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INSTRUCTOR’S SOLUTIONS MANUAL PROBABILITY AND STATISTICAL INFERENCE
INSTRUCTOR’S
SOLUTIONS MANUAL
PROBABILITY AND
STATISTICAL INFERENCE
NINTH EDITION
*/2%$/(‘,7,
ROBERT V. HOGG
University of Iowa
Elliot A. Tanis
Hope College
Dale L. Zimmerman
University of Iowa
Boston Columbus Indianapolis New York San Francisco Upper Saddle River
Amsterdam Cape Town Dubai London Madrid Milan Munich Paris Montreal Toronto
Delhi Mexico City São Paulo Sydney Hong Kong Seoul Singapore Taipei Tokyo
Copyright © 2016 Pearson Education Limited
All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted in any form or by any means, electronic, mechanical, photocopying, recording or otherwise without the prior written permission of the publisher.
Many of the designations used by manufacturers and sellers to distinguish their products are claimed as trademarks. Where those designations appear in this book, and the publisher was aware of a trademark claim, the designations have been printed in initial caps or all caps.
ISBN-13: ISBN-10:
iv Contents
6 Point Estimation 6 Descriptive Statistics 6 Exploratory Data Analysis 6 Order Statistics 6 Maximum Likelihood Estimation 6 A Simple Regression Problem Likelihood Estimators 6 Asymptotic Distributions of Maximum 6 Sufficient Statistics 6 Bayesian Estimation 6 More Bayesian Concepts
7 Interval Estimation 7 Confidence Intervals for Means 7 Confidence Intervals for the Difference of Two Means 7 Confidence Intervals For Proportions 7 Sample Size 7 Distribution-Free Confidence Intervals for Percentiles 7 More Regression 7 Resampling Methods
8 Tests of Statistical Hypotheses 8 Tests About One Mean 8 Tests of the Equality of Two Means 8 Tests about Proportions 8 The Wilcoxon Tests 8 Power of a Statistical Test 8 Best Critical Regions 8 Likelihood Ratio Tests
9 More Tests 9 Chi-Square Goodness-of-Fit Tests 9 Contingency Tables 9 One-Factor Analysis of Variance 9 Two-Way Analysis of Variance 9 General Factorial and 2kFactorial Designs 9 Tests Concerning Regression and Correlation 9 Statistical Quality Control
Preface v
Preface
4HISÅ SOLUTIONSÅ MANUALÅ PROVIDESÅ ANSWERSÅ FORÅ THEÅ EVEN
NUMBEREDÅ EXERCISESÅ INÅ 0ROBABILITYÅ ANDÅ
3TATISTICALÅ)NFERENCE ÅTHÅEDITIONÅ’LOBALÅEDITIONÅBYÅ2OBERTÅ6Å(OGGÅ%LLIOTÅ!Å4ANISÅANDÅ$ALEÅ,Å
:IMMERMANÅ #OMPLETEÅ SOLUTIONSÅ AREÅ GIVENÅ FORÅ MOSTÅ OFÅ THESEÅ EXERCISESÅ 9OUÅ THEÅ INSTRUCTORÅ MAYÅ
DECIDEÅHOWÅMANYÅOFÅTHESEÅSOLUTIONSÅANDÅANSWERSÅYOUÅWANTÅTOÅMAKEÅAVAILABLEÅTOÅYOURÅSTUDENTSÅ.OTEÅ
THATÅTHEÅANSWERSÅFORÅTHEÅODD
NUMBEREDÅEXERCISESÅAREÅGIVENÅINÅTHEÅTEXTBOOK
All of the figures in this manual were generated usingMaple, 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 math.hope/tanis/. Short descriptions of these procedures are pro- vided on the “Maple Card.” Complete descriptions of these procedures are given inProbability and Statistics: Explorations with MAPLE, second edition, 1999, written by Zaven Karian and Elliot Ta- nis, published by Prentice Hall (ISBN 0-13-021536-8). You can download a copy of this manual at math.hope/tanis/MapleManual.pdf. Our hope is that this solutions manual will be helpful to eachof you in your teaching. If you find an error or wish to make a suggestion, send these to Elliot Tanis,tanis@hope, and he will post corrections on his web page, math.hope/tanis/.
