = e−(y/5) 10/7

7

/

1 3/7

5

y3/7

y3/7e −(y/5)10/7 ,

0 < y < ∞.

510/7 (The reason for writing the pdf in that form is because Y has a Weibull distribution with α = 10/7 and β = 5.) =

5.1–12 X has cdf F given by x < −1,

0 x+1 F (x) =

4

−1 ≤ x ≤ 3

,

1,

3 < x.

The cdf G of Y is given by √ √ G(y) = P (Y ≤ y) = P (− y ≤ X ≤ y). √ √ Taking cases then gives G(y) = y/2 for 0 ≤ y ≤ 1 and G(y) = ( y + 1)/4 for 1 ≤ y ≤ 9.

Taking derivatives then gives the pdf

g(y) = ∫∞ 5.1–14

E(X)

= =

1 √ , 4 y 1 √ , 8 y

1 ≤ y < 9.

x

2 −∞ π(1 + x )

1

dx

1

lim a→−∞

=

0 < y < 1,

2π

2

ln(1 + x )

0

1

+ lim

a

lim {− ln(1 + a )} + lim a→−∞

b

2π

b→+∞

2

2π

ln(1 + x2)

0 2

ln(1 + b ) .

→b+∞

E(X) does not exist because neither of these limits exists.

5.1–16 The cdf of Y = |X| is given by G(y) =

P (Y ≤ y) = P (−y ≤ X ≤ y)

= Φ(y) − Φ(−y) = 2Φ(y) − 1,

0 < y < ∞,

where Φ is the standard normal cdf. Taking derivatives then gives the pdf g(y) = 2φ(y) = √

2 e 2π

− 2 y /2

,

0 < y < ∞.

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Chapter 5 Distributions of Functions of Random Variables

47

–18 (a) First note that the cdf of X is F (x) = (x2 + 2x + 1)/9 , −1 < x < 2. Then √ √ √ √ G(y) = P (X2 < y) = P (− y < X < y ) = F ( y ) − F (− y ) √ √ √ ( [(y + 2 y + 1)/9] − [y − 2 y + 1)/9] = 4 y/9, 0 < y < 1, √ √ = 1 ≤ y < 4. (y + 2 y + 1)/9 − 0 = (y + 2 y + 1)/9,

5.1

Thus g(y) = G′(y) =

2 √ , 9 y

0 < y < 1,

1 1 + √ , 9 9 y

1 ≤ y < 4;

(b) First note that the cdf of X is F (x) = (x2 + 4x + 4)/9 , −2 < x < 1. Then √ √ √ √ G(y) = P (X2 < y) = P (− y < X < y ) = F ( y ) − F (− y ) √ √ √ ( [(y + 4 y + 4)/9] − [(y −4 y + 4)/9] = 8 y/9, 0 < y < 1, = √ 1 − [(y − 4 y + 4)/9], 1 ≤ y < 4.

4 √ , 9 y 2 – 1 + √ , 9 9 y

Thus g(y) = G′(y) =

5.2

0 < y < 1, 1 ≤ y < 4.

Transformations of Two Random Variables

5.2-2 (a) The joint pdf of X1 and X2 is 1 r1/2−1 r2/2−1 −(x1+x2)/2 x2 e , r1 r2 2(r1+r2)/2 x1 Γ 2 Γ 2

f (x , 1x ) 2=

0 < x1 < ∞, 0 < x2 < ∞.

Let Y1 = (X1/r1)/(X2/r2) and Y2 = X2. The Jacobian of the transformation is (r1/r2)y2. Thus r y1 y1 2 r1/2−1

1 g(y1, y2) =

r2

r2

2(r1+r2)/2

2

r Γ 1 Γ 2

−(y2/2)(r1y1/r2+1) r2/2−1

y2

e

r y 1r 2

,

2

0 < y1 < ∞, 0 < y2 < ∞; ∫ ∞ g(y 1, y 2) dy2. (b) The marginal pdf of Y1 is g1 (y1 ) = 0

Make the change of variables w =

y2 r1y1 2

g1(y1) =

r1 + r2 2

Γ r

r Γ Γ 2 2 2

r1 r2 1+

1

5.2-4 (a) F0.05(9, 24) = 2.30; (b) F0.95(9, 24) =

1

F 0.05(24, 9)

=

r1/2 1 1

r r2y

1 2.90

+ 1 . Then

r2

y1r1/2−1 (r1+r2)/2

· 1,

0 < y1 < ∞ .

= 0.3448;

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48

Section 5.2 Transformations of Two Random Variables (c) P (W < 0.277) = P

1

1

>

1

=P

> 3.61 = 0.025; W 0.277 W P (0.277 ≤ W ≤ 2.70) = P (W ≤ 2.70) − P (W ≤ 0.277) = 0.975 − 0.025 = 0.95.

5.2-6

F (w)

= =

P

X1

′

=

xα−1xβ−1e−(x1+x2)/θ

2 dx2dx1 Γ(α)Γ(β)θα+β ∫ ∞ −xα−1[(1 − w)x /w]β−1 e −[x1 +(1−w)x1 /w]/θ −1

(1−w)x1/w

1

1

0

1 =

0<w<1

1

0

f (w) = F (w)

w≤ ,

X1 + X2 ∫ ∞∫ ∞

Γ(α)Γ(β)

Γ(α)Γ(β)θα+β ∫ ∞ α+β−1 −x1/θw x e (1 − w)β−1 1

wβ+1

0

θα+β

w2

x1 dx1

dx 1

Γ(α + β) (θw)α+β (1 − w)β−1 θα+β Γ(α + β) wβ+1 Γ(α + β) α−1 = w (1 − w)β−1, 0 < w < 1. Γ(α)Γ(β) ∫ 1 Γ(α + β) x 5.2-8 (a) E(X) = xα−1(1 − x)β−1 dx Γ(α)Γ(β) 0 Γ(α + β)Γ(α + 1) ∫ 1 Γ(α + 1 + β) = . xα+1−1(1 − x)β−1 dx Γ(α)Γ(α + β + 1) 0 Γ(α + 1)Γ(β) (α)Γ(α)Γ(α + β) = (α + β)Γ(α + β)Γ(α) α ; = α+β ∫ Γ(α + β)Γ(α + 2) 1 Γ(α + 2 + β) E(X2) = xα+2−1(1 − x)β−1 dx Γ(α)Γ(α + 2 + β) 0 Γ(α + 2)Γ(β) (α + 1)α . = (α + β + 1)(α + β) Thus α(α + 1) α2 αβ 2 σ = ; − = 2 (α + β + 1)(α + β) (α + β)2 (α + β + 1)(α + β) Γ(α + β) α−1 (b) f (x) = x (1 − x)β−1 Γ(α)Γ(β) Γ(α + β) f ′(x) = (α 1) xα−2 (1 x)β−1 (β 1) xα−1 (1 x)β−2 −. Γ(α)Γ(β) − − − − =

Set f ′(x) equal to zero and solve for x: Γ(α + β) α−2 x (1 — x)β−2 [(α − 1)(1− x) − (β − 1) x] =0 Γ(α)Γ(β) α − αx − 1 + x − βx + x = 0 (α + β − 2)x = α − 1 α−1 x = . α + β– 2

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Chapter 5 Distributions of Functions of Random Variables

49

5.2-10 Use integration by parts two times to show ∫ p p 6! y3(1 − y)2 dy = 6 y4(1 − y)2 + 6 y5(1 − y)1 + 6 y6(1 − y)0 5 6 4 0 3!2!

0 6 Σ n y p (1 − p)6−y. y y=4

= 5.2-12 (a) w1 = 2x1 and

dw1

= 2. Thus

dx1 2 f (x1) =

π(1 + 4x21)

,

−∞ < x1 < ∞;

(b) For x2 = y1 − y2, x1 = y2, | J | = 1. Thus g(y1, y2) = f (y2)f (y1 − y2), (c) g1(y1) = 1

∫∞ −∞

1

(d) g (y ) = = = =

−∞ < yi < ∞, i = 1, 2;

f (y 2)f (y 1− y )2dy ;2 ∫∞ 2 2 · h(y ) dy dy = π[1 + 4y22] π[1 + 4(y1 – y2)2] ∫ ∞ −∞

2

4 ∫∞

−∞

2

2

1 1 · dy2 − 2i(y 1− y )] π2∫ −∞ [1 + 2iy2][1 − 2iy2] [1 + 2i(y 1− y )][1 2 2 ∞ 4 1 1 −1 1 −1 −1 1 1 2 · y2 + i · 2i · y2 — (y —i ) · 2i · y — (y + i ) dy 2 π −∞ 2i · y — i · 2i 2 2 1 2 1 2 2 2 4(2πi) i i Res h(y ); y = + Res h(y ); y = y + 2 2 2 2 1 π2 2 2 8πi 1

1

1

1

1

1 1 · · π 16 i i − y1 −y1 y1 y1 + i i 1 1 1 1 y1 + i + y1 − i 1 = · 1 · + = 2π y1 y1 − i y1 + i 2π y1 (y1 − i)(y1 + i) 1 = . π(1 + y21) =

2

·

·

+

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50

Section 5.2 Transformations of Two Random Variables A Maple solution for Exercise 5.2-12: >f := x-> 2/Pi/(1 + 4*x^2); 1 f := x− > 2

π (1 + 4 x2) >simplify(int(f(y[2])*f(y[1]-y[2]),y[2]=-infinity..infinity)); 1 π (1 + y12) A Mathematica solution for Exercise 5.2-12: In[1]:= f[x_] := 2/(Pi*(1 + 4(x)^2)) g[y1_,y2_] := f[y2]*f[y1-y2] In[3]:= Integrate[g[y1,y2], {y2, -Infinity,Infinity}] Out[3]= 1 2 Pi + Pi y1 5.2-14 The joint pdf is x −(x+y)/5 e , 53

h(x, y) = z

=

x

=

x

, y zw,

0 < x < ∞, 0 < y < ∞;

w = y y

= w

The Jacobian is J

The joint pdf of Z and W is zw f (z, w) = 3 e−(z+1)w/5 w, 5

=

w z . . = w. . 0 1 .

0 < z < ∞, 0 < w < ∞;

The marginal pdf of Z is f Z(z)

=

∫ ∞ zw 53

0

= =

e−(z+1)w/5 w dw 3

Γ(3)z

5

53 2z

z+1

(z + 1)3

,

∫ ∞ 0

w3−1 Γ(3)(5/[z + 1])3

e−w/(5/[z+1]) dw

0 < z < ∞.

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Chapter 5 Distributions of Functions of Random Variables

51

5.2-16 Y = 1/W is a one-to-one transformation from S W = {w : w > 0} to SY = {y : y > 0}, with dw 1 inverse W = g−1(Y ) = 1/Y and derivative (of the inverse transformation) = − 2. dy y Thus, the pdf of Y is fY (y)

dw fW [g−1(y)] .. . .dy r1/2 r1 /2−1 1 Γ[(r (r1/r2) (1/y) 1 + r2)/2] .− . = Γ(r1 /2)Γ(r2/2) [1 + (r1 /r2 )(1/y)](r1+r2)/2 . y2 . ! Γ[(r1 + r 2)/2] rr11/2 (1/y)r1 /2−1(r +r )/2 (1/y2) = r1/2 [(r1 + r2y)/(r2y)] 1 2 Γ(r1/2)Γ(r2/2) r 2 ! r1/2 r2/2 (r +r )/2 ! r1/2 Γ[(r1+ r )2/2] r r 1 2 y r 2 2 1 = y−1−r1/2 rr1/2 Γ(r1/2)Γ(r2/2) (r1 + r2y)(r1+r2)/2 2 Γ[(r1+ r )2/2] ! = Γ(r /2)Γ(r /2) 1r /2 − − r2r2/2 r1/2)+(r1+r2)/2 ( 1 1 2 1 r y [1 + (r2 /r1 )y](r1+r2)/2rr11/2rr21/2

=

=

Γ[(r1 + r2)/2] (r2/r1)r2/2yr2/2−1 , Γ(r /2)Γ(r /2) [1 + (r /r )y](r1+r2)/2 1

2

2

y > 0,

1

which is the pdf of an F (r2, r1) distribution.

You could also note by Theorem 5.2-1 that the F (r1, r2) distribution of W is the distribution of (U/r1)/(V /r2) where U and V are independent chi-square variables with r1 and r2 degrees of freedom, respectively. It follows that 1/W is the distribution of (V /r2)/(U/r1).

By the same theorem, this is the F (r2, r1) distribution.

5.2-17 The pdf of W is f (w) =

(r1/r2)r1/2Γ[(r1 + r2)/2]wr1 /2−1 . Γ(r /2)Γ(r /2)[1 + (r w/r )](r1+r2)/2 1

2

1

2

Let α = r1/2 and β = r2/2. Then the pdf becomes f (w) =

(α/β)αΓ[(α + β]wα−1 Γ(α)Γ(β)[1 + αw/β](α+β)

.

We are given 1

z =

1 + αw/β β β 1 w = −1 = α z α dw β 1 = − dz α z2

1−z z

It follows that the pdf of Z is g(z) = =

Γ(α + β)

α

Γ(α)Γ(β)

β

Γ(α + β) Γ(α)Γ(β)

α

β 1−z α

(1 − z)α−1zβ−1,

α−1

zα+β

z

0 < z < 1.

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52

Section 5.3 Several Random Variables

5.3

Several Random Variables "

5.3–2 (a) P (X1 = 2, X2 = 4)

=

#"

1 2 1 1

3!

# 5!

1 4 1 1

2!1! 2 2 4!1! 2 2 15 15 ; = = 28 256 (b) {X1 + X2 = 7} can occur in the two mutually exclusive ways: {X1 = 3, X2 = 4} and {X1 = 2, X2 = 5}. The sum of the probabilities of the two latter events is " #" # " #" # 3! 1 3 5! 1 5 3! 1 3 5! 1 5 5+3 1 + = = .

5!0! 2 3!0! 2 4!1! 2 2!1! 2 28 32 ∫ 1.2

∫ 1.0 2e

5.3 –4 (a)

−2x1

2e−2x2 dx2

dx1

0.5

0.7

= (e −1 − e −2)(e −1.4 − e−2.4) =

(0.368 − 0.135)(0.247 − 0.091)

= (0.233)(0.156) = 0.036; (b) E(X1) = E(X2) = 0.5, E[X1(X2 − 0.5)2] = E(X1)Var(X2) = (0.5)(0.25) = 0.125. ∫ 1 5.3–6

E(X)

∫ 1 x6x(1− x )dx =

= 0

∫ 1 E(X2)

=

(6x2− 6x3) dx =

0

3

(6x3− 6x4) dx =

=

Y

6

x5

5

1 0

3 2

x4

1 0

=

1

;

2

= 3.

10

1

=

X

µ

x4 −

2

0

Thus µ

2x3 −

; σ2 = 3 − 1 = 1 , and X 2 10 4 20 1 1 1 2 + = 1; σ = + 1 = 1.

Y 2 2 20 20 10

5.3 –8 Let Y = max(X1, X2). Then

G(y) =

[P (X ≤ y)]2 ∫y 4

=

2

dx x5 1 2 , 1 − y4 1

=

1<y<∞

g(y) = G′(y) = 2 1− ∫∞ E(Y )

= 1 ∫ ∞

=

1

4

y4

y5

y·2 1−

1 < y < ∞;

, 1

4

y4

y5

dy

8 [y−4 − y−8] dy

1

=

32

(in 1000s of dollars).

21

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Chapter 5 Distributions of Functions of Random Variables

53 "

3

3

1

2#

3

= 27 ; 1024 (b) 3P (X1 = 3, X2 = 1, X3 = 1) + 3P (X1 = 2, X2 = 2, X3 = 1) =

5.3–10 (a) P (X1 = 1)P (X2 = 3)P (X3 = 1) =

4

4

27 1024

3 3 3 1 + · 4 4 4

3

3

15 . = 16 ∫∞ 5.3–12 P (1 < min Xi) = [P (1 < Xi)]3 = e−x dx (c) P (Y ≤ 2) =

4

4

+3

27 1024

=

162 ; 1024

3

= e−3 = 0.05.

1

5.3–14

P (Y > 1000) = P (X1 > 1000)P (X2 > 1000)P (X3 > 1000) = e−1e−2/3e−1/2 = e−13/6 = 0.1146.

5.3 –16 Denote the three lifetimes by X1, X2, X3 and let Y = X1 + X2 + X3.

E(Y ) = E(X1 + X2 + X3) = E(X1) + E(X2) + E(X3) = 3 · 2 · 2 = 12.

Var(X1 + X2 + X3) = Var(X1) + Var(X2) + Var(X3) = 3 · 2 · 22 = 24.

Cov(W, V ) 5.3 –18

ρ = = = = = =

σW σV E(WV ) − µW µV σW σV E(X2)E(Y ) − E(X)E(Y )E(X) σXY σX (σ2 + µ2 )(µY ) − µ2 µY X √ E(XX2Y 2) X [E(X)E(Y )]2 σX − µY σX √ (σ2X + µ2X)(σ2Y + µ2Y) − µ2Xµ2Y µY σX √ . σ2X σ2Y + σ2Xµ2Y + σ2Yµ2 X

5.3–20 Var( X ) =

Σ n i=1

=

σ2 n

1 n

2

σ2 + 2

ΣΣ

1 n

1 n

ρσ2 =

σ2 n

i<j

n(n − 1)

2 +

2

n

ρσ2

2

[1 + (n − 1)ρ].

Since 1 + (n − 1)ρ > 1, this variance is larger than the variance of X when the variables are uncorrelated.

5.4

The Moment-Generating Function Technique

5.4 –2 MY (t)

=

E[et(X1 +X2 )] = E[etX1 ]E[etX2 ]

=

(q + pet)n1 (q + pet)n2 = (q + pet)n1+n2 ,

−∞ < t < ∞.

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54

Section 5.4 The Moment-Generating Function Technique Thus Y is b(n1 + n2, p).

5.4–4

E[et(X1 +···+Xn )]

n

n

i=1

i=1

Y Y t E[etXi ] = eµi (e −1)

=

t −∞ < t < ∞, = e(µ1+µ2+···+µn)(e −1), the moment generating function of a Poisson random variable with mean µ1 + µ2 + · · · + µn.

5.4–6 (a)

E[etY ]

E[et(X1+X2+X3+X4+X5)]

= =

E[etX1 etX2 etX3 etX4 etX5 ]

=

E[etX1 ]E[etX2 ]E[etX3 ]E[etX4 ]E[etX5 ] (1/3)et (1/3)et (1/3)et ··· 1 − (2/3)et 1 − (2/3)et 1 − (2/3)et 5 (1/3)et

= = =

1 − (2/3)e t [(1/3)et]5 [1 − (2/3)et]5

,

t < − ln(1 − 1/3);

(b) So Y has a negative binomial distribution with p = 1/3 and r = 5.

5.4–8 E[etW ] =

E[et(X1+X2 +···+Xh )] = E[etX1 ]E[etX2 ] · · · E[etXh ]

= [1/(1 − θt)]h = 1/(1 − θt)h,

t < 1/θ,

the moment generating function for the gamma distribution with mean hθ.

5.4 –10

(a) E[etX ] = (1/4)(e0t + e1t + e2t + e3t), tY

(b) E[e ] = (1/4)(e (c)

E[e

tW

] =

0t

E[e

+e

4t

+e

8t

12t

+ e ),

−∞ < t < ∞; −∞ < t < ∞;

t(X+Y )

]

= E[etX ]E[etY ] = (1/16)(e0t + e1t + e2t + e3t)(e0t + e4t + e8t + e12t) =

(1/16)(e0t + e1t + e2t + e3t + · · · e15t),

(d) P (W = w) = 1/16,

−∞ < t < ∞;

w = 0, 1, 2, . . . , 15.

5.4–12 First die has 0 on four faces and 2 on four faces; second die has faces numbered 0, 1, 4, 5, 8, 9, 12, 13.