R.V. E.A. D.L.
Chapter 1 Probability 1
Chapter 1
Probability
1 Properties of Probability
1-
1-4 (a)
(b) (i) /, (ii) 0, (iii) /, (iv) /, (v) /, (vi) /, (vii) /. 1-6 (a) P (A ∪ B) = 0. 4 + 0. 5 − 0. 3 = 0 .6; (b) A = ( A∩B∀)∪(A∩B) P(A) = P(A∩B∀) + P(A∩B) 0. 4 = P(A∩B∀) + 0. 3 P(A∩B∀) = 0 .1;
(c)P(A∀∪B∀) = P[(A∩B)∀] = 1 −P(A∩B) = 1 − 0 .3 = 0 .7.
1-8 LetA={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 + 0. 53 −P(A∩B) P(A∩B) = 0 .41 + 0. 53 − 0 .79 = 0. 15.
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).
1-12 (a) /; (b) /; (c) 0; (d) /.
0ROBABILITYÅOFÅINSURINGÅEXACTLYÅÅCARÅ 0 ! ÅÅÅ
0ROBABILITYÅOFÅINSURINGÅMOREÅTHANÅÅCARÅ 0 ” ÅÅÅ
0ROBABILITYÅOFÅINSURINGÅAÅSPORTSÅCARÅ 0 # ÅÅÅ
# 0 _!_ c
0″ 0! # 0″ # 0 0 _!_ #
#
2 Section 1 Methods of Enumeration
1-14P(A) = 2[r−r(
√
3 /2)]
2 r
= 1 −
√
3
2
.
1-16Note that the respective probabilities arep 0 , p 1 =p 0 / 4 , p 2 =p 0 / 42 ,···. ∀∞
k=
p 0 4 k
= 1
p 0 1 − 1 / 4
= 1
p 0 = 3 4 1 −p 0 −p 1 = 1 −
15
16 =
1
16.
1 Methods of Enumeration
1-2 (a) .UMBERÅ OFÅ EXPERIEMENTSÅ Å ÅÑÅÅÑÅ ÅÅ
1-4 (a) 4
6
3
= 80;
(b)4(2 6 ) = 256;
(c) (4−1 + 3)! (4−1)!3!
= 20.
1-6S={DDD, DDFD, DFDD, FDDD, DDFFD, DFDFD, FDDFD, DFFDD,
FDFDD, FFDDD, FFF, FFDF, FDFF, DFFF FFDDF, FDFDF,
1-
1-
− 1
r
+
n− 1 r− 1
=
(n−1)! r!(n− 1 −r)!
+
(n−1)! (r−1)!(n−r)!
=
(n−r)(n−1)! +r(n−1)! r!(n−r)! =
n! r!(n−r)!=
n r
.
1-12 0 = (1 −1)n=
∀n
r=
n r
(−1)r(1)n−r=
∀n
r=
(−1)r
n r
.
2 n = (1 + 1) n=
∀n
r=
n r
(1)r(1)n−r=
∀n
r=
n r
.
1-
10 −1 + 36
36
=
45!
36!9!
= 886, 163 ,135.
1-16 (a)
19
3
52 − 19
6
52
9
= 102 , 486
351 , 325 = 0 .2917;
(b)
19
3
10
2
7
1
3
0
5
1
2
0
6
2
52
= 7 , 695
1 , 236 , 664 = 0. 00622.
9
(b)
n
DFFDF, FDDFF, DFDFF, DDFFF } so there are 20 possibilities.
Total number of varieties of pizzas = 4 u 3 u (2) 16 = 786432.
Number o f experiments with each factor at 4 levels = ( 4 ) u (4) u ( 4 ) = 64.
4 Section 1 Independent Events
1-
1 Independent Events
1-2 (a) P(A∩B) = P(A)P(B) = (0 .3)(0) = 0; 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)
P(B) =
0
0. 6 = 0.
1-4Proof of(b): P(A∀∩B) = P(B)P(A∀|B) = P(B)[1−P(A|B)] = P(B)[1−P(A)] = P(B)P(A∀).
Proof of(c): P(A∀∩B∀) = 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∀).