5.4–14 Let X1, X2, X3 be the number of accidents in weeks 1, 2, and 3, respectively. Then Y = X1 + X2 + X3 is Poisson with mean λ = 6 and P (Y = 7) = 0.744 − 0.606 = 0.138.

5.4–16 Let X1, X2, X3, X4 be the number of sick days for employee i, i = 1, 2, 3, 4, respectively.

Then Y = X1 + X2 + X3 + X4 is Poisson with mean λ = 8 and P (Y > 10) = 1 − P (Y ≤ 10) = 1 − 0.816 = 0.184.

5.4–18 Let Xi equal the number of cracks in mile i, i = 1, 2, . . . , 40. Then Y =

40 Σ

Xi

is Poisson with mean λ = 20.

i=1

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Chapter 5 Distributions of Functions of Random Variables It follows that P (Y < 15) = P (Y ≤ 14) =

55

14

20ye−20 = 0.1049. y! y=0

5.4 –20 Y = X1 + X2 + X3 + X4 has a gamma distribution with α = 6 and θ = 10. So

∫ ∞

P (Y > 90) = 90

5.4 –22 (a)

E(etY )

1 Γ(6)106

y6−1e−y/10 dy = 1 − 0.8843 = 0.1157.

=

E(etX1 etX2 )

=

E(etX1 )E(etX2 ) 1 E(etX2 ) (1 − 2t)r1/2 1 ; (1 − 2t)(r−r1)/2

1 (1 − 2t)r/2

=

E(etX2 )

=

(b) X2 is χ2(r−r1).

5.4 –24 (a)

MX (t) =

4 1 [1 − 2(t/4)]r/2

=

1 [1 − (1/2)t]2r

,

t < 2;

(b) Gamma with α = 2r and θ = 1/2.

5.5

Random Functions Associated with Normal Distributions

5.5 –2

0.4

0.3 n = 36 0.2 n=9 0.1 n=1 40

50

60

Figure 5.5–2: X is N (50, 36), X is N (50, 36/n), n = 9, 36 5.5 –4 (a) P (X < 6.0171) = P (Z < −1.645) = 0.05;

(b) Let W equal the number of boxes that weigh less than 6.0171 pounds. Then W is b(9, 0.05) and P (W ≤ 2) = 0.9916; 6.035 − 6.05 (c) P ( X ≤ 6.035) = P Z ≤ 0.02/3 = P (Z ≤ −2.25) = 0.0122.

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56

Section 5.5 Random Functions Associated with Normal Distributions –6 (a) X is N (µ, 4/100) and Y is N (µ, 9/n). Equating 4/100 to 9/n yields n = 225; (b) Y − X is N [µ − µ, (4/100) + (9/225)], i.e, N (0, 0.08). So 0.2 Z>√ = P (Z > 0.71) = 0.2389.

P ( Y − X > 0.2) = P 0.08 5.5 –8 X − Y is N (184.09 − 171.93, 39.37 + 50.88);

X − Y − 12.16 0 − 12.16 √ > 9.5 90.25

P (X > Y ) = P

= P (Z > −1.28) = 0.8997.

5.5–10 Let Y = X1 + X2 + · · · + Xn. Then Y is N (800n, 1002n). Thus P (Y ≥ 10000) = 0.90 Y – 800n 10000 – 800n P √ ≥ √ = 0.90 100 n 100 n 10000 − 800n −1.282 = √ 100 n √ 800n − 128.2 n − 10000 = 0. √ Either use the quadratic formula to solve for n or use Maple to solve for n. We find that √ n = 3.617 or n = 13.08 so use n = 14 bulbs.

5.5 –12 (a) The joint pdf of X1 and X2 is

f (x , x ) = 1

2

1

√

2

1

x1/2

Γ(r/2)2r/2

e−

2π

√ x2/r,

y1 =

x1/

x1

y1√y2/r,

=

−∞ < x < ∞, 0 < x < ∞;

xr/2−1e−x2/2,

1

2

y2 =

x2

x2

y2

=

2

The Jacobian is J

=

.

√ y2/r

.

0

The joint pdf of Y1 and Y2 is 1 —2 g(y1, y2) = √e 2

1

2π

y1 y /2r Γ(r/2)2r/2

y1( 12 )y−21/2/

√

1

r/2−1

y 2

−y2/2

e

r . .

=

√ y2 √ , r

√ y2/r.

−∞ < y1 < ∞, 0 < y2 < ∞;

(b) The marginal pdf of Y1 is ∫ ∞ √y 1 1 r /2−1 g1(y1) = √ e −y12y2/2r e−y2/2 √ 2 dy2 y r Γ(r/2)2r/2 2 0 2π ∫ ∞ 1 Γ [(r + 1)/2] (r+1)/2−1 − 2(y /2)(1+y2 /r) 1 = √ y e . πr Γ(r/2) 0 Γ[(r + 1)/2]2(r+1)/2 2 u dy2 1 Let u = y (1 + y2/r). Then y = and = . So 2 1 2 1 + y21/r du 1 + y12/r

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Chapter 5 Distributions of Functions of Random Variables g (y ) 1

=

1

=

√

Γ[(r + 1)/2]

πr Γ(r/2)(1 + y12/r)(r+1)/2 Γ[(r + 1)/2] √ , πr Γ(r/2)(1 + y2/r)(r+1)/2

57 ∫ ∞ 0

1

u(r+1)/2−1 e−u/2

Γ[(r + 1)/2]2(r+1)/2 −∞ < y1 < ∞.

1

5.5 –14 Because Z is N (0, 1), E(Z) = 1 and E(Z2) = 1. U is χ2(r) so it follows that

√ E[1/ U] =

= =

∫ ∞ 1 √

1 r/2−1 e−u/2 du r/2 u u Γ(r/2)2 0 ∫ Γ[(r − 1)/2] ∞ 1

√ 2 Γ(r/2) Γ[ (r − 1)/2] √ .

2 Γ(r/2)

0

Γ[(r − 1)/2]2(r−1)/2

−u/2

(r−1)/2−1

u e

du

Note that the last integral is equal to one because the integrand is the pdf of a χ2(r−1) random variable.

To find E(1/U ) we have ∫ ∞ 1 1 ur/2−1 e−u/2 du E[1/U ] = u Γ(r/2)2r/2 0 ∫ Γ[(r − 2)/2] ∞ 1 = u(r−2)/2−1 e−u/2 du (r−2)/2 2 Γ(r/2) Γ[(r − 2)/2]2 0 1 Γ(r/2 − 1) . = = r−2 2Γ(r/2 − 1) (r/2 − 1) Note that the last integral is equal to one because the integrand is the pdf of a χ2(r−2) random variable. " # Z E(T ) = E √ U/r √ = E[Z]E[1/ U/r] √ r√ Γ[(r − 1)/2] = 0 provided r ≥ 2; = 0 2 Γ(r/2) Var(T ) = E(T 2) − 02 = =

5.5 –16 T =

E[Z2]E[r/U ] r , provided r ≥ 3. r− 2

X− µ √ is t with r = 9 − 1 = 8 degrees of freedom.

S/ 9

(a) t0.025(8) = 2.306.

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58

Section 5.6 The Central Limit Theorem ≤

−t

(b)

0.025

S −t0.025 √ n S −X − t0.025 √ n S X − t0.025 √ n

5.6

X− µ √

≤

0.025

S/ n

S ≤ t0.025 √ n

≤ X −µ ≤

−µ

≤

µ

t

S ≤ −X + t0.025 √ n S ≤ X + t0.025 √ . n

The Central Limit Theorem

5.6 –2 If

(3/2)x2, −1 < x < 1, ∫1 E(X) = x(3/2)x2 dx = 0; −1 ∫ 1 3 5 1 (3/2)x4 dx = x Var(X) = 10 −1 −1 f (x)

=

= 3.

5

−0.3 − 0 ≤ √Y − 0 ≤ √1.5 − 0 √ 15(3/5) 15(3/5) 15(3/5)

Thus P (−0.3 ≤ Y ≤ 1.5) = P

!

≈ P (−0.10 ≤ Z ≤ 0.50) = 0.2313.

39.75 − 40 –4

P (39.75 ≤ X ≤ 41.25)

≤ √ ≤√ (8/32) (8/32) (8/32) ≈ P (−0.50 ≤ Z ≤ 2.50) = 0.6853.

=

P

√

∫2 5.6–6 (a)

µ

x2

x(1− x/2) dx =

=

σ2 =

0

2

3

2 3 4 2 3 x 4 2 x — = ; − 8 9 9 3 0

x2(1 − x/2) dx −

=

(b) P

−

2

0

∫ 2

2

5 ≤X≤

6

!

41.25 − 40

X − 40

=

–

2

6

= 2−

0

4

2 = ; 3 3

2

X− 2

2

5

3

3 9 P q 2

x3 2

9

3

≤ q2

≤ q

/18 /18 ≈ P (0 ≤ Z ≤ 1.5) = 0.4332.

9 2

3

/18

5.6–8 (a) E(X) = µ = 24.43; σ2 2.20 (b) Var(X) = = = 0.0733; n 30 (c) P (24.17 ≤ X ≤ 24.82)

≈

P

24.√ 17 − 24.43 24.√ 82 − 24.43 ≤Z ≤ 0.0733 0.0733

= P (−0.96 ≤ Z < 1.44) = 0.7566.

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Chapter 5 Distributions of Functions of Random Variables 5.6 –10

E(X + Y ) = Var(X + Y )

59

30 + 50 = 80; σ2 + σ2 + 2ρσX σY

=

X

Y

= 52 + 64 + 28 = 144; Z

25 Σ

=

(Xi + Yi) is approximately N (25 · 80, 25 · 144).

i=1

Thus P (1970 < Z < 2090)

=

P

1970 − 2000 Z − 2000 2090 − 2000 < < 60 60 60

≈ Φ(1.5) − Φ(−0.5) = 0.9332 − 0.3085 = 0.6247.

–12 Let Xi equal the time between sales of ticket i − Σ 1 and i, for i = 1, 2, . . . , 10. Each Xi has 10 a gamma distribution with α = 3, θ = 2. Y = i=1 Xi has a gamma distribution with parameters αY = 30, θY = 2.

5.6

(a) Thus P (Y ≤ 60) =

∫ 60 0

1 y30−1e−y/2 dy = 0.52428 using Maple; Γ(30)230

(b) The normal approximation is given by P –14 We are given that Y =

5.6

Y − 60 6 0 − 60 √ ≤ √ 120 120

≈ Φ(0) = 0.5000.

Σ20 i=1 Xi has mean 200 and variance 80. We want to find y so that

P (Y > y) = P

Y − 200 y − 200 √ > √ < 0.20.

80 80

We have that y − 200

= 0.842

80 y

5.7

= 207.5 ↑ 208 days.

Approximations for Discrete Distributions 5.7 –2 (a) P (2 < X < 9) = 0.9532 − 0.0982 = 0.8550;

!

X − 25(0.2) 8.5 − 5 = P ≤√ ≤ 2 2 25(0.2)(0.8) ≈ P (−1.25 ≤ Z ≤ 1.75) 2.5 − 5

(b)

P (2 < X < 9)

= 0.8543.

5.7 –4 P (35 ≤ X ≤ 40)

≈

34.5 − 36 ≤ Z ≤ 40.5 − 36 3 3 = P (−0.50 ≤ Z ≤ 1.50) = 0.6247.

P

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60

Section 5.7 Approximations for Discrete Distributions 5.7–6 µX = 84(0.7) = 58.8, Var(X) = 84(0.7)(0.3) = 17.64, 52.5 − 58.8 P (X ≤ 52.5) ≈ Φ = Φ(−1.5) = 0.0668.

4.2 5.7–8 (a) P (X < 20.857)

=

X − 21.37

< 20.857 − 21.37 0.4 0.4 = P (Z < −1.282) = 0.10; P

(b) The distribution of Y is b(100, 0.10). Thus P (Y ≤ 5) = P

Y 100(0.10) √ − ≤ 5.5 − 10 100(0.10)(0.90) 3

≈ P (Z ≤ −1.50) = 0.0668;

21.31 − 21.37 ≤ Z ≤ 21.39 − 21.37 0.4/10 0.4/10 = P (−1.50 ≤ Z ≤ 0.50) = 0.6247.

(c) P (21.31 ≤ X ≤ 21.39)

≈

P

5.7–10 The distribution of Y is b(1000, 18/38). Thus P (Y > 500) ≈ P

!

500.5 − 1000(18/38) Z≥ √ 1000(18/38)(20/38)

! = P (Z ≥ 1.698) = 0.0448.

5.7–12 (a) E(X) = 100(0.1) = 10, Var(X) = 9, P (11.5 < X < 14.5)

14.5 − 10

11.5 − 10 –Φ 3 3 = Φ(1.5) − Φ(0.5) = 0.9332 − 0.6915 = 0.2417;

≈

Φ

(b) P (X ≤ 14) − P (X ≤ 11) = 0.917 − 0.697 = 0.220; 14

(c) x=12

100 (0.1)x(0.9)100−x = 0.2244. x

5.7–14 (a) E(Y ) = 24(3.5) = 84, Var(Y ) = 24(35/12) = 70, 85√.5 − 84 = 1 − Φ(0.18) = 0.4286; 70 (b) P (Y < 85.5) ≈ 1 − 0.4286 = 0.5714; P (Y ≥ 85.5) ≈ 1 − Φ

(c) P (70.5 < Y < 86.5) ≈ Φ(0.30) − Φ(−1.61) = 0.6179 − 0.0537 = 0.5642.

5.7–16 (a) See Figure 5.7-16. (b) (i) When p = 0.1, 1.5 P (−1.5 < Y − 10 < 1.5) ≈ Φ 3 (ii) When p = 0.5,

–Φ

1.5

–Φ

P (−1.5 < Y − 50 < 1.5) ≈ Φ (iii) When p = 0.8,

5

−1.5

= 0.6915 − 0.3085 = 0.3830;

3 −1.5

= 0.6179 − 0.3821 = 0.2358;

5

P (−1.5 < Y − 80 < 1.5) ≈ Φ 1.5 − Φ −1.5 4 4

= 0.6462 − 0.3538 = 0.2924.

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Chapter 5 Distributions of Functions of Random Variables

0.12

12

0.10

10

0.08

8

0.06

6

0.04

4

0.02

2 10

20

30

40

50

60

70

80

90 100

0

61

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

Figure 5.7–16: Normal approximations of the pdfs of Y and Y/100, p = 0.1, 0.5, 0.8 –18 X is N (0, 0.52). The probability that one item exceeds 0.98 in absolute value is

5.7

P (|X| > 0.98) =

1 − P (−0.98 ≤ X ≤ 0.98) = 1 − P −0.98 − 0 ≤X − 0 ≤ 0.98 − 0 0.5 0.5 0.5 = 1 − P (−1.96 ≤ Z ≤ 1.96) = 1 − 0.95 = 0.05

If we let Y equal the number out of 100 that exceed 0.98 in absolute value, Y is b(100, 0.05). (a) Let λ = 100(0.05) = 5.

(b)

P (Y ≥ 7)

P (Y ≥ 7) = 1 − P (Y ≤ 6) = 1 − 0.762 = 0.238; !

Y−5 6.5 − 5 P √ ≥ 2.179 100(0.05)(0.95)

=

≈ P (Z ≥ 0.688) = 1 − 0.7543 = 0.2447; (c) P (Y ≥ 7) = 1 − P (Y ≤ 6) = 1 − 0.7660 = 0.2340.

5.8

Chebyshev’s Inequality and Convergence in Probability 5.8 –2 Var(X) = 298 − 172 = 9.

(a)

P (10 < X < 24) =

because k = 7/3;

P (10 − 17 < X − 17 < 24 − 17) 9 40 = P ( |X − 17| < 7) ≥ 1 − = , 49 49

(b) P ( |X − 17| ≥ 16) ≤ 5.8 –4 (a) P

9 162

= 0.035, because k = 16/3.

. Y − 0.5. < 0.08 ≥ 1 − (0.5)(0.5) = 0.609; . 100

.

100(0.08)2

√ because k = 0.08/ (0.5)(0.5)/100;

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62

Section 5.9 Limiting Moment-Generating Functions (b) P

. Y − 0.5. < 0.08 ≥ 1 − (0.5)(0.5) = 0.922; . 500

.

500(0.08)2 √ because k = 0.08/ (0.5)(0.5)/500; (c) P

(0.5)(0.5) 2 .. Y ..

1000 − 0.5 < 0.08 ≥ 1 − 1000(0.08)= 0.961,

√ because k = 0.08/ (0.5)(0.5)/1000.

5.8 –6 P (75 < X < 85)

=

P (75 − 80 < X − 80 < 85 − 80) 60/15 = 0.84, = P ( |X − 80| < 5) ≥ 1 − 25

√ because k = 5/ 60/15 = 5/2.

5.9

Limiting Moment-Generating Functions

5.9 –2 Using Table III with λ = np = 400(0.005) = 2, P (X ≤ 2) = 0.677.

5.9 –4 Let Y =

n Σ

Xi, where X1, X2, . . . , Xn are mutually independent χ2(1) random variables.

i=1

Then µ = E(Xi) = 1 and σ2 = Var(Xi) = 2, i = 1, 2, . . . , n. Hence Y − nµ Y − n √ = √ 2n nσ2 has a limiting distribution that is N (0, 1).

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Chapter 6 Point Estimation

63

Chapter 6

Point Estimation 6.1

Descriptive Statistics

6.1 –2 x = 3.58; s = 0.5116.

6.1 –4 (a)

Class Interval (303.5, 307.5) (307.5, 311.5) (311.5, 315.5) (315.5, 319.5) (319.5, 323.5) (323.5, 327.5) (327.5, 331.5) (331.5, 335.5)

Class Limits (304, 307) (308, 311) (312, 315) (316, 319) (320, 323) (324, 327) (328, 331) (332, 335)

Frequency fi 1 5 6 10 11 9 7 1

Class Mark,ui 305.5 309.5 313.5 317.5 321.5 325.5 329.5 333.5

*

*

(b) x = 320.1, s = 6.7499; (c) h(x) 0.06 0.05 0.04 0.03 0.02 0.01

*

*

*

303.5 3 07.5 311.5 315.5 319 .5 323.5 32 7.5 331.5 335.5

x

Figure 6.1–4: Melting points of metal alloys There are 31 observations within one standard deviation of the mean (62%) and 48 observations within two standard deviations of the mean (96%).

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64

Section 6.1 Descriptive Statistics

64.1

–6 (a) With the class boundaries 0.5, 5.5, 17.5, 38.5, 163.5, 549.5, the respective frequencies are 11, 9, 10, 10, 10; h(x)

(b)

0.04

0.03

0.02

0.01 x 100

200

300

400

500

Figure 6.1–6: Mobil home losses (c) This is a skewed to the right distribution.

6.1 –8 (a) With the class boundaries 3.5005, 3.5505, 3.6005, . . . , 4.1005, the respective class fre-

quencies are 4, 7, 24, 23, 7, 4, 3, 9, 15, 23, 18, 2; (b)

h(x) 3.5 3.0 2.5 2.0 1.5 1.0 0.5 3.6

3.7

3.8

3.9

4.0

4.1

x

Figure 6.1–8: Weights of mirror parts (c) This is a bimodal histogram.

6.1 –10 (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 Player J’s batting average is higher than Player H’s on both grass and artificial turf, Player H’s is higher over all. Note the different numbers of at bats on grass and artificial turf and how this affects the batting averages.

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Chapter 6 Point Estimation

6.2

65

Exploratory Data Analysis

6.2 –2 (a) Baby carrots: 1.02, 1.03, 1.04, 1.04, 1.06;

Regular-size carrots: 1.00, 1.15, 1.21, 1.26, 1.43; (b)

1

1.1

1.2

1.3

1.4

Figure 6.2–2: (b) Box-and-whisker diagrams of small and regular-size carrots (c) Regular-size packages tend to be heavier.

6.2 –4 (a) The five-number summary is: min = 1, q˜1 = 6.75, m̃ = 32, q˜3 = 90.75, max = 527;

0 100 200 300 400 500 Figure 6.2–4: (a) Box-and-whisker diagram of mobile home losses (b) IQR = 90.75 − 6.75 = 84. The inner fence is at 216.75 and the outer fence is at 342.75;

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66

Section 6.2 Exploratory Data Analysis (c)

0 100 200 300 400 500 Figure 6.2–4: (c) Box-and-whisker diagram of losses with fences and outliers 6.2 –6 (a) Stems 0• 1∗ 1• 2∗ 2• 3∗ 3• 4∗ 4• 5∗ 5•

Leaves 55555555555555555556666666666666677777778888888999999 0000001111111222334 5555666677889 0111133444 5 4 5 5 5

Freq 53 19 13 10 1 1 1 0 1 0 1

(b) The five-number summary is: min = 5, q˜1 = 6, m̃ = 9, q˜3 = 15, max = 55.

10 20 30 40 50 Figure 6.2–6: (b) Box-and-whisker diagram of maximum capital

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Chapter 6 Point Estimation

67

(c) IQR = 15 − 6 = 9. The inner fence is at 28.5 and the outer fence is at 42.

10 20 30 40 50 Figure 6.2–6: (d) Box-and-whisker diagram of maximum capital with outliers and fences (e) The 90th percentile is 22.8.

6.2 –8 (a)

Stems 101 102 103 104 105 106 107 108 109 110

Leaves

Frequency

7 000

89 133667788 379 8 139 022

1 3 0 0 2 9 3 1 3 3

(Multiply numbers by 10−1.)

Table 6.2–8: Ordered stem-and-leaf diagram of weights of indicator housings

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68

Section 6.2 Exploratory Data Analysis (b) min = 101.7, q˜1 = 106.0, m˜= 106.7, q˜3 = 108.95, max = 110.2; (c) The interquartile range in IQR = 108.95 − 106.0 = 2.95. The inner fence is located at 106.0 − 1.5(2.95) = 101.575 so there are no suspected outliers.

102

104

106

108

110

Figure 6.2–8: (b) Weights of indicator housings

6.2 –10 (a) With the class boundaries 2.85, 3.85, . . . , 16.85 the respective frequencies are 1, 0, 2,

4, 1, 14, 20, 11, 4, 5, 0, 1, 0, 1; h(x) 0.30 0.25 0.20 0.15 0.10 0.05

3.85

5.85

7.85

9.85

11.85

13.85

15.85

x

Figure 6.2–10: (a) Lead concentrations

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Chapter 6 Point Estimation

69

(b) x = 9.422, s = 2.082; (c) h(x) 0.30 0.25 0.20 0.15 0.10 0.05

3.85

*5.85

*7.85

*9.85

*

1 1.85 1 3.85

15.85

x

Figure 6.2–10: (c) Lead concentrations showing x, x ± s, x ± 2s There are 48 (48/64 = 75%) within one standard deviation of the mean and 60 (60/64 = 93.75%) within two standard deviations of the mean; (d) 1976 Leaves

Stems

1 92 9 9943200 98875544443221100000 986632210 766554331100 975320 961 2

2 3 4 5 6 7 8 9 10 11 12 13 14 15 17

1

1977 Leaves 9

07 3568 3 0112223677788899 1123333444455678889999 22345579 0469 0346 8 7

Multiply numbers by 10−1 Table 6.2–10: (d) Back-to-Back Stem-and-Leaf Diagram of Lead Concentrations

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70

Section 6.3 Order Statistics (e)

1977

1976

2

4

6

8

10

12

14

16

Figure 6.2–10: (e) Box-and-whisker diagrams of 1976 and 1977 lead concentrations

6.3

Order Statistics 6.3 –2 (a) The location of the median is (0.5)(17 + 1) = 9, thus the median is

m̃ = 5.2.

The location of the first quartile is (0.25)(17 + 1) = 4.5. Thus the first quartile is q˜ 1 = (0.5)(4.3) + (0.5)(4.7) = 4.5.

The location of the third quartile is (0.75)(17 + 1) = 13.5. Thus the third quartile is ˜q 3 = (0.5)(5.6) + (0.5)(5.7) = 5.65; (b) The location of the 35th percentile is (0.35)(18) = 6.3. Thus π̃ 0 . 3 5 = (0.7)(4.8) + (0.3)(4.9) = 4.83.

The location of the 65th percentile is (0.65)(18) = 11.7. Thus Σ

π̃ 0 . 6 5 = (0.3)(5.6) + (0.7)(5.6) = 5.6.

5

6.3 –4

6! k−1 6−k k!(6 − k)! (k)[F (y)] f (y)[1 − F (y)] 6! + [F (y)]k(6 − k)[1 − F (y)]6−k−1[−f (y)] + 6[F (y)]5f (y) k!(6 − k)!

6!

6! = [F (y)]2f (y)[1 − F (y)]3 − [F (y)]3[1 − F (y)]2f (y) 2!3!

3!2!

6!

6! + [F (y)]3f (y)[1 − F (y)]2 − [F (y)]4[1 − F (y)]1f (y) 3!2!

4!1!

6!

6! + [F (y)]4f (y)[1 − F (y)]1 − [F (y)]5[1 − F (y)]0f (y) + 6[F (y)]5f (y) 4!1!

5!0!

6! = [F (y)]2[1 – F (y)]3f (y), a < y < b.

2!3!

g(y) =

k=3

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Chapter 6 Point Estimation 6.3 –6 (a)

(b)

71

F (x)

=

x,

0 < x < 1. Thus

g1(w)

=

n[1 − w]n−1(1),

0 < w < 1;

gn(w) = n[w]n−1(1), 0 < w < 1; ∫1 E(W1) = (w)(n)(1 − w)n−1 dw 0

1

−w(1 − w)n −

=

n+1

∫ 1

1

= 0

n n n+1 w +1

(w)(n)wn−1dw =

E(Wn) =

1

(1 − w)n+1

0

1

=

n+1 n

0

.

n+1

;

(c) The pdf of Wr is n! g (w ) = [w ]r−1[1 − w ]n−r · 1 r r r r (r − 1)!(n − r)! =

Γ(r + n − r + 1) r−1 w (1 − wr )n−r+1−1. Γ(r)Γ(n − r + 1) r

Thus Wr has a beta distribution with α = r, β = n − r + 1. ∫1 n!

2 w2 6.3 –8 (a) E(W r ) = wr−1(1 − w)n−rdw (r − 1)!(n − r)!

0 ∫1 r(r + 1) (n + 2)! wr+1(1 − w)n−rdw = (n + 2)(n + 1) 0 (r + 1)!(n − r)! r(r + 1) = (n + 2)(n + 1) since the integrand is like that of a pdf of the (r + 2)th order statistic of a sample of size n + 2 and hence the integral must equal one; r(r + 1) r2 r(n − r + 1) – = . (b) Var(Wr) = (n + 2)(n + 1)2 (n + 1)2 (n + 2)(n + 1) 6.3 –10 g(yi, yj )dyidyj is the probability that i − 1 of the sample variables are less than yi, one variable is in the infinitesimal interval of length dyi about yi, j − i− 1 variables are between yi and yj , one variable is in the infinitesimal interval of length dyj about yj , and n − j variables are greater than yj. By independence and combinatorics, the probability is n!

[F (y )]i−1f (y )dy [F (y )−F (y )]j−1+1f (y )dy [1−F (y )]n−j. (i − 1)!1!(j − i − 1)!1!(n − j)!

i

Thus, for −∞ < yi < yj < ∞, n! g(y , y ) = i

j

(i − 1)!(j − i − 1)!(n − j)!

i

i

j

i

j

j

j

[F (y )]i−1[F (y )−F (y )]j−1+1[1−F (y )]n−jf (y )f (y ). i

j

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i

j

i

j

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72

Section 6.3 Order Statistics 6.3–12

k 1 2 3 4 5

Strengths p = k/10 7.2 8.9 9.7 10.5 10.9

z1−p

k

−1.282 −0.842 −0.524 −0.253 0.000

0.10 0.20 0.30 0.40 0.50

Strengths p = k/10 6 7 8 9

11.7 12.9 13.9 15.3

0.60 0.70 0.80 0.90

z1−p 0.253 0.524 0.842 1.282

2

1

0

8

10

12

14

16

–1

–2 Figure 6.3–12: q-q plot of N (0, 1) quantiles versus data quantiles It seems to be an excellent fit.

6.3 –14 (a)

X

Y

133 134 135 136 137 138 139 Figure 6.3–14: q-q plot of N (0, 1) quantiles versus data quantiles (b) It looks like an excellent fit.

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Chapter 6 Point Estimation

6.4

73

Maximum Likelihood Estimation

6.4 –2 The likelihood function is

L(θ) =

1

"

n/2

exp −

2πθ

n Σ

# (xi − µ) /2θ ,

0 < θ < ∞.

2

i=1

The logarithm of the likelihood function is n Σ ln L(θ) = − n(ln 2π) − n (ln θ) − 1 (xi — µ)2.

2 2 2θ i=1

Setting the first derivative equal to zero and solving for θ yields n

d ln L(θ) dθ

n

+ (xi — µ)2 = 0 2θ2 i=1 2θ 1 n Σ (x i — µ)2 . n

= − θ =

Σ 1

i=1

Thus θ ^ =

1Σn (Xi — µ)2. n i=1

To see that θ^ is an unbiased estimator of θ, note that ! n (X — µ)2 σ2 2 i σ2 Σ E(θ^) = E n · n = σ , 2 σ = n i=1 since (Xi −µ)2/σ2 is χ2(1) and hence the expected value of each of the n summands is equal to 1.

For the unbiased property, it could also be noted that θ^ is the sample mean from a random sample of size n from the distribution of (X− µ)2. The sample mean is always an unbiased estimator of the distribution mean, which in this case is σ2.

6.4 –4 The joint probability of {3, 2, 3, 1} when θ = −1 is

P ({3, 2, 3, 1}; θ = −1) =

3

2

3

1

6

6

6

6

=

18

.

64

In similar fashion we find that the joint probability of {3, 2, 3, 1} when θ = 0 is P ({3, 2, 3, 1}; θ = 0) =

2 6

2 6

2 6

16 2 = , 6 64

1 6

3 6

and the joint probability of {3, 2, 3, 1} when θ = 1 is P ({3, 2, 3, 1}; θ = 1) =

1 6

2 6

=

6

.

64

Since the likelihood function is largest when θ = −1, the maximum likelihood estimate is θ^= −1.

6.4 –6 θ^1 = µ ^ = 33.4267; θ^2 = σ^2 = 5.0980.

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74

Section 6.4 Maximum Likelihood Estimation

6.4 –8 (a)

L(θ)

=

ln L(θ) = d ln L(θ) dθ

=

θ^ =

!1/θ−1

n

1

xi

θn

0<θ<∞

,

i=1 n

1

−n ln θ +

− 1 ln xi θn i=1 Y −n − 1 ln x = 0 i θ2 θ 1

n

Y

i=1

− ln xi n i=1 n

i 1Σ = − n i=1ln x ;

(b) We first find E(ln X): ∫ 1 ln x(1/θ)x1/θ−1 dx.

E(ln X) = 0

Using integration by parts, with u = ln x and dv = (1/θ)x1/θ−1dx, h i1 E(ln X) = lim x1/θ ln x − θx1/θ = −θ. a→0

Thus

a

1 E( θ^) = − n

n

Σ (−θ) = θ.

i=1

a) x 1 /bXe r =o Σ 6.4–10 (b p̃ = e q 1u /a pl s stoh ep̃ n=u m f nsn/uc c eis=s1e sX, in;, divided by the number of Bernoulli trials, Σn i=1 Xi; (c) 20/252 = 0.0794.

6.4–12 (a) E( X ) = E(Y )/n = np/n = p; (b) Var( X ) = Var(Y )/n2 = np(1 − p)/n2 = p(1 − p)/n; 2

(c) E[ X(1 − X )] = [E( X ) − E( X )] = {p − [p2 + p(1 − p)/n]} = [p(1 − 1/n) − p2(1 − 1/n)] = (1 − 1/n)p(1 − p)/n = (n − 1)p(1 − p)/n; (d) From part (c), the constant c = n/(n − 1). ( 1/2 ) cσ (n − 1)S2 6.4 –14 (a) E(cS) = E √ σ2 n− 1 cσ ∫ ∞ v1/2v(n−1)/2−1e−v/2 dv = √n − 1 n−1 0 2(n−1)/2 Γ 2 √ cσ 2 Γ(n/2) = √ , n − 1 Γ[(n − 1)/2)] √ n — 1 Γ[(n − 1)/2] √ ; so c = 2 Γ(n/2) √ √ √ (b) When n = 5, c = 8/(3 2π ) and when n = 6, c = 3 5π/(8 2 );

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Chapter 6 Point Estimation (c)

75

c 1.18 1.16 1.14 1.12 1.10 1.08 1.06 1.04 1.02 1.00 0.98 0.96

n 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 Figure 6.4–14: c as a function of n

We see that lim c = 1.

n→∞

6.4 –16 The experiment has a hypergeometric distribution with n = 8 and N = 64. From the

sample, x = 1.4667. Using this as an estimate for µ we have 1.4667 = 8

N1

implies that Ñ = 11.73.

1

64

A guess for the value of N1 is therefore 12.

6.4–18 L=

kn/2

1 2πσ2

n

k+1

k

exp −Σ Σ

xij − c − d j −

2

/(2σ2)

.

2

i=1 j=1

To find ^ c and d^ try to minimize S=

nΣ Σ k

xij − c −d j −

k n Σ ∂S = Σ 2

∂c Because

d j− j=1

∂S = ∂d

k+1

i=1 j=1

k

n

xij − c − d j − Σn

k+1 2

= 0,

i=1

c^=

.

2

i=1 j=1

We have

2

k+1

(−1) = 0.

2 Σk

xij

j=1

kn

k

ΣΣ i=1 j=1

= x, say. k+1

2 xij − c − d j −

2

k+1 — j−

= 0.

2

Thus Σn Σk (x − x)(j − {k + 1}/2) Σni=1 Σk xij(j − {k + 1}/2) d^= i=1 j=1 ij j=1 . = Σn Σ k Σk 2 2 i=1 j=1 (j − {k + 1}/2) j=1 n(j − {k + 1}/2)

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76

Section 6.5 A Simple Regression Problem ∫y 6.4 –20 (a) F (y) =

(1/θ) dx = [(x/θ)]y 0= y/θ, so the desired pdf is

0

gn(y) = n(y/θ)n−1(1/θ) = nyn−1/θn, ∫ θ n

(b) E(Yn) =

n

(ny /θ ) dy = 0

nyn+1 (n + 1)θn

θ

nθ

=

;

n+1

0

0 < y < θ;

(c) c = (n + 1)/n, since E{[(n + 1)/n]Yn} = [(n + 1)/n] · [nθ/(n + 1)] = θ; ∫ θ θ nθ2 , so nyn+1 dy = nyn+2 (d) E(Y 2) = = n θn n+2 (n + 2)θn 0 0 # " nθ 2 nθ2 2 Var(cYn) = c − n+1 n+2 =

(n + 1)2 2 θ n2

n

−

n+2

n2 (n + 1)2

θ2

=

n(n + 2)

;

(e) Var(2X ) = (4/n)Var(X1) = (4/n)(θ2/12) = θ2/(3n).

Since 3n < n(n + 2) for n > 1, Var(2X ) is larger.

6.5

A Simple Regression Problem nY

6.5 –2 (a)

2

L(β, σ )

(y − βx )2

1

√ exp – 2πσ2 i=1

=

Σn

n/2

1 =

i

exp −

2πσ2

2σ2

i=1

i

(yi – βx i)

2

Maximizing L or equivalently maximizing n 2

ln L(β, σ ) =

2

H′(β)

=

2

n Σ

Σn

ln(2πσ2) +

requires selecting β to minimize H(β) =

.

2σ2

Σn i=1

i=1

(y – βx )2 i

i

2σ2

(yi − βxi)2.

(yi − βxi)(−xi) = 0

i=1

Σn xiyi Σi=1 n x2

β^ =

i=1

i

=

n 1Σ (yi — β^x i ) 2 ; n

Σ which is a point of minimum since H ′′(β^) = 2 ni=1 x2i > 0. Σn ∂ ln L(β, σ2) n (y − βx )2 i i = 0 = 2σ2 − i=1 ∂(σ2) 2(σ2)2 ^2 σ

(b) β^ =

n

i=1

!−1 n

xiYi. Since β^ is a linear combination of independent normal

2

xi i=1

i=1

random variables, it is normal with mean

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Chapter 6 Point Estimation

77 !−1 n

n

E ( β^ ) =

2 xi

n

xiE(Yi) i=1

i=1

!−1 n 2 xi

= i=1

xi(βxi) i=1

!−1

n

!

n

2

=

2 xi

β

xi i=1

i=1

= β and variance

!−2 n

n

Var(β^) =

2 xi i=1

x2i Var(Yi) i=1

!−2 n

n 2 xi

= i=1

x2i σ2 i=1

n

,Σ σ2 x2i.

=

i=1

Also n Σ

(Yi − βxi)2

=

n Σ

[Yi − β^xi + (β^− β)xi]2

i=1 n

i=1

Σ

=

n

(Yi − β^x i )2 + (β^ − β)2

i=1

Σ x2.i i=1

Σ 2 2 n Because Y ~ independent N (βx , σ ) and β^ ∼ N (β, σ i

i

Σn

(β^ − β)2

(Y − βx )2 i

i=1

σ2

i

2

~ χ (n)

x2), i=1 i

and

Σn

x2 i=1

σ2

i

2

~ χ (1).

By Theorem 9.3-1, n Σ

(Yi − β^xi)2 ∼ χ2(n−1);

i=1

(c) β^ =

(1)(1) + (2)(1) 3 ^2 [1 − (3/5)(1)]2 + [1 − (3/5)(2)]2 1 = = ; = , σ 2 2 1 +2 5 2 10

(d) y1 − y ^1 = 1−(3/5) = 2/5, y2 −y2 ^ = 1−(6/5) = −1/5, and their sum is (2/5)−(1/5) =1/5 /= 0.

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78

Section 6.5 A Simple Regression Problem 6.5–4 (a) x

y

x2

xy

y2

(y − y^)2

2.0 3.3 3.7 2.0 2.3 2.7 4.0 3.7 3.0 2.3

1.3 3.3 3.3 2.0 1.7 3.0 4.0 3.0 2.7 3.0

4.00 10.89 13.69 4.00 5.29 7.29 16.00 13.69 9.00 5.29

2.60 10.89 12.21 4.00 3.91 8.10 16.00 11.10 8.10 6.90

1.69 10.89 10.89 4.00 2.89 9.00 16.00 9.00 7.29 9.00

0.361716 0.040701 0.027725 0.009716 0.228120 0.206231 0.006204 0.217630 0.014900 0.676310

29.0

27.3

89.14

83.81

80.65

1.849254

α ^ = y = 27.3/10 = 2.73; β^ =

83.81 − (29.0)(27.3)/10 4.64 = = 0.9206; 89.14 − (29.0)(29.0)/10 5.04

y^ =

2.73 + (4.64/5.04)(x − 2.90);

(b)

y 4.0 3.5 3.0 2.5 2.0 1.5 x 1.5

2.0

2.5

3.0

3.5

4.0

Figure 6.5–4: Earned grade (y) versus predicted grade (x) (c) σ ^2 = 6.5–6 (a)

1.849254

α ^ = β^ = y^ =

10 395

= 0.184925.

= 26.333,

15 9292 − (346)(395)/15 180.667 = = 0.506, 8338 − (346)2/15 356.933 180.667 346 26.333 + x− 356.933 15

= 0.506x + 14.657;

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Chapter 6 Point Estimation

79

(b)

y 32 30 28 26 24 22 20 18 16

x 16 18 20 22 24 26 28 30 32

Figure 6.5–6: ACT natural science (y) versus ACT social science (x) scores (c)

26.33, β^ = 0.506, 3952 2 n σ^ = 10, 705 − 15 — 0.5061636(9292) + 0.5061636(346)(395)/15 = 211.8861, 211.8861 ^ 2 = 14.126. σ = 15 α ^ =

6.5 –8 (a) y = 3.575 + 1.225x;

(b) 35 34 33 32 31 30 29 28 27 26 18 19

20

21

22

23

24

25

26

27

Figure 6.5–8: (b) Highway mpg (y) versus City mpg (x)

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80

Section 6.6 Asymptotic Distributions of Maximum Likelihood Estimators (c) y = 55.6745 − 0.0075x.

35 34 33 32 31 30 29 28 27 26 3100 3200 3300 3400 3500 3600 3700 3800 3900

Figure 6.5–8: (c) Highway mpg (y) versus Curb Weight (x)

6.5 –10 (b) The least squares regression line for y = a + b versus b is y^= 1.360 + 1.626b;

(c) y = φx = 1.618x is added on the right figure below;

280 260 240 220 200 180 160 140 120 100 80 60 40

280 260 240 220 200 180 160 140 120 100 80 60 40 20

40

60

80

100 120 140 160 180

20

40

60

80

100 120 140 160 180

Figure 6.5–10: Scatter plot of a + b versus b with least squares regression line and with y = φx

(d) The sample mean of the points (a+b)/b is 1.647 which is close to the value of φ = 1.618.

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Chapter 6 Point Estimation

6.6

81

Asymptotic Distributions of Maximum Likelihood Estimators

6.6 –2 (a)

f (x; p)

=

px(1 − p)1−x,

x = 0, 1

ln f (x; p) = x ln p + (1 − x) ln(1 − p) x x−1 + p 1−p x x−1 = − 2 + (1 − p)2 ∂p2 p X p p−1 E − X−1 = − = p2 (1 − p)2 p2 (1 − p)2 p(1 − p) Rao-Cram´er lower bound = ; n p(1 − p)/n = 1. (b) p(1 – p)/n ∂ ln f (x; p) ∂p ∂2 ln f (x; p)

6.6 –4 (a)

=

ln f (x; θ) = −2 ln θ + ln x − x/θ ∂ ln f (x; θ) 2 x = − + ∂θ θ θ2 2x ∂2 ln f (x; θ) = 2 − ∂θ2 θ2 θ3 2 2X 2 + 2(2θ) = 2 E − + = − 2 3 θ θ θ2 θ3 θ2 2 θ ; Rao-Cram´er lower bound = 2n

(b) Very similar to (a); answer = (c)

θ2 ; 3n

1−θ ln f (x; θ) = − ln θ + ln x θ ∂ ln f (x; θ) 1 1 = − – ∂θ θ θ2 ln x 2 ∂ ln f (x; θ) 1 2 = + 2 2 3 ln x ∂θ ∫θ 1 lnθx 1 (1 θ)/θ E[ ln X ] = θ x − dx. Let y = ln x, dy = x dx.

0 ∫∞ y −y(1−θ)/θ −y = − e e dy = −θ Γ(2) = −θ. θ 0 Rao-Cram´er lower bound =

6.7

1 . p(1 − p)

1 1 2 n − + θ2 θ2

=

θ2 . n

Sufficient Statistics

6.7 –2 The distribution of Y is Poisson with mean nλ. Thus, since y = Σ xi,

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82

Section 6.7 Sufficient Statistics P (X1 = x1, X2 = x2, · · · , Xn = xn | Y = y)

= =

which does not depend on λ.

6.7 –4 (a) f (x; θ) = e(θ−1) ln x+ln θ,

(λΣ xi e−nλ)/(x1! x2! · · · xn!) (nλ)y e−nλ/y! y! , y x1! x2! · · · xn ! n

0 < θ < ∞;

0 < x < 1,

so K(x) = ln x and thus Y =

n Σ

ln Xi = ln(X1X2 · · · Xn)

i=1

is a sufficient statistic for θ; (b)

L(θ) =

θn(x1x2 · · · xn)θ−1

n ln θ + (θ − 1) ln(x1x2 · · · xn) n = + ln(x1x2 · · · xn) = 0. θ

ln L(θ) = d ln L(θ) dθ Hence

θ^ = −n/ ln(X1X2 · · · Xn),

which is a function of Y ; (c) Since θ^is a single valued function of Y with a single valued inverse, knowing the value of ^ θ is equivalent to knowing the value of Y , and hence it is sufficient. (x1x2 · · · xn)α−1e−Σ xi/θ 6.7–6 (a) f (x , x , . . . , x ) = 1

2

n

[Γ(α)]nθαn =

e−Σ xi/θ

(x1x2 · · · xn)α−1

θαn

[Γ(α)]n

.

The se c oΣ n d factor is free of θ. The first factor is a function of the xis through n only, so i=1 xi is a sufficient statistic for θ; Σni=1 x/θ − ln[Γ(α)]n − αn ln θ (b) ln L(θ) = ln(x 1x 2 · · · xn)α−1 − i n d ln L(θ) Σ 2 = i=1 xi/θ − αn/θ = 0 dθ Σn xi αnθ = i=1 Σ α1n n

^ θ = Σ

Σn

i=1 xi

i

i=1 X .

n

Y =

Xi has a gamma distribution with parameters αn and θ. Hence

i=1

1 E(θ^) = (αnθ) = θ. αn 6.7–8

∫ ∞ E(e tZ ) = −∞

···

∫ ∞ −∞

√

1 2πθ

n

" # n Σn Σ x2 exp t exp − Σi=1 ai xi dx1dx2 . . . dxn. i n 2θ xi i=1

i=1

√ √ Let xi/ θ = yi, i = 1, 2, . . . , n. The Jacobian is ( θ )n. Hence

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Chapter 6 Point Estimation ∫ ∞

83

" n # Σn Σ y2i exp − exp t Σi=1 aiyi dy1dy2 . . . dyn. θ) 2 n yi 2πθ −∞ −∞ i=1 i=1 Σ n # n aiyi ∫∞ ∫∞ n i " t 1 Σ y2 √ dy1dy2 . . . dyn exp – exp Σi=1 ··· n = 2 i=1 yi −∞ −∞ 2π i=1 Σn which is free of θ. Since the distribution of Z is free of θ, Z and Y = i=1 X2,i the sufficient statistics, are independent.

E(etZ ) =

···

∫ ∞√ (

n

√

1

n

6.7–10 The joint pdf can be written

#θ "n Γ(2θ)2 Y xi(1 − xi) {Γ(θ)}

n

1 Qn x (1 − x )] ; [ i=1 i i

i=1

so by definition

n Y

Xi(1 − Xi)

i=1

is a sufficient statistics for θ.

6.7 –12 The joint pdf of Y1, Y2, . . . , Yn, letting

c= is

Σn

i=1[yi

c · exp

Σn = c · exp

,

2πσ2

– α − β(xi – x)]2 2σ2

[y2i+ α2 + β2(xi − x)2 − 2αyi − 2β(xi − x)yi + 2αβ(xi − x)] 2σ2 Σn Σn Σn 2 y − 2α y − 2β y (x − x) + quantities that don’t involve the y ’s i=1 i

i=1

i

Σn y2 − 2(α + βx) = c · exp

i=1 i

n

yi, i=1

n

1

i=1

= c · exp

= φ

√

n Σ i=1

yi2,

i=1

Σ n

yi − 2β

i=1 n Σ

i

i

i

Σ

2σ2 yixi + quantities that don’t involve the yi’s

n

i=1

2σ2

! xiyi; α, β, σ2

· 1.

i=1

The result follows immediately by the Factorization Theorem.

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84

Section 6.8 Bayesian Extimation

6.8

Bayesian Estimation

6.8–2 (a)

g(τ | x1, x2, . . . , xn)

∝

(x1x2 · · · x n)α − 1 τnατα0 − 1e−τ/θ0 e−Σxi/(1/τ ) [Γ(α)]n Γ(α )θα0 0

∝ θ0

which is Γ nα + α , 0

(b)

ταn + α0 − 1e−(1/θ0 + Σxi)τ

0

;

1 + θ0Σxi

E(τ | x1, x2, . . . , xn ) =

(nα + α0)

θ0

1 + θ0Xn αnθ0 + 1 + θ0nX 1 + nθ0X nα + α0 ; 1/θ0 + nX α0θ0

= =

(c)

The posterior distribution is Γ(30 + 10, 1/[1/2 + 10 x ]). Select a and b so that P (a < τ < b) = 0.95 with equal tail probabilities. Then ∫ b a

(1/2 + 10 x )40 40−1 −w(1/2 + 10 x) w e dw = Γ(40)

∫

b(1/2+10x)

a(1/2+10x)

1

z39e−zdz,

Γ(40)

making the change of variables w(1/2 + 10 x ) = z. Let v0.025 and v0.975 be the quantiles for the Γ(40, 1) distribution. Then a=

v0.025

;

1/2 + 10 x b=

v0.975

.

1/2 + 10 x It follows that P (a < τ < b) = 0.95.

6.8–4 3 3 g(θ | x1 , x2, . . . , x n ) ∝ (3θ) n(x 1x 2 · · · x n )2e−θΣx i · θ4 − 1e−4θ ∝ θn + 3e−(4 + Σx )θi

which is Γ n + 4,

1

. Thus 3

4 + Σxi E(θ | x1, x2, . . . , xn) =

n+4 4 + Σ x3i

.

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Chapter 6 Point Estimation

85

g(y | θ)

=

g(y | θ)h(θ)

∝

k(θ | y)

=

c

=

6.8–6 (a)

nyn−1 θn 1

,

θn+β+1 c θn+β+1

0<y<θ ,

max(α, y) < θ < ∞

,

max(α, y) < θ < ∞ ∫∞

(n + β)[max(α, y)]n+β by

k(θ | y) dθ = 1

max(α,y)

∫∞

E(θ | y) =

(θ)

[max(α, y)]n+β(n + β)

n+β max(α, y) = w(y) n+β−1 n+β max(α, Y ); n+β −1

= w(Y )

dθ

θn+β+1

max(α,y)

=

(b) With n = 4, α = 1, β = 2, [max(1, y)]6(6)

k(θ | y) =

,

θ7

max(1, y) < θ.

Since Loss = |θ − w(Y )|, we want w(y)

=

1

=

median ∫ w(y)

[max(1, y)]6(6)

max(1,y)

2

= 1−

dθ

θ7

[max(1, y)]6 [w(y)]6

so w(Y ) = 6.8–8 L=

1

√ 2πσ2

(

n

exp

–

n Σ

√ 6 2 max(1, Y ).

) [yi − α − β(xi − x)]2/2σ2 .

i=1

The summation in the exponent can be written as follows: n

Σ

i=1

n

[yi − α − β(xi − x)] = n ( α^ − α)2 + (β^ − β)2

Σ

n

(xi − x)2 +

i=1

Σ

[yi − α ^ − β^(xi − x)]2.

i=1

Note: Let’s say σ2 is known in the exercise. Then α has prior N (α1, σ2);1 so posterior of α (eliminating factors not involving α) is

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86

Section 6.8 Bayesian Extimation

posterior of α ∝

−(α − α1)2 exp

2πσ21

∝ exp − =

!

√ 1

−n(α − α^ ) 2 exp

2σ21

2σ2

(nα2 − 2n α ^ α) (α2 − 2αα1) — + terms not involving α 2 2 2σ1 2σ 1

exp − α2

n

+

2

− 2α

2

2σ1

α1

+ 2

2σ

2σ1

n α^

+

2σ2

terms not involving α 1 = exp − 2σ2 + n 2σ2 1

2 ^/(2σ 2 ) α2 − 2αα1/(2σ1 ) + nα + 2 1/(2σ1 ) + n/(2σ2)

terms not involving α " = exp −

1 2σ12

+

n 2σ2

α−

1α /(2σ2) + nα ^ /(2σ 2 ) 1

1/(2σ12) + n/(2σ2)

#

2

+

terms not involving α.

This is normal with posterior mean equal to α1 n α^ + σ12 σ2 .

1 n + σ12 σ2 Now β has prior N (β0, σ2); so posterior of β is 0 "(β − β^)2 Σn (xi − x)2 posterior of β

∝

exp −

" ∝ exp −

∝ exp −

(β − β0)2

#

– 2σ02 2σ2 ! # Σ Σn (x − 2 β^ n (x − x)2 x) 1 β2 + 2 i=1 i i=1 i + β02 β + 2 2σ0 2σ2 2σ0 2σ2 Σn )2 ( Σn 2 2 2 + β /σ 2 1 i=1(x i − x) 0 0 β^ i=1 (x i − x) /σ . β − Σn + 2 2 2 2 2σ2 2σ 0 (x — x) /σ + 1/σ i i=1 0 i=1

This is normal with posterior mean 2 2 2 Σn β^ i=1 (xi − x) /σ + β0/σ 0 . Σn 2 2 2 i=1 (xi — x) /σ + 1/σ 0 Hence the posterior mean of α + β(x − x) is Σn α1 n α^ β^ i=1(xi − x)2 β0 + 2 + 2 σ0 σ2 σ12 σ (x − x).

2 1 + n + Σni=1(xi — x) + 1 2 2 2 2 σ σ0 σ1 σ

Chapter 7 Interval Estimation

87

Chapter 7

Interval Estimation 7.1

Confidence Intervals for Means

7.1–2 (a) [77.272, 92.728]; (b) [79.12, 90.88]; (c) [80.065, 89.935];

(d) [81.154, 88.846].

7.1 –4 (a) x = 56.8;

√ √ (b) 56.8 − 1.96(2/ 10 ), 56.8 + 1.96(2/ 10 = [55.56, 58.04] ; 52 − 56.8 (c) P (X < 52) = P Z <

7.1 –6

11.95 − 1.96

11.80 √ 37

2

= P (Z < −2.4) = 0.0082.

, 11.95 + 1.96

11.80 √ 37

= [8.15, 15.75].

If more extensive t-tables are available or if a computer program is used, we have 11.80 11.80 , 11.95 + 2.028 √ = [8.016, 15.884].

11.95 − 2.028 √ 37 37 7.1 –8 (a) x = 46.42;

(b) 46.42 ± 2.132s/

√ 5

or

[40.26, 52.58].

7.1 –10 (a) x = 3.580;

(b) s = 0.512; (c) [0, 3.580 + 1.833(0.512/

√ 10 ] = [0, 3.877].

7.1 –12 (a) x = 245.80, s = 23.64, so a 95% confidence interval for µ is

√ √ [245.80 − 2.145(23.64)/ 15 , 245.80 + 2.145(23.64)/ 15 ] = [232.707, 258.893];

88

Section 7.2 Confidence Intervals for the Difference of Two Means (b)

220

240

260

280

Figure 7.1–12: Box-and-whisker diagram of signals from detectors (c) The standard deviation is quite large. √ √ √ 7.1 –14 (a) ( x + 1.96σ/ 5 ) − ( x − 1.96σ/ 5 ) = 3.92σ/ 5 = 1.753σ; √ √ √ (b) ( x + 2.776s/ 5 ) − ( x − 2.776s/ 5 ) = 5.552s/ 5. √ √ 2 Γ(5/2)σ 3 πσ = From Exercise 6.4–14 with n = 5, E(S) = √ = 0.94σ, so that 25/2 4 Γ(4/2) √ E[5.552S/ 5 ] = 2.334σ.

7.1 –16

7.2

P (X − 1 < µ < X + 1) = 0.90 P (−1 < µ − X < 1) = 0.90 −1 X−µ 1 P √ < √ < √ 3/ n = 0.90 3/ n 3/ n √ n = 1.645 3 √ n = 3(1.645) n = 24.35 3 25.

Confidence Intervals for the Difference of Two Means

7.2 –2 x = 539.2, s2 = 4,948.7, y = 544.625, s2 = 4,327.982, sp = 67.481, t0.05 (11) = 1.796, x

y

so the confidence interval is [−74.517, 63.667].

7.2 –4 (a) x − y = 1511.714 − 1118.400 = 393.314;

(b) s2 = 49, 669.905, s2 = 15, 297.600, r = [8.599] = 8, t0.025(8) = 2.306, so the x

y

confidence interval is [179.148, 607.480].

7.2–6 (a) x = 2.584, y = 1.564, s2 = 0.1042, s2 = 0.0428, sp = 0.2711, t0.025(18) = 2.101. x

y

Thus a 95% confidence interval for µX − µY is [0.7653, 1.2747];

Chapter 7 Interval Estimation

89

(b)

X

Y

1.5

2.0

2.5

3.0

Figure 7.2–6: Box-and-whisker diagrams, wedge on (X) and wedge off (Y ) (c) Yes. 7.2 –8 From (a), (b), and (c), we know s

X − Y − (µX − µY ) (d) ÷ s 2 2 σ dσ Y Y nX + nY

, (n − 1 ) XS 2 (n − 1)YS2 Y X 2 + dσ 2 σ Y Y

(nX + nY − 2)

has a t(nX +nY −2) distribution. Clearly, this ratio does not depend upon σ2Y; so s 2 2 d 1 x − y ± t (n +n −2) (nX − 1)s X /d + (nY − 1)s Y Y α/2 X nX + nY nX + nY – 2 provides a 100(1 − α)% confidence interval for µX − µY .

7.2 –10 (a) d = 0.07875;

√ (b) [ d − (1.714)(0.25492)/ 24, ∞) = [−0.0104, ∞); (c) not necessarily.

7.2 –12 (a) x = 136.61, y = 134.87, s2 = 3.2972, s2 = 1.0957; x

y

(b) Using Welch with r = 18 degrees of freedom, the 95% confidence interval is [0.436, 3.041].

Assuming equal variances with r = 20 degrees of freedom, the 95% confidence interval is [0.382, 3.095];

90

Section 7.3 Confidence Intervals for Proportions (c) The five-number summary for the X observations is 133.30, 135.625, 136.95, 137.80, 139.40. The five-number summary for the Y observations is 132.70, 134.15, 134.95, 135.825, 136.00;

X

Y

133 134 135 136 137 138 139 Figure 7.2–12: Hardness of hot (X) and cold (Y ) water (d) The mean for hot seems to be larger than the mean for cold.

7.3

Confidence Intervals For Proportions

7.3 –2 (a) p ^=

"

142 200

= 0.71; r

# r (0.71)(0.29) (0.71)(0.29) (b) 0.71 − 1.645 = [0.657, 0.763]; , 0.71 + 1.645 200 200 √ 0.71 + 1.6452/400 ± 1.645 0.71(0.29)/200 + 1.6452·/(4 2002) (c) = [0.655, 0.760]; 1 + 1.6452/200 142 + 1.6452/2 142 + 1.3530 (d) p˜ = 200 + 1.6452 = 200 + 2.7060 = 0.7072; " r # (0.7072)(0.2928) 0.7072 1.645 = [0.6546, 0.7598]; 202.7060 ± " r # (0.71)(0.29) (e) 0.71 − 1.282 , 1 = [0.669, 1].

200 " # r r (0.70)(0.30) (0.70)(0.30) 0.70 − 1.96 7.3–4 = [0.674, 0.726]. , 0.70 + 1.96 1234 1234 " # r r (0.26)(0.74) (0.26)(0.74) 7.3–6 0.26 − 2.326 = [0.247, 0.273]. , 0.26 + 2.326 5757 5757 388

7.3 –8 (a) p^ =

1022

= 0.3796;

r (b) 0.3796 ± 1.645

(0.3796)(0.6204)

7.3 –10 (a) p^1 = 28/194 = 0.144;

1022

or

[0.3546, 0.4046].

Chapter 7 Interval Estimation

91

√ (b) 0.144 ± 1.96 (0.144)(0.856)/194 or [0.095, 0.193]; (c) p^ − 1^p1/ =16228=/109. 0476; 1 2 " # r (0.144)(0.856) (0.068)(0.932) (d) 0.076 − 1.645 or + , 1 194 162

[0.044, 1].

7.3 –12 (a) pA^ = 170/460 = 0.37, pB^ = 141/440 = 0.32,

r

(0.37)(0.63) (0.32)(0.68) + 460 440 (b) yes, the interval includes zero.

0.37 − 0.32 ± 1.96

7.4

or [−0.012, 0.112];

Sample Size (1.96)2(169)

7.4 –2 n =

(1.5)2

= 288.5 so the sample size needed is 289.

(1.96)2(34.9)

7.4 –4 Approximately n =

(0.5)2 (1.96)2(33.7)2

7.4–6 Approximately n = 7.4 –8 n =

= 537, rounded up to the nearest integer. = 175, rounded up to the nearest integer.

52

(1.645)2(0.394)(0.606) (0.04)2 z p∗(1 − p∗) 2

α/2

= 404, rounded up to the nearest integer.

(1.6452)(0.1)(0.9)

= = 2435; 0.012 ε2 2 z2 ^p (1 − p^ ) (1.645 )(0.16)(0.84) (b) n = α/2 ε2 = = 3637; 0.012 2 2 z 1.645 = (c) n = α/2 = 6765.

4ε2 (4)(0.012)

7.4–10 (a) n =

7.4 –12 m =

(1.96)2(0.5)(0.5)

(a) n =

(0.04)2 601 1 + 600/1500

= 601, rounded up to the nearest integer. = 430;

(b) n =

601 = 578; 1 + 600/15,000

(c) n =

601 = 587.

1 + 600/25,000

7.4 –14 For the difference of two proportions with equal sample sizes

ε = z α/2 or

z2 n=

For unknown p∗,

α/2

r ∗ p (1 − p∗) 1

1

n

p∗(1 − p∗) +

2

2

n

[p∗(1 − p∗) + p∗(1 − p∗)] 1

1

2

2

ε2

z2 z2 [0.25 + 0.25] n = α/2 = α/22 .

2ε ε2

.

92

Section 7.5 Distribution-Free Confidence Intervals for Percentiles 1.2822 So n =

7.5

2(0.05)2

= 329, rounded up.

Distribution-Free Confidence Intervals for Percentiles

7.5 –2 (a) (y3 = 5.4, y10 = 6.0) is a 96.14% confidence interval for the median, m;

(b) (y1 = 4.8, y7 = 5.8); P (Y1 < π0.3 < Y7)

=

6 Σ 12 k=1

(0.3)k(0.7)12−k

k

= 0.9614 − 0.0138 = 0.9476, using Table II with n = 12 and p = 0.30.

7.5 –4 (a) (y4 = 80.28, y11 = 80.51) is a 94.26% confidence interval for m;

(b) (y6 = 80.32, y12 = 80.53); 11 Σ k=6

k

14 (0.6)k (0.4)14−k

=

8 Σ k=3

14 (0.4)k(0.6)14−k k

= 0.9417 − 0.0398 = 0.9019.

The interval is (y6 = 80.32, y12 = 80.53).

7.5–6 (a)

Stems Leaves 20f 5 20s 6677 20• 88999 21∗ 0000011 21t 2222233333333 21f 444444455555555 21s 66666666666667777777777 21• 8888889999 22∗ 000 (Multiply numbers by 10−1.)

Frequency 1 4 5 7 13 15 23 10 3

21.2 + 21.2 y20 + y21 = = 21.2 2 2 q2 = m = y41 = 21.5 21.7 + 21.7 y61 + 762 = q3 = = 21.7; 2 2 (ii) π0.60 = 0.8y49 + 0.2y50 = 21.6; (iii) π0.15 = 0.7y12 + 0.3y13 = 21.0; (c) (i) We first find i and j so that P (Yi < π0.25 < Yj) ≈ 0.95. Let the distribution of W be b(81, 0.25). Then P (Yi < π0.25 < Yj) = P (i ≤ W ≤ j − 1)

(b) (i) q1

=

i − 0. 5 − 20.25 j − 1 + 0.5− 20.25 . √ ≤Z≤ √ 15.1875 15.1875 If we let i − 20.75 j − 20.75 √ = 1.96 √ = −1.96 and 15.1875 15.1875 we find that i ≈ 13 and j ≈ 28. Furthermore P (13 ≤ W ≤ 28 − 1) ≈ 0.9453. Also note that the point estimate of π0.25, ≈ P

π̃ 0 .2 5 = (y20 + y21)/2

Stuvia.com - The Marketplace to Buy and Sell your Study Material

Chapter 7 Interval Estimation

93

falls near the center of this interval. So a 94.53% confidence interval for π0.25 is (y13 = 21.0, y28 = 21.3); (ii) Let the distribution of W be b(81, 0.5). Then P (i ≤ W ≤ 81 − i)

P (Yi < π0.5 < Y82−i) =

81 − i + 0.5− 40.5 i − 0.5 − 40.5 √ . √ ≤Z ≤ 20.25 20.25 If i − 41 = − 1.96, 4.5 then i = 32.18 so let i = 32. Also 81 − i − 40 = 1.96 4.5 implies that i = 32. Furthermore ≈

P

P (Y32 < π0.5 < Y50) = P (32 ≤ W ≤ 49) ≈ 0.9544.

So an approximate 95.44% confidence interval for π0.5 is (y32 = 21.4, y50 = 21.6); (iii) Similar to part (a), P (Y54 < π0.75 < Y69) ≈ 0.9453. Thus a 94.53% confidence interval for π0.75 is (y54 = 21.6, y69 = 21.8).

7.5 –8 A 95.86% confidence interval for m is (y6 = 14.60, y15 = 16.20).

P (b(25, 0.5) ≤ 15) − P (b(25, 0.5) ≤ 8) = 0.8852 − 0.0539 = 0.8313; (b) Want P (Yr < π0.90) ≥ 0.95, that is, P (b(25, 0.9) ≥ r) ≥ 0.95. Now,

7.5 –10 (a) P (Y9 < π0.50 < Y16)

=

P (b(25, 0.9) ≥ 21) = 0.9020,

and P (b(25, 0.9) ≥ 20) = 0.9666.

So the desired r is 20, and y20 = 190, and the desired interval is [190, ∞). Because this interval straddles the water quality standard, it does not provide strong evidence of a violation. Σ 8 8 (0.7)k(0.3)8−k = 0.2553; 7.5 –12 (a) P (Y7 < π0.70) = k=7 k 7 Σ

(b) P (Y5 < π0.70 < Y8) = k=5

7.6

8 (0.7)k(0.3)8−k = 0.7483. k

More Regression

7.6 –2 The mean of the observations is

Y

∗

1 Σ m

=

Y ∗,

m i=1 i ∗ where Y ∗, Y ∗, . . . Y ∗ are the future observations at X = x∗. Since Y is a linear combina1

2

m

tion of independent and normally distributed random variables, it is normally distributed with mean E

1 m

!

m

Y i∗

=

m Σ 1 E(Yi∗) m i=1 m

i=1

=

1 Σ m i=1

[α + β(x∗ − x)]

= α + β(x∗ – x)

94

Section 7.6 More Regression and variance m !

1Σ ∗ Y Var m i

m 1 Σ

=

i=1

=

m2 i=1

Var(Yi∗)

m σ 2 1 Σ σ2 = . m m2 i=1

Furthermore,

∗ Y −α ^ − β^(x∗ − x) is a linear combination of independent and normally distributed random variables, so it is normally distributed with mean

E[Y − α ^ − β^(x∗ − x)] = α + β(x∗ − x) − α − β(x∗ − x) = ∗

(x∗ − x)2σ2 + + Σ i — x)2 . (x m n ∗ 2 ^ ^ Since Y , α^, and β are independent of σ , we can form the t-statistic ∗ Y −α ^ − β^(x∗ − x) 0and variance

σ2

σ T=

s

σ2

+ 1 m

1 + Σ((x∗x −—x)2 2 n x) i ∼ t(n−2). s 2 ^ nσ n−2

We can then obtain the following expression for the endpoints of the desired prediction interval: α ^ + β^(x∗ − x) ± htα/2(n−2) where s 1 + 1 + Σ(x∗ − x)2 2 . h=^ σr n (xi — x) n−2 m n 7.6 –4 (a) In Exercise 6.5–4 we found that β^ = 4.64/5.04,

10 Σ

n σ^2 = 1.84924,

(xi − x)2 = 5.04.

i=1

So the endpoints for the confidence interval are givenrby r 1.8493 1 4.64 (x − 2.90) ± 2.306 5.04 x = 2 : [1.335, 2.468], 2.73 +

8

10

+

(x − 2.90)2 5.04

,

x = 3 : [2.468, 3.176], x = 4 : [3.096, 4.389]; (b) The endpoints for the prediction interval are given by r r 1 4.64 (x − 2.90)2 1.8493 1 + + , 2.73 + (x − 2.90) ± 2.306 10 5.04 8 5.04 x = 2 : [0.657, 3.146], x = 3 : [1.658, 3.986], x = 4 : [2.459, 5.026].

Chapter 7 Interval Estimation

95

y

y

4.0

4.5 4.0

3.5

3.5 3.0

3.0 2.5

2.5

2.0

2.0

1.5 1.5 x 2.0

2.5

3.0

3.5

1.0

4.0

x 2.0

2.5

3.0

3.5

Figure 7.6–4: A 95% confidence interval for µ(x) and a 95% prediction band for Y

7.6 –6 (a) For these data, 10 Σ

10 Σ

xi = 55,

i=1

10 Σ

i=1

10 Σ

x 2i = 385,

yi = 9811, i=1

xiyi = 65,550,

i=1

Σ10 y i2= 11,280, 031. i=1

Thus α ^ = 9811/10 = 981.1 and 65,550 − (55)(9811)/10 11589.5 β^ = = = 140.4788.

385 − (55)2/10 82.5 The least squares regression line is y^ = 981.1 + 140.4788(x − 5.5) = 208.467 + 140.479x; (b)

y 1600 1400 1200 1000 800 600 400

x 1

2

3

4

5

6

7

8

9

10

Figure 7.6–6: Number of programs (y) vs. year (x) (c) 1753.733 ± 160.368 or [1593.365, 1914.101].

4.0

96

Section 7.6 More Regression

7.6–8 Let K(β1, β2, β3) =

n Σ

(yi − β1 − β2x1i − β3x2i)2. Then

i=1

∂K ∂β1 ∂K ∂K2 ∂β3

=

n Σ

2(y i− β 1 — β2x1i — β3x2i)(−1) = 0;

i=1 n

Σ — β x — β x )(−x ) = 0; = Σ 2(y i− β 1 2 1i 3 2i 1i i=1 n 2(y − β — β x — β x )(−x ) = 0.

2 1i 3 2i 2i i 1

=

i=1

Thus, we must solve simultaneously the three equations ! ! n n nβ1 + x1i β2 + x2i β3 = i=1

!

n

x1i

!

n

β1 +

!

i=1

i=1

n

x2i β1 +

x1ix2i

i=1

!

n

β2 +

i=1

yi i=1

Σ

β3 =

x1ix2i

x 1i β2 + i=1 ! n

i=1

!

n

2

n

2

x2i β3 = i=1

n

i=1 x1iyi n Σ

x2iyi.

i=1

We have 12β1 + 4β2 + 4β3 = 23 4β1 + 26β2 + 5β3 = 75 4β1 + 5β2 + 22β3 = 37 so that β^1 =

4373 5956

β^2 =

= 0.734,

1430

3852 1489

= 2.587,

β^3 =

1489

= 0.960.

10 8

5

4 0

0

–4 –5

–8 –2

–1

0 1

2

3

–2

–1

y

1

2

3

–10

–2

–1

0 x

1

2

3

Figure 7.6–8: Two views of the points and the regression plane 7.6–10 (c) and (d)

(e) Linear regression is not appropriate. Finding the least-squares quadratic regression line using the raw data yields y^ = −1.88 + 9.86x − 0.995x2;

0 y

Chapter 7 Interval Estimation

97

y 24 22 20 18 16

8

14 12

0

6 4 2 1

2

3

4

5

6

7

8

–2 –4 –6

10 8 6 4 2

–8 x 1

2

3

4 5 6 7 8 9 Figure 7.6–10: (y) versus (x) with linear regression line and residual plot

(f) and (g)

y 1.8

24 22 20 18 16 14 12 10 8 6 4 2

1.4 1.0 0.6 0.2 –0.2

2

4

6

–0.6 –1.0 1

2

3

4

5

6

7

8

9

10

x –1.4

Figure 7.6–10: (y) versus (x) with quadratic regression curve and residual plot

7.6–12 The x is part of the exponent of e. However ln µ(x) = ln β1 + β2x is linear in x.

7.6–14 Recall that

√

n( α − α) ^ α −α σ T = T2 = s s= nσ^2 σ (n − 2) 2

has a t distribution with n − 2 degrees of freedom. Thus

^ σ2 n−2

8

9

98

Section 7.6 More Regression α^ − α ≤ t γ/2 (n−2) P − t γ / 2 (n−2) ≤ q ^ s σ2/(n − 2) P − α^ − tγ/2 ^2 ≤ −α ≤ − α^ +t s σ σ ^2 γ/2

n−2

σ2 ^ + tγ/2 ≤ α≤α n^ −2

Pα ^ − tγ/2

= 1− γ

n−2 s

s

= 1−γ

^2 σ

=

n−2

1 − γ.

It follows that the endpoints for a 100(1 − γ)% confidence interval for α are

7.6–16 Recall that α^= 2.73, confidence interval are

s σ ^2 . α ^ ± tγ/2(n−2) n − 2 β^ = 4.64/5.04, σ^2 = 0.184925, n = 10. The endpoints for the 95% r

0.184925 or [2.379, 3.081] for α; 8 1.84925 s 4.64/5.04 2.306 or [0.4268, 1.4145] for β; 8(5.04) ± 1.84925 1.84925 , = [0.105, 0.848] for σ2.

17.54 2.180 2.73 ± 2.306

7.6 –18 (a) The given facts, plus the result from Exercise 6.5-2 that β^ has a N (β, σ2/

Σn

distribution, imply that

i=1

x2i)

β^− β √Σ n 2 ^ σ2/ i=1 xi β−β =q s 2 2 ^2 /[(n − 1) Σn x2] ^ nσ n σ /σ i=1 i n−1

has a t(n−1) distribution. Thus, for γ ∈ (0, 1), β^− β P −tγ/2 (n−1) ≤ q Σn ^2 /[(n − 1) nσ i=1

x2]

≤ tγ/2 (n−1) = 1 − γ,

i

and algebraic manipulation of the inequality to get β alone in the middle yields the 100(1 − γ)% confidence interval for β^ as s nσ ^2 Σn ; β^ ± tγ/2(n−1) (n − 1) i=1 xi2 n (b) Yn+1 is N (βxn+1, σ2) and β^ is [as noted in the solution to N (β, σ2/ Σ x2), Σnpart (a)] 2 2 2 2 ^ , σ + x σ / so Y −Y i=1 – βx is N (βx 2 i n+1 n+1 n+1 x ), that is, N [0, σ (1 + n+1 xn+1/ Σ i=1 xi )]. This, plus the facts given in part (a), imply that 2

n

i=1

2

n+1

i

Y^n+1 – YΣ n+1 =q n q 2 s 2 2 Σn 2 ^ − Yn+n1+1 x ) Y i=1 i / σ (1 + x n^ i=1 xi ) +1 n+1 nσ2/σ2 [nσ^2 /(n − 1)](1 + x2 / n−1

Chapter 7 Interval Estimation

99

has a t(n−1) distribution. Thus, for γ ∈ (0, 1), P −t

Y^n+1 – Yn+1

(n−1) ≤ q

γ/2

≤ tγ/2 (n−1) = 1 − γ,

Σ /

[nσ^2 /(n − 1)](1 + x2n+1

in=1

x2i)

and algebraic manipulation of the inequality to get Yn+1 alone in the middle yields the 100(1 − γ)% prediction interval for Yn+1 as ‚ s n ^2 . nσ ,1 + x2 / Y^n+1 ± tγ/2(n−1) x2. n+1

n−1

i

i=1

7.6–20 (a) W =

^ β + tγ/2(n−2)

s (n − 2)

s =

2tγ/2(n−2)

(n − 2)

n^ σ2 Σn

i=1 (xi — x)

nσ ^2 Σn

i=1 (xi — x̄ )

2

2

^ β − tγ/2(n−2)

–

^2 nσ

2

E(W ) =

Σn

n^ σ2

;

(b) 2

s

!

Σn 4tγ/2(n−2) (n – 2) i=1 (xi — x)2 nσ2 2 ; Σ n = 4tγ/2(n−2) (x i — x)2 (n − 2) E

i=1

(x − x)2) = 24 so E(W 2) = [4t2 (4)] · [6σ ]/[(4)(24)] = γ/2 i=1 i γ/2 Σ (1/4)t2γ/2(4), while for the second data set, ni=1(x −x)2) = 8 so E(W 2) = [4t2 · i (4)] γ/2 [6σ2]/[(4)(8)] = (3/4)t2 (Σ4 )n. The expe2 cted squared width is smallest for the first data set because its value of i=1 (xi − x) is larger.

(c) For the first data set,

7.7

2

Resampling Methods

7.7 –2 (a) > read ‘C:\\HTZ-CD\\Maple Examples\\stat.m‘:

with(plots): read ‘C:\\HTZ-CD\\Maple Examples\\HistogramFill.txt‘: read ‘C:\\HTZ-CD\\Maple Examples\\ScatPlotCirc.txt‘: read ‘C:\\HTZ-CD\\Maple Examples\\Chapter_07.txt‘: XX := Exercise_7_7_2; XX := [12.0, 9.4, 10.0, 13.5, 9.3, 10.1, 9.6, 9.3, 9.1, 9.2, 11.0, 9.1, 10.4, 9.1, 13.3, 10.6] > Probs := [seq(1/16, k = 1 .. 16)]: XXPDF := zip((XX,Probs)-> (XX,Probs), XX, Probs): > for k from 1 to 200 do X := DiscreteS(XXPDF, 16): Svar[k] := Variance(X): od: Svars := [seq(Svar[k], k = 1 .. 200)]: > Mean(Svars); 1.972629584

100

Section 7.7 Resampling Methods > xtics := [seq(0.4*k, k = 1 .. 12)]: ytics := [seq(0.05*k, k = 1 .. 11)]: P1 := plot([[0,0],[0,0]], x = 0 .. 4.45, y = 0 .. 0.57, xtickmarks=xtics, ytickmarks=ytics, labels=[‘‘,‘‘]): P2 := HistogramFill(Svars,0 .. 4.4, 11): display({P1, P2}); The histogram is shown in Figure 7.7-2(ab). (b) > theta := Mean(XX) - 9; for k from 1 to 200 do Y := ExponentialS(theta,21): Svary[k] := Variance(Y): od: Svarys := [seq(Svary[k], k = 1 .. 200)]: θ := 1.31250000 > Mean(Svarys); 1.747515570 > xtics := [seq(0.4*k, k = 1 .. 14)]: ytics := [seq(0.05*k, k = 1 .. 15)]: P3 := plot([[0,0],[0,0]], x = 0 .. 5.65, y = 0 .. 0.62, xtickmarks=xtics, ytickmarks=ytics, labels=[‘‘,‘‘]): P4 := HistogramFill(Svarys,0 .. 5.6, 14): display({P3, P4}); 0.6

0.5

0.5 0.4

0.4

0.3

0.3

0.2

0.2

0.1

0.1 0.4 0.8 1.2 1.6 2.0 2.4 2.8 3.2 3.6 4.0 4.4

0.4

1.2

2.0

2.8

3.6

4.4

5.2

2

Figure 7.7–2: (ab) Histogram of S s: Resampling on Left, From Exponential on Right

>

> Svars := sort(Svars): Svarys := sort(Svarys): xtics := [seq(k*0.5, k = 1 .. 18)]: ytics := [seq(k*0.5, k = 1 .. 18)]: P5 := plot([[0,0],[5.5,5.5]], x = 0 .. 5.4, y = 0 .. 7.4, color=black, thickness=2, xtickmarks=xtics, ytickmarks=ytics, labels=[‘‘,‘‘]): P6 := ScatPlotCirc(Svars,Svarys): display({P5, P6});

Chapter 7 Interval Estimation

101

7 6 5 4 3 2 1 1

2

3

4

5

Figure 7.7–2: (c) q-q Plot of Exponential S2s Versus Resampling S2s Note that the variance of the sample variances from the exponential distribution is greater than the variance of the sample variances from the resampling distribution.

7.7–4 (a) > with(plots): read ‘C:\\HTZ-CD\\Maple Examples\\stat.m‘: read ‘C:\\HTZ-CD\\Maple Examples\\HistogramFill.txt‘: read ‘C:\\HTZ-CD\\Maple Examples\\Chapter_07.txt‘: XX := Example_7_7_4; XX := [261, 269, 271, 274, 279, 280, 283, 284, 286, 287, 292, 293, 296, 300, 304, 305, 313, 321, 322, 329, 341, 343, 356, 364, 391, 417, 476]; Probs := [seq(1/27, k = 1 .. 27)]: XXPDF := zip((XX,Probs)-> (XX,Probs), XX, Probs); XXPDF := [261, 1/27, 269, 1/27, 271, 1/27, 274, 1/27, 279, 1/27, 280, 1/27, 283, 1/27, 284, 1/27, 286, 1/27, 287, 1/27, 292, 1/27, 293, 1/27, 296, 1/27, 300, 1/27, 304, 1/27, 305, 1/27, 313, 1/27, 321, 1/27, 322, 1/27, 329, 1/27, 341, 1/27, 343, 1/27, 356, 1/27, 364, 1/27, 391, 1/27, 417, 1/27, 476, 1/27]; for j from 1 to 500 do X := DiscreteS(XXPDF, 27): Y := sort(X): YY[j] := Y[7]: od: Y7 := [seq(YY[k], k = 1 .. 500)]: Y7sorted := sort(Y7): Y7sorted[1]; 271 Y7sorted[500]; 304 >xtics := [seq(273 + 5*k, k = 0 .. 7)]: ytics := [seq(0.01*k, k = 1 .. 7)]: P1 := plot([[265,0],[265,0]], x = 268 .. 306, y = 0 .. 0.0771, xtickmarks=xtics, ytickmarks=ytics, labels=[‘‘,‘‘]): P2 := HistogramFill(Y7,270.5 .. 304.5, 7): display({P1, P2});

102

Section 7.7 Resampling Methods

0.07 0.06 0.05 0.04 0.03 0.02 0.01 273

278

283

288

293

298

303

Figure 7.7–4: (a) A histogram of N = 500 seventh order statistics (b)

(c)

>Here is an estimate of the first quartile.

Mean(Y7); 141443 = 282.886; 500 >Here is an 82% confidence interval for the first quartile using, in an ordered Y7, the 44th number and the 206th number.

[279, 283];

(d) In Exercise 7.5-2, the 82% confidence interval for π0.25 was [274, 287].

7.7–6 (a) > with(plots): read ‘C:\\HTZ-CD\\Maple Examples\\stat.m‘: read ‘C:\\HTZ-CD\\Maple Examples\\ScatPlotPoint.txt‘: read ‘C:\\HTZ-CD\\Maple Examples\\EmpCDF.txt‘: read ‘C:\\HTZ-CD\\Maple Examples\\HistogramFill.txt‘: read ‘C:\\HTZ-CD\\Maple Examples\\ScatPlotCirc.txt‘: read ‘C:\\HTZ-CD\\Maple Examples\\Chapter_07.txt‘: Pairs := Exercise_7_7_6; Pairs := [[2.500, 72], [4.467, 88], [2.333, 62], [5.000, 87], [1.683, 57], [4.500, 94], [4.500, 91], [2.083, 51], [4.367, 98], [1.583, 59], [4.500, 93], [4.550, 86], [1.733, 70], [2.150, 63], [4.400, 91], [3.983, 82], [1.767, 58], [4.317, 97], [1.917, 59], [4.583, 90], [1.833, 58], [4.767, 98], [1.917, 55], [4.433, 107], [1.750, 61], [4.583, 82], [3.767, 91], [1.833, 65], [4.817, 97], [1.900, 52], [4.517, 94], [2.000, 60], [4.650, 84], [1.817, 63], [4.917, 91], [4.000, 83], [4.317, 84], [2.133, 71], [4.783, 83], [4.217, 70], [4.733, 81], [2.000, 60], [4.717, 91], [1.917, 51], [4.233, 85], [1.567, 55], [4.567, 98], [2.133, 49], [4.500, 85], [1.717, 65], [4.783, 102], [1.850, 56], [4.583, 86],[1.733, 62]]: > r := Correlation(Pairs); r := .9087434803 >

xtics := [seq(1.4 + 0.1*k, k = 0 .. 37)]: ytics := [seq(48 + 2*k, k = 0 .. 31)]:

Chapter 7 Interval Estimation

103

P1 := plot([[1.35,47],[1.35,47]], x = 1.35 .. 5.15, y = 47 .. 109, xtickmarks = xtics, ytickmarks=ytics, labels=[‘‘,‘‘]): P2 := ScatPlotCirc(Pairs): display({P1, P2});

100 90 80 70 60 50 1.5

2.0

2.5

3.0

3.5

4.0

4.5

5.0

Figure 7.7–6: (a) Scatterplot of the 50 Pairs of Old Faithful Data (b) > Probs := [seq(1/54, k = 1 .. 54)]: EmpDist := zip((Pairs,Probs)-> (Pairs,Probs), Pairs, Probs): > for k from 1 to 500 do Samp := DiscreteS(EmpDist, 54); RR[k] := Correlation(Samp): od: R := [seq(RR[k], k = 1 .. 500)]: rbar := Mean(R); rbar := .9079354926 (c) > xtics := [seq(0.8 + 0.01*k, k = 0 .. 20)]: ytics := [seq(k, k = 1 .. 25)]: > P3 := plot([[0.79, 0],[0.79,0]], x = 0.79 .. 1.005, y = 0 .. 23.5, xtickmarks=xtics, ytickmarks=ytics, labels=[‘‘,‘‘]): P4 := HistogramFill(R, 0.8 .. 1, 20): display({P3, P4}); The histogram is plotted in Figure 7.7-6 ce;

104

Section 7.7 Resampling Methods (d) Now simulate a random sample of 500 correlation coefficients, each calculated from a sample of size 54 from a bivariate normal distribution with correlation coefficient r = 0.9087434803. > for k from 1 to 500 do Samp := BivariateNormalS(0,1,0,1,r,54): RR[k] := Correlation(Samp): od: RBivNorm := [seq(RR[k], k = 1 .. 500)]: AverageR := Mean(RBivNorm); AverageR := .9073168034 > P5 := plot([[0.79, 0],[0.79,0]], x = 0.79 .. 1.005, y = 0 .. 18.5, xtickmarks=xtics, ytickmarks=ytics, labels=[‘‘,‘‘]): P6 := HistogramFill(RBivNorm, 0.8 .. 1, 20): display({P5, P6}); (e)

20

15

15 10 10 5 5

0.80

0.85

0.90

0.95

1

0.80

0.85

0.90

0.95

Figure 7.7–6: (ce) Histograms of Rs: From Resampling on Left, From Bivariate Normal on Right

1

Chapter 7 Interval Estimation

105

(f) > R := sort(R): RBivNorm := sort(RBivNorm): xtics := [seq(0.8 + 0.01*k, k = 0 .. 20)]: ytics := [seq(0.8 + 0.01*k, k = 0 .. 20)]: P7 := plot([[0.8, 0.8],[1,1]], x = 0.8 .. 0.97, y = 0.8 ........0.97, color=black, thickness=2, labels=[‘‘,‘‘], xtickmarks=xtics, ytickmarks=ytics): P8 := ScatPlotCirc(R, RBivNorm): display({P7, P8});

0.96 0.94 0.92 0.90 0.88 0.86 0.84 0.82 0.82 0.84 0.86 0.88 0.90 0.92 0.94 0.96 Figure 7.7–6: (f) q-q Plot of the Values of R from Bivariate Normal Versus from Resampling

> StDev(R); StDev(RBivNorm); .01852854051 .02461716901 The means are about equal but the standard deviation of the values of R from the bivariate normal distribution is larger than that of the resampling distribution.

Chapter 8 Tests of Statistical Hypotheses

107

Chapter 8

Tests of Statistical Hypotheses 8.1

Tests About One Mean

8.1–2 (a) The critical region is z=

x − 1 3.0 x − 13.0 √= √ ≤ −1.96; 0.2/ n 0.2/ 25

(b) The observed value of z, z=

12.9 − 13.0 0.04

= – 2.5,

is less that -1.96 so we reject H0. (c) The p-value of this test is P (Z ≤ −2.50) = 0.0062. | x − 7.5 | √ 8.1 –4 (a) |t| = ≥ t0.025(9) = 2.262. s/ 10 0.4

0.3

T, r = 9 d.f.

0.2

0.1  =  –3

–2

 =  –1

1

2

3

Figure 8.1–4: The critical region is | t | ≥ 2.262 | 7.55 − 7.5| = 1.54 < 2.262, do not reject H0. √ 0.1027/ 10 (c) A 95% confidence interval for µ is 0.1027 0. 1 0 2 7 √ , 7.55 + 2.262 7.55 − 2.262 √ 10 10

(b) |t| =

= [7.48, 7.62].

108

Section 8.1 Tests About One Mean Hence, µ = 7.50 is contained in this interval. We could have obtained the same conclusion from our answer to part (b).

8.1 –6 (a) H0: µ = 3.4;

(b) H1: µ > 3.4; (c) t = (x − 3.4)/(s/3); (d) t ≥ 1.860; 0.4

0.3

T, r = 8 d.f.

0.2

0.1  =  –3

–2

–1

1

2

3

Figure 8.1–6: The critical region is t ≥ 1.860 (e) t = 3.556 − 3.4 = 2.802 ; 0.167/3 (f) 2.802 > 1.860, reject H0; (g) 0.01 < p-value < 0.025, p-value = 0.0116. | x − 125 | √ 8.1 –8 (a) | t | = ≥ t0.025(14) = 2.145; s/ 15 0.4

0.3

T, r = 14 d.f.

0.2

0.1  = 

 = 

–3 –2 –1 1 2 3 Figure 8.1–8: The critical region is | t | ≥ 2.145 (b) | t | =

| 127.667 − 125 | √ = 1.076 < 2.145, do not reject H .

9.597/ 15

0

Chapter 8 Tests of Statistical Hypotheses

109

8.1 –10 (a) The test statistic and critical region are given by

x − 5. 7 0 ≥ 1.895; √ s/ 8 (b) The observed value of the test statistic is 5.869 − 5 . 7 0 t= √ = 2.42; 0.19737/ 8 t=

(c) The p-value is a little less than 0.025. Using the computer, the p-value = 0.023.

–3

–2

–1

1

2

–3

3

–2

–1

1

2

3

Figure 8.1–10: A T (7) pdf showing the critical region on the left, p-value on the right 8.1 –12 The critical region is

d−0 ≥ 1.746 = t0.05(16). √ s d/ 17 Because d = 4.765 and sd = 9.087, t = 2.162 > 1.746 and we reject H0. t=

8.2

Tests of the Equality of Two Means

8.2 –2 (a) t = s

(b) t =

x− y

1 15sX2 + 12sy Y 2 1 + 16 13 27 415.16 − 347.40 s

≥ t0.01 (27) = 2.473;

= 5.570 > 2.473, reject H0.

1

1

15(1356.75) + 12(692.21) + 27 16 13 1356.75 (c) c = = 0.662, 1356.75 + 692.21 1 0.6622 0.3382 = + = 0.0387, r 15 12 r = 25.

The critical region is therefore t ≥ t0.01(25) = 2.485. Because t = 5.570 > 2.485, we again reject H0.

110

Section 8.2 Tests of the Equality of Two Means 8.2 –4 (a) t = s

x− y 2

12s + 15s2 X

1

≤ −t0.05(27) = −1.703;

1

Y

27 (b) t = s

+ 13 16 72.9 − 81.7 1

(12)(25.6)2 + (15)(28.3)2 27

13

+

1

= −0.869 > −1.703, do not reject H0;

16

(c) 0.10 < p-value < 0.25.

8.2 –6 (a) Assuming σ2 = σ2 , X

Y

|x− y| |t| = s ≥ t0.025(18) = 2.101; 2 2 9sX + 9sY 1 + 1 10 10 18 (b) | − 2.151 | > 2.101, reject H0; (c) 0.01 < p-value < 0.05; using a computer, p-value = 0.045; (d) Yes. X

Y

110

120

130

140

150

Figure 8.2–6: Box-and-whisker diagram for stud 3 (X) and stud 4 (Y ) forces x−y

8.2 –8 t = s 2

= 3.380 > 2.400 = t0.01(52),

2

24sX + 28sY 52

1 + 1 25 29

reject µX = µY .

4.1633 − 5.1050 8.2–10 t = r

r

= −1.648. Because −1.330 < −1.648 < −1.734, + 1

11(0.91426) + 7(2.59149) 1 18 12 8 0.05 < p-value < 0.10. In fact, p-value = 0.058. We would reject H0 at an α = 0.05

significance level and fail to reject H0 at an α = 0.10 significance level. y−x 8.2–12 (a) s 2 > 1.96; 2 sY + sX nY nX (b) 9.01 > 1.96, reject µX = µY ;

Chapter 8 Tests of Statistical Hypotheses

111

(c) Yes. X

Y

5

6

7

8

9

10

11

Figure 8.2–12: Lengths of male (X) and female (Y ) green lynx spiders y = 4.3921, sX2 = 6.0191, sY2 = 1.9776. Using the number of degrees of freedom given by Equation 7.2-1 (Welch) we have that r = [28.68♩ = 28. We have 5.9947 − 4.3921 t√= = 2.47 > 6.0191/19 + 1.9776/19 2.467= t (28)0.01 so we reject H0.

8.2 –14 (a) For these data, x = 5.9947,

(b) Beech

Maple

2 4 6 8 10 Figure 8.2–14: Tree dispersion distances in meters

8.3

Tests for Variances

8.3 –2 (a) Reject H0 if χ2 =

10s2Y 5252

≤ χ20.95 (10) = 3.940.

The observed value of the test statistic, χ2 = 4.223 > 3.940, so we fail to reject H0. s2 113108.4909 = 0.9718 so we clearly accept the equality of the variances. (b) F = 2X = sY 116388.8545 (c) The critical region is | t | ≥ t0.025(20) = 2.086. x — y− 0 3385.909 − 3729.364 = = – 2.378. t= √ 2 2 s X/11 + s Y/11 144.442 Because| − 2.378|> 2.086, we reject the null hypothesis. The p-value for this test is 0.0275.

112

Section 8.2 Tests of the Equality of Two Means 8.3 –4 (a) The critical region is

19s2

2

χ =

(0.095)2

≤ 10.12.

The observed value of the test statistic, 2

χ =

19(0.065)2 (0.095)2

= 8.895,

is less than 10.12, so the company was successful. (b) Because χ20.975 (19) = 8.907, p-value ≈ 0.025.

8.3 –6 (a) Because

2

χ 0.975 (22) = 10.98

and

χ20.025 (22) = 36.78,

H0 will be rejected if 22s2

2

χ =

100

22s2

2

≤ 10.98

or

χ =

= 49.91

or

s2 ≥ c =

100

≥ 36.78

or, equivalently, if s2 ≤ c1 =

100(10.98)

100(36.78)

2

22

22

(b)

0.06

2(22)

0.04

0.02 /2 = 0.025

2 = 32.52

10

20

30

/2 = 0.025 40

Figure 8.3–6: Test about the variance of IQ scores (c) Because 49.91 < s2 = 147.82 < 167.18, H0 is not rejected.

100 2 100 22S2 (d) Var(S2) = Var = · (2)(22) = 10,000/11.

22 100 22 (e) The 95% confidence interval for σ2, (22)(147.82) (22)(147.82) , = [88.42, 296.18], 36.78 10.98 contains σ 2 = 100.

= 167.18.

Chapter 8 Tests of Statistical Hypotheses 8.3 –8 (a)

(b)

113

s2X 9.88 = = 2.42 < 3.28 = F0.05(12, 8), so fail to reject H0. s2Y 4.08 9.88/4.08 9.88 9.88/4.08 9.88 ,F (8, 12) = , (2.85) = [0.738, 6.901].

0.05 F0.05(12, 8) 4.08 3.28 4.08

12.92 s2X = 3.30 > 3.07 = F0.05(9−1, 11−1), so we reject H0. = s2Y 7.12 93 − 132 = −8.12 < −1.645 so H0 is clearly rejected. A modification (b) z = √ 12.92/9 + 7.12/11 of this test would be to use Welch’s t with 11 degrees of freedom. (See Equation 7.2-1.)

This also leads to rejection of H0.

8.3 –10 (a) F =

9201 8.3 –12 F = = 1.895 < 3.37 = F0.05(6, 9), so we fail to reject H0.

4856

8.4

Tests about Proportions C = {x : x = 0, 1, 2 }; (b) α = P (X = 0, 1, 2; p = 0.6) = (0.4)4 + 4(0.6)(0.4)3 + 6(0.6)2(0.4)2 = 0.5248;

8.4 –2 (a)

β

= P (X = 3, 4; p = 0.4) = 4(0.4)3(0.6) + (0.4)4 = 0.1792.

OR (a′) C = {x : x = 0, 1 }; (b′) α = P (X = 0, 1; p = 0.6) = (0.4)4 + 4(0.6)(0.4)3 = 0.1792; β

= P (X = 2, 3, 4; p = 0.4) = 6(0.4)2(0.6)2 + 4(0.4)3(0.6) + (0.4)4 = 0.5248.

8.4 –4 Using Table II in the Appendix,

(a) (b)

α = P (Y ≥ 13; p = 0.40) = 1 − 0.8462 = 0.1538; β = P (Y ≤ 12; p = 0.60) = P (25 − Y ≥ 25 − 12) where 25 − Y is b(25, 0.40) = 1 − 0.8462 = 0.1538.

8.4 –6 The value of the test statistic is

0.70 − 0.75 = −2.280. z=√ (0.75)(0.25)/390 (a) Because z = −2.280 < −1.645, reject H0. (b) Because z = −2.280 > −2.326, do not reject H0. (c) p-value ≈ P (Z ≤ −2.280) = 0.0113. Note that 0.01 < p-value < 0.05.

8.4 –8 Test H0: p = 0.334 against H1: p > 0.334. The computed test statistic is

Z = √

(179/480) − 0.334 0.334(1 – 0.334)/480

= 1.81,

which exceeds the critical value of z0.05 = 1.645. So H0 is rejected at the 0.05 level of significance, indicating that the educational requirement is met. y/n − 0.65 ≥ 1.96; 8.4 –10 (a) z = √ (0.65)(0.35)/n

114

Section 8.5 The Wilcoxon Tests 414/600 − 0.65 = 2.054 > 1.96, reject H0 at α = 0.025; (b) z = √ (0.65)(0.35)/600 (c) Because the p-value ≈ P (Z ≥ 2.054) = 0.0200 < 0.0250, reject H0 at an α = 0.025 significance level; (d) A 95% one-sided confidence interval for p is √ [0.69 − 1.645 (0.69)(0.31)/600 , 1] = [0.659, 1].

z =|^ | 8.4–12 (a) √

| p − 0.20|

8.4–14 (a) z = √

p^1 − p^2

≥ 1.96; (0.20)(0.80)/n (b) Only 5/54 for which z = −1.973 leads to rejection of H0, so 5% reject H0; (c) 5%; (d) 95%; 219/1124 − 0.20 = −0.43, so fail to reject H0. (e) z = √ (0.20)(0.80)/1124

(b) z = √

p^(1 − ^ p)(1/n1 + 1/n2) 0.15 − 0.11

≥ 1.645;

(0.1325)(0.8675)(1/900 + 1/700) (c) z = 2.341 > 2.326, reject H0; (d) The p-value ≈ P (Z ≥ 2.341) = 0.0096.

204/300 − 0.73 −0.05 8.4 –16 z = √ = = −1.95; (0.73)(0.27)/300 0.02563

= 2.341 > 1.645, reject H0;

p-value ≈ P (Z < −1.95) = 0.0256 < α = 0.05 so we reject H0. That is, the test indicates that there is progress.

8.5

Some Distribution-Free Tests

8.5–2 In the following display, those observations that were negative are underlined. |x| :

1

2

2

2

2

3

4

4

4

5

6

6

Ranks :

1

3.5

3.5

3.5

3.5

6

8

8

8

10

12

12

|x| : Ranks :

6 12

7 14.5

7 14.5

8 16

11 17

12 18

13 19

14 20.5

14 20.5

17 22

18 23

21 24

The value of the Wilcoxon signed rank statistic is w

= −1 − 3.5 − 3.5 − 3.5 + 3.5 − 6 − 8 − 8 − 8 − 10 − 12 + 12 + 12 + 14.5 + 14.5 + 16 + 17 + 18 + 19 − 20.5 + 20.5 + 22 + 23 + 24

= 132.

For a one-sided alternative, the approximate p-value is, using the one-unit correction, !

W− 0 132 − 1 − 0 P (W ≥ 132) = P √ ≥ 70 24(25)(49)/6 ≈ P (Z ≥ 1.871) = 0.03064.

For a two-sided alternative, p-value = 2(0.03064) = 0.0613 > 0.05. Thus, we do not reject H0.

Chapter 8 Tests of Statistical Hypotheses

115

8.5–4 In the following display, those observations that were negative are underlined. |x| : 0.0790 Ranks :

0.5901

0.7757

1.0962

1.9415

2

3

4

5

3.8545 7

5.9848 8

9.3820 9

74.0216 10

1

|x| : 3.0678 Ranks : 6

The value of the Wilcoxon signed rank statistic is w = −1 + 2 − 3 − 4 − 5 − 6 + 7 − 8 + 9 − 10 = −19.

Because

—19 + 1 − 0 = 0.917 < 1.96, |z| = . √ . . 10(11)(21)/6 .

we do not reject H0.

8.5–6 (a) In the following display, those differences that were negative are underlined. |xi − 50| :

2

2

2.5

3

4

4

4.5

6

7

Ranks : 1.5

1.5

3

4

5.5

5.5

7

8

9

|xi − 50| : 7.5 Ranks : 10

8 11.5

8 11.5

14.5 13

15.5 14

21 15

The value of the Wilcoxon signed rank statistic is w = 1.5 − 1.5 + 3 + 4 + 5.5 − 5.5 − 7 − 8 + 9 + 10 + 11.5 + 11.5 − 13 + 14 + 15 = 50.

Because 50 − 1 z=√ = 1.392 < 1.645, 15(16)(31)/6 we do not reject H0; (b) The approximate p-value is, using the one-unit correction, p-value =

P (W ≥ 50)

50 − 1

!

= P (Z ≥ 1.3915) = 0.0820. ≈ P Z≥√ 15(16)(31)/6 8.5–8 The 24 ordered observations, with the x-values underlined and the ranks given under each observation are: 0.7494 1

0.7546 2

0.7565 3

0.7613 4

0.7615 5

0.7701 6

0.7712

0.7719

0.7719

0.7720

0.7720

0.7731

7

8.5

8.5

10.5

10.5

12

0.7741

0.7750

0.7750

0.7776

0.7795

0.7811

Ranks :

13

14.5

14.5

16

17

18

Ranks :

0.7815 19

0.7816 20

0.7851 21

0.7870 22

0.7876 23

0.7972 24

Ranks : Ranks :

116

Section 8.5 The Wilcoxon Tests (a) The value of the Wilcoxon rank sum statistic is w = 1 + 2 + 3 + 5 + 6 + 7 + 8.5 + 8.5 + 10.5 + 13 + 14.5 + 16 = 95.

Thus p-value = P (W

95) ≤ √P

Z

!

95.5 − 150

≈

12(12)(25)/12

≤

= P (Z ≤ −3.15) < 0.001,

so that we clearly reject H0; (b)

y 0.80 0.79 0.78 0.77 0.76 0.75

x 0.75

0.76

0.77

0.78

0.79

0.80

Figure 8.5–8: q-q plot of pill weights, (good, defective) = (x, y)

8.5 –10 The ordered combined sample with the x observations underlined are:

Ranks:

67.4 1

69.3 2

72.7 3

73.1 4

75.9 5

77.2 6

77.6 7

82.5

83.2

83.3

84.0

84.7

86.5

87.5

Ranks:

9

10

11

12

13

14

15

Ranks:

87.6 16

88.3 17

88.6 18

90.2 19

90.4 20.5

90.4 20.5

92.7 22

78.9 8

94.4 95.0 23 24

The value of the Wilcoxon rank sum statistic is w = 1 + 2 + 3 + 5 + 6 + 7 + 10 + 11 + 12 + 16 + 19 + 20.5 = 112.5.

Because p-value = P (W ≤ 112.5) ≈ P we reject H0.

!

113 – (12)(25)/2 Z≤ √ 12(12)(25)/12

= P (Z ≤ −2.136) < 0.05,

Chapter 8 Tests of Statistical Hypotheses

117

8.5 –12 The ordered combined sample with the 48-passenger bus values underlined are:

Ranks:

104 1

184 2

196 3

197 4

248 5

253 6

260 7

279 8

300

308

323

331

355

386

393

396

Ranks:

9

10

11

12

13

14

15

16

Ranks:

414 17

432 18

450 19

452 20

The value of the Wilcoxon statistic is w = 1 + 6 + 9 + 10 + 11 + 12 + 16 + 17 + 20 = 102.

Because p-value = P (W ≥ 102) ≈ P

!

101.5 – (9)(21)/2 Z> √ (9)(11)(21)/12

= P (Z > 0.532) > 0.05,

we do not reject H0.

8.5 –14 (a) Here is the two-sided stem-and-leaf display.

Young Subjects 97 33 98 88 85 68 50 04 86 81 79 77 76 69 65 61 96

Stems 3• 4∗ 4• 5∗ 5• 6∗ 6• 7∗ 7• 8∗ 8• 9∗

Older Subjects

63 38 49 73 88 90 94 26 27 53 74 28 18 35 63 88 36

(b) The value of the Wilcoxon rank sum statistic, the sum of the ranks for the younger subjects, is w = 1 + 2 + 3 + 5 + 6 + 7 + 8 + 9 + 12 + 13 + 14 + 16 + 17 + 18 + 19 + 20 + 28 = 198.

Because !

198.5 – (17)(35)/2 p-value = P (W ≤ 198) ≈ P Z ≤ √ = P (Z ≤ −3.410) < 0.05, 17(17)(35)/12 we clearly reject H0. (c) The t test leads to the same conclusion.

8.5

–16 (a) Using the Wilcoxon rank sum statistic, the sum of the ranks for the normal air is w = 2 + 3 + 4 + 5 + 6 + 7 + 8 + 10 + 12 + 13 + 15 + 17 = 102.

118

Section 8.6 Power of a Statistical Test Because p-value = P (W ≤ 102) ≈ P

!

102.5 – (12)(21)/2 Z≤ √ 12(8)(21)/12

= P (Z ≤ −1.813) = 0.035,

we reject the null hypothesis. (b) Using a t statistic, we failed to reject the null hypothesis at an α = 0.05 significance level. (c) For these data, the results are a little different with the Wilcoxon rank sum statistic leading to rejection of the null hypothesis while the t test did not reject H0.

8.6

Power of a Statistical Test

8.6 –2 (a)

K(µ)

=

P ( X ≤ 354.05; µ)

05 − µ ; µ = P Z ≤ 354. √ 2/ 12 354.0 5 µ − = Φ √ ; 2/ 12 (b) α = K(355) = Φ

354.05 355 − = Φ(−1.645) = 0.05; √ 2/ 12

(c) K(354.05) = Φ(0) = 0.5; K(353.1)

= Φ(1.645) = 0.95;

K() 1.00

0.75

0.50

0.25  353

353.5

354

355 √ Figure 8.6–2: K(µ) = Φ([354.05 − µ]/[2/ 12 ]) (d) x = 353.83 < 354.05, reject H0; (e) p-value

= P ( X ≤ 353.83; µ = 355) = P (Z ≤ −2.03) = 0.0212.

354.5

Chapter 8 Tests of Statistical Hypotheses

119

Σ10 8.6–4 (a) α = P ( i =X 1 i ≤ 14; p = 0.4) = P (10 ≤ Y ≤ 14) where Y is distributed as negative binomial with parameters r = 10 and p = 0.4. So 9 10 11 α = 0.4100.60 + 0.4100.61 + 0.4100.62 9 9 9 12 13 + 0.4100.63 + 0.4100.64 9 9 = 0.0175; (b) β = 1 —P (Y ≤ 14) where Y is distributed as negative binomial with parameters r = 10 and p = 0.7. So 9 10 11 β = 1− 0.7100.30 − 0.7100.31 − 0.7100.32 9 9 9 12 13 10 3 – 0.7 0.3 − 0.7100.34 9 9 = 0.4158.

P (S2 < c | σ2 = 10) (n − 1)S2 (n − 1)c 2 σ| = 10 = P < 2 σ2 σ 23c = P Y < , 10 23c where Y is distributed as χ2(23). So = χ2 (23), i.e., c = (10/23)13.09 = 5.69; 0.95 10 2 2 (b) 0.9 = P (S < c | σ = 2) 23c (n − 1)S2 (n − 1)c 2 Y < σ = 2 = P | = P < σ2 2 σ2 23c where Y is distributed as χ2(23). So = χ2 (23), i.e., c = (2/23)32.01 = 2.78.

0.10 2 668.94 − µ ≤ ;µ 8.6–8 (a) K(µ) = P ( X 668.94; µ) = P ≤Z 140/5 668.94 − µ = Φ ; 140/5 668.94 − 715 (b) α = K(715) = Φ 140/5 = Φ(−1.645) = 0.05; 8.6–6 (a) 0.05

=

(c) K(668.94) = Φ(0) = 0.5; K(622.88) = Φ(1.645) = 0.95;

120

Section 8.6 Power of a Statistical Test (d)

K() 1.00

0.75

0.50

0.25

600

620

640

660

680

700

720

 740

Figure 8.6–8: K(µ) = Φ([668.94 − µ]/[140/5]) (e)

x = 667.92 < 668.94, reject H0;

(f) p-value

= P ( X ≤ 667.92; µ = 715) = P (Z ≤ −1.68) = 0.0465.

8.6 –10 (a) and (b)

K(p) 1.00

K(p) 1.00

0.75

0.75

0.50

0.50

0.25

0.25 p 0.3

0.4

0.5

0.6

0.7

p 0.2

0.3

0.4

0.5

0.6

Figure 8.6–10: Power functions corresponding to different critical regions and different sample sizes

8.6 –12 Let Y =

8 Σ

Xi. Then Y has a Poisson distribution with mean µ = 8λ.

i=1

(a) α

(b) (c)

=

P (Y ≥ 8; λ = 0.5) = 1 − P (Y ≤ 7; λ = 0.5)

= 1 − 0.949 = 0.051.

7 (8λ)ye−8λ Σ .

K(λ) = 1 − y! y=0 K(0.75)

=

1 − 0.744 = 0.256,

K(1.00) = 1 − 0.453 = 0.547, K(1.25)

=

1 − 0.220 = 0.780.

0.7

Chapter 8 Tests of Statistical Hypotheses

121

K() 1.00

0.75

0.50

0.25

0.5

1.0

1.5

2.0

Figure 8.6–12: K(λ) = 1 − P (Y ≤ 7; λ) 8.6 –14

K(θ) =

P (X − (Y ) ≥ c; θ)

! √ 4X00−/nY +−2θ25/n ≥ 25c/– θn; θ √ − 0 = 1.645 K(0) = 0.05 ⇒ 25/c √ n c − 10 √ = −1.282 K(10) = 0.90 ⇒ 25 / n √ c − 0 = 1.645(25/ n) √ c − 10 = −1.282(25/ ) n √ 10 = 2.927(25/ n) n = 53.55 3 54 √ c = 1.645(25/ 54) = 5.60. = P

8.7

Best Critical Regions 8.7 –2 (a)

L(4) L(16) 3Σ 2 – xi 32 n

=

√ (1/2 2π )n exp[−Σ x2/8 ] i √ (1/4 2π )n exp[−Σ x2i/32 ]

=

2n exp[−3Σ x2i/32 ] ≤ k

≤ ln k − ln 2n

i=1

n Σ

x2 i=1

i

(b) 0.05 = P

− ≥ − 3 (ln k ln 2n) = c; !

15 Σ X2i ≥ c; σ2 = 4 32

i=1

Σ15

= P

i=1

4

!

2

Xi

c ≥ 4 ; σ2 = 4

 2.5

122

Section 8.7 Best Critical Regions c

Thus

= χ2

(15) = 25 and c = 100.

0.05

4

!

15

(c)

β =

P i=1

X2i < 100; σ2 = 16 !

Σ

15 i=1

= P 8.7 –4 (a)

100 = 6.25 < 16

X2i

16

L(0.9) (0.9)Σxi (0.1)n−Σxi = L(0.8) (0.8)Σxi (0.2)n−Σxi Σn x

9 8

2 1

!

≤ k n

1 2

≤ k

ln(9/4)

≤

ln k + n ln 2

≤

ln k + n ln 2 = c. ln(9/4)

i

1

n Σ xi

≈ 0.025.

i=1

n

Σ y=

xi i=1

Recall that the distribution of the sum of Bernoulli trials, Y , is b(n, p). (b) 0.10 = P [ Y ≤ n(0.85); p = 0.9 ] " Y n(0.9) # n(0.85) − n(0.9) √ − ; p = 0.9 . = P ≤ √ n(0.9)(0.1) n(0.9)(0.1) n(−0.05) It is true, approximately, that √ = −1.282 n(0.3) n = 59.17 or n = 60. " # Y − 60(0.8) 51 − 48 > √ (c) β = P [ Y > n(0.85) = 51; p = 0.8 ] = P √ ; p = 0.8 60(0.8)(0.2) 9.6 ≈ (d) Yes. 8.7–6 (a) 0.05

=

P

=

1

Thus c1 − 80 3/4 c1 Similarly, c2 − 80 3/4

X − 80 c1 − 80 ≥ 3/4 3/4 c1 − 80 .

– Φ 3/4

= 1.645 = 81.234.

= −1.645

c2 c3 3/4

= 78.766;

c3

= 1.47.

= 1.96

P (Z ≥ 0.97) = 0.166.

Chapter 8 Tests of Statistical Hypotheses (b)

123

K1(µ) = 1 − Φ([81.234 − µ]/[3/4]); K2(µ) =

Φ([78.766 − µ]/[3/4];

K3(µ) =

1 − Φ([81.47 − µ]/[3/4]) + Φ([78.53 − µ]/[3/4]).

K() 1.00

0.75

0.50

0.25

K3  77

78

79

80

81

82

83

Figure 8.7–6: Three power functions 8.7 –8 Let

n Y L(θ) = θ (1 − xi)θ−1; n

i=1

thus

Y

n

n − L(θ ) = θθ0 1

0

i=1

(1 − x ) = 1,0 because θ0 = 1. So when θ > 1, a best critical region is L(θ0) 1 Q L(θ) = θn n (1 − x )θ−1 ≤ k. i

i=1

That is,

" θn

n

(1 − xi)

#θ−1

i=1

or

Y n (1 − xi) ≥ i=1

1 kθn

≥1 k

1/(θ−1)

= c.

This is true for all θ > 1; so it is a uniformly most powerful test.

8.7

–10 By the Neyman-Pearson Theorem, the desired critical region is of the form L(1) ≤ k L(2) Σ (1/2)n exp (− n |xi|) i=1 Σ (1/4)n exp ( − n |xi|/2) ≤ k i=1 Σn n exp(− i=1 |xi|/2) ≤ (1/2) k Σn ≥ −2 log[(1/2)nk] = c. i=1 |xi|

124

Section 8.8 Likelihood Ratio Tests

8.8

Likelihood Ratio Tests

8.8 –2 (a) If µ ∈ ω (that is, µ ≥ 10.35), then µ^ = x if x ≥ 10.35, but µ^ = 10.35 if x < 10.35.

Thus λ = 1 if x ≥ 10.35; but,Σif x < 10.35, then [1/(0.3)(2π)]n/2 exp[− n (x − 10.35)2/(0.06)] i Σ1 n1 λ= n/2 i 2 [1/(0.3)(2π)] exp[− h( x − x) /(0.06)] i n exp − (x − 10.35)2 0.0n6 2 – (x − 10.35) 0.06 x – 10.35 √ 0.03/n

≤ k ≤ k ≤

ln k

≤ −

√

−2 ln k

= −z0.05 = −1.645.

10.31 − 10.35 = −1.633 > −1.645; do not reject H0. √ (b) 0.03/50 (c) p-value = P (Z ≤ −1.633) = 0.0513. Σ [1/(2πσ2)]n/2 exp[− n (xi – µ0)2/(2σ2)] 0 0 i=1 Σni=1 i 02 n/2 2 2 0 [1/(2πσ (x − x)Σ/(2σ ) ] Σn)] (x exp[− 8.8 –4 (a) λ = − x + x − µ )2 n i 0 = exp – i=1 + i=1 (xi − x)2 2σ02 2σ02 2 —n(x − µ 0 ) = exp ≤ k 2σ02 −n(x − µ0)2 2σ02 |x − µ0| √ σ0 / n

≤

ln k

≥ c

x 59 |z| = | −√ | ≥ 1.96.

15/ n | 56.13 − 59 | (b) |z| = = | − 1.913| < 1.96, do not reject H ; 0 15/10 (c) p-value = P (|Z| ≥ 1.913) = 0.0558.

324.8 − 335 √ 8.8–6 t = = −1.051 > −1.337, do not reject H .

40/ 17 8.8–8 In Ω, ^µ = x. Thus, Σ n n (Σ 1/θ λ= 1 ) exp[−≤ k (1/x)n exp[− n x /x] 1 n

0

i

x exp[−n(x/θ0 − 1)] ≤ k.

Plottinθg0 λ as a function of w = x/θ0, we see that λ = 0 when x/θ0 = 0, it has a maximum when x/θ0 = 1, and it approaches 0 as x/θ0 becomes large. Thus λ ≤ k when x ≤ c1 or x ≥ c2.

Chapter 8 Tests of Statistical Hypotheses

125

Σ n Because the distribution of2 Xi is χ2(2n) when H0 is true, we could let the critical θ0 i=1

region be such that we reject H0 if n Σ Xi ≤ χ2 2 (2n) 1−α/2 θ0

or

i=1

n Σ 2 Xi ≥ χ2 (2n). α/2 θ0 i=1

1.00

0.75

0.50

0.25

0.5

1.0

1.5

2.0

2.5

3.0

Figure 8.8–8: Likehood functions: solid, n = 5; dotted, n = 10 8.8–10 Referring to Exercise 6.4-19, we test H0: γ2 = 1 against H1: γ2 /= 1. n " Σ 2 (yi —^µ)# n 1 !

Y √ exp − 2x2 2πx2 λ = ≤ k Yq 1 exp − ^π2γ2 ^2 2 2 i=1 i=1 2 γ x i 2 Σ i (yi − µ^ ) " # n n x 2 Σ 2 Σ (yi − µ (y − µ ^ ) i ^ ) + ≤ k 2 2 ^ 2 = ( γ^2 )n/2 exp – 2 γ x 2x i=1

i

i=1

i

2 )n/2 exp[−n γ 2^ = ( γ^ / 2 + n/2] ≤ k

=

2 )n/2 exp[n(1 − γ^ 2 )/2] ≤ k. ( γ^

This is a function of where

Σ (yi − µ^ ) 2 γ^2 = 1 n x2 i i=1 Σ n y /x2 i. i Σi=1 n µ ^= 1/x2 n

i=1

i

Chapter 9 More Tests

127

Chapter 9

More Tests 9.1

Chi-Square Goodness-of-Fit Tests

9.1–2 q4 =

(224 − 232)2

232 = 3.784.

+

(119 − 116)2 116

+

(130 − 116)2 116

+

(48 − 58)2 58

+

(59 − 58)2 58

The null hypothesis will not be rejected at any reasonable significance level. Note that E(Q4) = 4 when H0 is true.

9.1–4 q3 =

(124 − 117)2

(30 − 39)2

(43 − 39)2

(11 − 13)2

+ + + 117 39 39 13 2 = 0.419 + 2.077 + 0.410 + 0.308 = 3.214 < 7.815 = χ0.05 (3).

Thus we do not reject the Mendelian theory with these data.

9.1–6 Using Table II in Appendix B with p = 0.30, the hypothesized probabilities are p0 = P (X = 0) = 0.343, p1 = P (X = 1) = 0.441, p2 = P (X = 2) = 0.189, p3 = P (X = 3) = 0.027. Thus the respective expected values are 68.6, 88.2, 37.8, and 5.4. The value of the chi-square goodness of fit statistic is: (57 − 68.6)2 (95 − 88.2)2 (38 − 37.8)2 (10 − 5.4)2 q = + + + 68.6 88.2 37.8 5.4 = 6.405 < 7.815 = χ20.05 (3).

Do not reject the hypothesis that X is b(3, 0.30) at a 5% significance level. Limits for the p-value are 0.05 < p-value < 0.10 because χ20.10 (3) = 6.251 < 6.405 < χ20.05 (3) = 7.815.

We find that p^ = 201/600 = 0.335. Thus a 95% confidence interval for p is √ 0.335 ± 1.96 (0.335)(0.665)/600 or [0.297, 0.373].

The pennies that were used were minted 1998 or earlier. See Figure 9.1-6. Repeat this experiment with similar pennies or with newer pennies and compare your results with those obtained by these students.

128

Section 9.1 Chi-Square Goodness-of-Fit Tests

0.4 0.3 0.2 0.1

0

1

2

3

Figure 9.1–6: The b(3, 3/10) probability histogram and the relative frequency histogram (shaded)

9.1 –8 (a) Estimate λ by x = 50/40 = 1.25. Then

q3 =

[15 − (0.2865)(40)]2

+

(0.2865)(40) [7 − (0.2238)(40)]2

[12 − (0.3581)(40)]2

+ (0.2238)(40) = 1.999 < χ20.05 (2) = 5.991; do not reject H0;

+ (0.3581)(40) [6 − (0.1316)(40)]2 (0.1316)(40)

(b) Estimate p by 1/y = 1/2.25 = 4/9. Then [15 − (4/9)(40)]2 [12 − (20/81)(40)]2 q3 = + + (4/9)(40) (20/81)(40) [7 − (100/729)(40)]2 [6 − (125/729)(40)]2 + (100/729)(40) (125/729)(40) 2 = 1.415 < χ 0.05 (2) = 5.991; do not reject H0.

9.1 –10 The respective probabilities and expected frequencies are 0.050, 0.149, 0.224, 0.224, 0.168,

0.101, 0.050, 0.022, 0.012 and 15.0, 44.7, 67.2, 67.2, 50.4, 30.3, 15.0, 6.6, 3.6. The last two cells could be combined to give an expected frequency of 10.2. From Exercise 3.5–12, the respective frequencies are 17, 47, 63, 63, 49, 28, 21, and 12 giving q7 =

(17 − 15.0)2 15.0

+

(47

44.7)2 (12 − 10.2)2 – = 3.841. +··· + 10.2 44.7

Since 3.841 < 14.07 = χ2 0.05 (7), do not reject. The sample mean is x = 3.03 and the sample variance is s2 = 3.19 which also supports the hypothesis. The following figure compares the probability histogram with the relative frequency histogram of the data.

Chapter 9 More Tests

129

0.20 0.15 0.10 0.05

1

2

3

4

5

6

7

8

Figure 9.1–10: The Poisson probability histogram, λ = 3, and relative frequency histogram (shaded) 9.1 –12 (a) For the infected snails, x = 84.74, sx = 64.79;

For the control snails, y = 113.16, sy = 87.02; (b)

X

Y

0

50

100

150

200

250

300

Figure 9.1–12: Box-and-whisker diagrams for infected (X) and control (Y )

(c) We shall use 5 classes with equal probability for the control snails.

Ai

Observed

Expected

q

[0, 25.25) [25.25, 57.81) [57.81, 103.69) [103.69, 182.13) [182.13, ∞)

4 9 5 6 7

6.2 6.2 6.2 6.2 6.2

0.781 1.264 0.232 0.006 0.103

31

31.0

2.386

The p-value for 5 — 1− 1 = 3 degrees of freedom is 0.496 so we fail to reject the null hypothesis;

130

Section 9.2 Contingency Tables (d) We shall use 10 classes with equal probability for the infected snalis.

Ai

Observed

Expected

q

[0, 22.34) [22.34, 34.62) [34.62, 46.09) [46.09, 57.81) [57.81, 70.49) [70.49, 84.94) [84.94, 102.45) [102.45, 125.76) [125.76, 163.37) [163.37, ∞)

3 3 8 4 2 4 4 4 2 5

3.9 3.9 3.9 3.9 3.9 3.9 3.9 3.9 3.9 3.9

0.208 0.208 4.310 0.003 0.926 0.003 0.003 0.003 0.926 0.310

39

39.0

6.900

The p-value for 10 − 1 = 9 degrees of freedom is 0.648 so we fail to reject the null hypothesis.

9.2

Contingency Tables

9.2 –2 10.18 < 20.48 = χ20.025 (10), accept H0.

9.2 –4 In the combined sample of 45 observations, the lower third includes those with scores of 61

or lower, the middle third have scores from 62 through 78, and the higher third are those with scores of 79 and above. low

middle

high

Totals

Class U

9 (5)

4 (5)

2 (5)

15

Class V

5 (5)

5 (5)

5 (5)

15

Class W

1 (5)

6 (5)

8 (5)

15

Totals

15

15

15

45

Thus q = 3.2 + 0.2 + 1.8 + 0 + 0 + 0 + 3.2 + 0.2 + 1.8 = 10.4.

Since q = 10.4 > 9.488 = χ20.05 (4), we reject the equality of these three distributions. (p-value = 0.034.)

9.2 –6 q = 8.410 < 9.488 = χ20.05 (4), fail to reject H0. (p-value = 0.078.)

9.2 –8 q = 4.268 > 3.841 = χ20.05 (1), reject H0. (p-value = 0.039.)

9.2–10 q = 7.683 < 9.210 = χ20.01 (2), fail to reject H0. (p-value = 0.021.)

(2), reject H0. (p-value = 0.012.)

9.2 –12 q = 8.792 > 7.378 = χ2 0.025

Chapter 9 More Tests

131

9.2 –14 The contingency table is:

Eyes

A1

A2

A3

A4

Totals

Blue

22 (20)

17 (20)

21 (20)

20 (20)

80

Brown

14 (17.50)

20 (17.50)

20 (17.50)

16 (17.0)

70

Other

14 (12.50)

13 (12.50)

9 (12.50)

14 (12.50)

50

Totals

50

50

50

50

200

Because q = 3.603 < 12.59 = χ20.05 (6), do not reject the hypothesis of independent attributes. (p-value = 0.730.)

9.3

One-Factor Analysis of Variance 9.3 –2

Source

SS

DF

MS

F

p-value

Treatment Error Total

388.2805 316.4597 704.7402

3 12 15

129.4268 26.3716

4.9078

0.0188

F = 4.9078 > 3.49 = F0.05(3, 12), reject H0.

9.3 –4 (a)

Source

SS

DF

MS

F

p-value

Treatment Error Total

184.8 102.0 286.8

2 17 19

92.4 6.0

15.4

0.00015

F = 15.4 > 3.59 = F0.05(2, 17), reject H0; (b)

A

B

C 74

76

78

80

82

84

86

Figure 9.3–4: Box-and-whisker diagrams of strengths of different beams

132

Section 9.3 One-Factor Analysis of Variance

9.3 –6 (a) F ≥ F0.05(3, 24) = 3.01;

(b) Source

SS

DF

MS

F

p-value

Treatment Error Total

12,280.86 28,434.57 40,715.43

3 24 27

4,093.62 1,184.77

3.455

0.0323

F = 3.455 > 3.01, reject H0; (c) 0.025 < p-value < 0.05. (d)

X1

X2

X3

X4 140

160

180

200

220

240

260

280

Figure 9.3–6: Box-and-whisker diagrams for cholesterol levels 9.3 –8 (a) F ≥ F0.05(4, 30) = 2.69;

(b) Source

SS

DF

MS

F

p-value

Treatment Error

0.00442 0.01157

4 30

0.00111 0.00039

2.85

0.0403

Total

0.01599

34

F = 2.85 > 2.69, reject H0;

Chapter 9 More Tests (c)

133

X1 X2 X3 X4 X5 1.02

1.04

1.06

1.08

1.10

Figure 9.3–8: Box-and-whisker diagrams for nail weights 92.143 − 103.000

9.3–10 (a) t = s

6(69.139) + 6(57.669) 12

= −2.55 < −2.179, reject H0.

1+1 7 7

412.517 F=

= 6.507 > 4.75, reject H0.

63.4048 The F and the t tests give the same results since t2 = F ; 86.3336 (b) F = = 0.7515 < 3.55, do not reject H0.

114.8889 9.3 –12 (a) Source SS DF MS F p-value Treatment Error Total

122.1956 860.4799 982.6755

2 30 32

61.0978 28.6827

2.130

0.136

F = 2.130 < 3.32 = F0.05(2, 30), fail to reject H0; (b) D6

D7

D 22

165 170 175 180 185 Figure 9.3–12: Box-and-whisker diagrams for resistances on three days

134

Section 9.4 Two-Way Analysis of Variance

9.3 –14 (a)

Source

SS

DF

MS

F

p-value

Worker Error Total

1.5474 17.2022 18.7496

2 24 26

0.7737 0.7168

1.0794

0.3557

F = 1.0794 < 3.40 = F0.05(2, 24), fail to reject H0; (b) A

B

C

1 1.5 2 2.5 3 3.5 4 4.5 Figure 9.3–14: Box-and-whisker diagrams for workers A, B, and C The box plot confirms the answer from part (a).

9.4

Two-Way Analysis of Variance

9.4–2

µ + βj

6 10 8 8

3 7 5 5

7 11 9 9

8 12 10 10

µ + αi 6 10 8 µ=8

So α1 = −2, α2 = 2, α3 = 0 and β1 = 0, β2 = −3, β3 = 1, β4 = 2.

9.4 –4

a b Σ Σ

( Xi· − X··)(Xij − Xi· − X·j + X··)

i=1 j=1

a

=

b

Σ Σ ( Xi· − X··) [(Xij − Xi·) − ( X·j − X··)]

i=1

j=1 b

a

– =

i=1 a

( Xi·

X··) j=1 (Xij − Xi·) −

j=1 ( X·j − X··)

= Σ ( Xi· − X··)(0 − 0) = 0; a b Σ Σ i=1 j=1 a

b

i=1 j=1

i=1

( X·j − X··)(Xij − Xi· − X·j + X··) = 0, similarly; ( a ) b Σ Σ ( Xi· − X··)( X·j − X··) = ( Xi· − X··) ( X·j − X··) i=1

j=1

= (0)(0) = 0 .

Chapter 9 More Tests

135

9.4–6

µ + βj

6 10 8 8

7 3 5 5

7 11 9 9

12 8 10 10

µ + αi 8 8 8 µ=8

So α1 = α2 = α3 = 0 and β1 = 0, β2 = −3, β3 = 1, β4 = 2 as in Exercise 9.4–2. However, γ11 = −2 because 8 + 0 + 0 + (−2) = 6. Similarly we obtain the other γij’s : —2 2 0

2 —2 0

−2 2 2 −2 0 0

9.4 –8

Source

SS

DF

MS

F

p-value

Row (A) Col (B) Int(AB) Error Total

5,103.0000 6,121.2857 1,056.5714 28,434.5714 40,715.4286

1 1 1 24 27

5,103.0000 6,121.2857 1,056.5714 1,184.7738

4.307 5.167 0.892

0.049 0.032 0.354

(a) Since F = 0.892 < F0.05(1, 24) = 4.26, do not reject HAB ; (b) Since F = 4.307 > F0.05(1, 24) = 4.26, reject HA; (c) Since F = 5.167 > F0.05(1, 24) = 4.26, reject HB .

9.5

General Factorial and 2k Factorial Designs

9.5 –4 (a)

[A] = −28.4/8 = −3.55, [B] = −1.45, [C] = 3.2, [AB] = −1.525, [AC] = −0.525, [BC] = 0.375, [ABC] = −1.2;

(b)

Identity of Effect [A] [AB] [B] [ABC] [AC] [BC] [C]

Ordered Effect −3.550 −1.525 −1.450 −1.200 −0.525 0.375 3.20

Percentile 12.5 25.0 37.5 50.0 62.5 75.0 87.5

Percentile from N(0,1) −1.15 −0.67 −0.32 0.00 0.32 0.67 1.15

The main effects of temperature (A) and concentration (C) are significant.

136

Section 9.6 Tests Concerning Regression and Correlation

3 2 1 [ABC]

–3

–2[B]

–1

1

[AB]

[A]

[C]

[BC]

[AC]

2

3

–1 –2 –3

Figure 9.5–4: q-q plot

9.6

Tests Concerning Regression and Correlation

9.6 –2 The critical region is t1 ≥ t0.025(8) = 2.306. From Exercise 6.5–4,

β^ =

10 Σ

4.64/5.04 and n σ^2 = 1.84924; also

(xi − x)2 = 5.04, so

i=1

t1 =

s 4.64/5.04 = 0.9206 = 4.299.

1.84924 0.2142 8(5.04)

Since t1 = 4.299 > 2.306, we reject H0.

9.6–4 For these data, r = −0.413. Since |r| = 0.413 < 0.7292, do not reject H0.

9.6 –6 Following the suggestion given in the hint, the expression equals

(n − 1)S Y−

2RsX SY

2

9.6–8

2 sX

(n − 1)RsX SY +

u(R) ≈

R2s2 S2 (n − 1)s2 X

Y

s 4X

X

=

(n − 1)S2Y (1 − 2R2 + R2)

= (n − 1)S2Y(1 − R2).

u(ρ) + (R − ρ)u′(ρ),

Var[u(ρ) + (R − ρ)u′(ρ)] = [u′(ρ)]2Var(R) = [u′(ρ)]

22 2 (1 − ρ )

= c, which is free of ρ, n k/2 k/2 + , u′(ρ) = 1−ρ 1+ρ k k k 1+ρ u(ρ) = − ln(1 − ρ) + ln(1 + ρ) = ln 1−ρ 2 2 2 Thus, taking k = 1, 1 1+R u(R) = ln 2 1−R has a variance almost free of ρ.

.

Chapter 9 More Tests

137

9.6–10 (a) r = −0.4906, | r | = 0.4906 > 0.4258, reject H0 at α = 0.10; (b) | r | = 0.4906 < 0.4973, fail to reject H0 at α = 0.05.

9.6 –12 (a) r = 0.339, | r | = 0.339 < 0.5325 = r0.025(12), fail to reject H0 at α = 0.05;

(b) r = −0.821 < −0.6613 = r0.005(12), reject H0 at α = 0.005; (c) r = 0.149, | r | = 0.149 < 0.5325 = r0.025(12), fail to reject H0 at α = 0.05.

9.7

Statistical Quality Control

9.7 –2 (a) x = 67.44, s = 2.392, R = 5.88;

(b) UCL = x + 1.43(s) = 67.44 + 1.43(2.392) = 70.86; LCL = x − 1.43(s) = 67.44 − 1.43(2.392) = 64.02; (c) UCL = 2.09(s) = 2.09(2.392) = 5.00; LCL = 0;

UCL

5

74

4

72 UCL 70

3 _ x

68 66

_ s

2 1

64

LCL 2

4

6

8

10 12 14 16 18 20

LCL 2

4

6

8

10 12 14 16 18 20

Figure 9.7–2: (b) x-chart using s and (c) s-chart

138

Section 9.7 Statistical Quality Control (d) UCL = x + 0.58(R) = 67.44 + 0.58(5.88) = 70.85; LCL = x − 0.58(R) = 67.44 − 0.58(5.88) = 64.03; (e) UCL = 2.11(R) = 2.11(5.88) = 12.41; LCL = 0; UCL

74

12

72 UCL

10 8

70 _ x

68

6 4

66

_ R

2 64

LCL 2

4

6

8

LCL 2

10 12 14 16 18 20

4

6

8

10 12 14 16 18 20

Figure 9.7–2: (d) x-chart using R and (e) R-chart (f) Quite well until near the end. √ 9.7–4 With p = 0.0254, UCL = p + 3 p(1 − p)/100 = 0.073; √ with p = 0.02, UCL = p + 3 p(1 − p)/100 = 0.062; In both cases we see that problems are arising near the end.

0.08

UCL

0.08 UCL

0.06

0.06

0.04

0.04

_ p

0.02

_ p 0.02

2

4

6

LCL 8 10 12 14 16 18 20 22 24

2

4

6

LCL 8 10 12 14 16 18 20 22 24

Figure 9.7–4: p-charts using p = 0.0254 and p = 0.02

Chapter 9 More Tests

139

9.7–6 (a) UCL = c + 3

√ √ c = 1.80 + 3 1.80 = 5.825; LCL = 0; UCL

6 5 4 3

_ c

2 1

LCL

2

4

6

8

10 12 14 16 18 20

Figure 9.7–6: c-chart (b) The process is in statistical control.

5101 9.7–8 (a) x =

= 34.0067, s = 2.8244, R = 7; 150 (b) UCL = x + 1.43(s) = 38.046 for the x chart; LCL = x − 1.43(s) = 29.968 for the x chart; (c) UCL = 2.09(s) = 5.903 for the s chart; LCL = 0(s) for the s chart; UCL

38

UCL

6 5

36 4

_ x

_ s

3 2

32

1 LCL

30 0

5

10

15

20

25

30

0

5

10

Figure 9.7–8: (b) x-chart using s and (c) s-chart

15

20

25

LCL 30

140

Section 9.7 Statistical Quality Control (d) UCL = x + 0.58(R) = 38.0667; LCL = x − 0.58(R) = 29.9467; (e) UCL = 2.11(R) = 2.11(7) = 14.77; LCL = 0(R) = 0; UCL

38

UCL 14 12

36

10

_ x

34

_ r

8 6

32

4 LCL

30 0

5

10

15

20

25

30

2 5

10

Figure 9.7–8: (d) x-chart using R and (e) R-chart (f) Yes. √ 9.7–10 LCL = 1.4 − 3 1.4 < 0 so LCL = 0; √ UCL = 1.4 + 3 1.4 = 4.9496; (a) If λ = 3, P (X ≥ 5) = 1 − P (X ≤ 4) = 1 − 0.815 = 0.185; (b) Y is b(10, 0.185). Thus P (at least 1) = 1 − P (Y = 0) = 1 − p0q10 = 1 − 0.81510 = 0.871.

15

20

25

LCL 30

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