1-6 P[A∩(B∩C)] = P[A∩B∩C] = P(A)P(B)P(C) = P(A)P(B∩C). P[A∩(B∪C)] = P[(A∩B)∪(A∩C)] = P(A∩B) + P(A∩C)−P(A∩B∩C) = P(A)P(B) + P(A)P(C)−P(A)P(B)P(C) = P(A)[P(B) + P(C)−P(B∩C)] = P(A)P(B∪C). P[A∀∩(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∀) 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)−P A )P(B)P(C) = [1 −P(A)][1−P(B)][1−P(C)] = P(A∀)P(B∀)P(C∀).
Probability that an employee has a college degree and works in sales = 0. 6 u 0 = 0. 0 6. Probability that an employee does not have a college degree and works in sales = 0 u (1 ñ 0) = 0. Probability that an employee chosen at random works in sales = 0 + 0 = 0.
(d)P(A 1 |B 2 ) = 11 /41; (e)P(B 1 |A 3 ) = 13 /29.
Chapter 1 Probability 5
1-10 (a) 3 4
· 3
4
= 9
16
;
(b) 1 4
· 3
4
+ 3
4
· 2
4
= 9
16
;
(c)
2
4 ·
1
4 +
2
4 ·
4
4 =
10
16.
1-12 (a)
(b)
(c)
(d)
1-14 (a) 1 −(0) 3 = 1 − 0 .064 = 0; (b) 1 −(0) 8 = 1 − 0 .00065536 = 0.
1-16 (a)
∀∞
k=
1
5
4
5
2 k = 5 9
;
(b) 1 5
+ 4
5
· 3
4
· 1
3
+ 4
5
· 3
4
· 2
3
· 1
2
· 1
1
= 3
5
.
1-18 (a)7;(b)(1/2) 7 ;(c)63;(d)No! (1/2) 63 = 1 / 9 , 223 , 372 , 036 , 854 , 775 ,808.
1-20No.
1 Bayes’ Theorem
1-2 (a)
(b)
1-4 Let eventBdenote an accident and letA 1 be the event that age of the driver is 16–25. Then
P(A 1 |B) =
(0)(0)
(0)(0) + (0)(0) + (0)(0) + (0)(0)
=
50
50 + 110 + 60 + 60
=
50
280
= 0. 179.
1-6 LetB be the event that the policyholder dies. LetA 1 , A 2 , A 3 be the events that the deceased is standard, preferred and ultra-preferred, respectively. Then
(! ) 0′ ( |! ) 0 ( ) ” 0′ ( | ” ) 0 0 0 0 0.
0’ ( ) 0 uu
) ) 0 0. ) ) 0.
0 (! | ‘ 0 ( !’ ) 0 ( ‘ |! ) 0 (! 0′ ( 0’ (
1 28 1 28
1 28
8
8! 1 4! 4! 2
1-8 Probability that exactly 2 of the 3 dice came up orange 2 3 2 4 3 4 2 6 6 6 6 6 6 6 6 6 18 18 18 18
3 4 1 4 2 7
Chapter 2 Discrete Distributions 7
Chapter 2
Discrete Distributions
2 Random Variables of the Discrete Type
2-2 (a)
f(x) =
§
̈
©
, x= 1 , , x = , 1 , x = , (b)
Figure 2–2: Line graph.
2-4 (a)f(x) =
1
10 , x= 0 , 1 , 2 ,···,9;
(b) N({ 0 })/150 = 11/150 = 0; N({ 5 })/150 = 13/150 = 0; N({ 1 })/150 = 14/150 = 0; N({ 6 })/150 = 22/150 = 0; N({ 2 })/150 = 13/150 = 0; N({ 7 })/150 = 16/150 = 0; N({ 3 })/150 = 12/150 = 0; N({ 8 })/ 150 = 18/150 = 0; N({ 4 })/150 = 16/150 = 0; N({ 9 })/150 = 15/150 = 0.
8 Section 2 Random Variables of the Discrete Type
(c)
x
f(x), h(x)
1 2 3 4 5 6 7 8 9
Figure 2–4: Michigan daily lottery digits
2-6 (a)f(x) = 6 −| 7 −x| 36
, x = 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10 , 11 ,12. (b)
x
f(x)
1 2 3 4 5 6 7 8 9 10 11 12
Figure 2–6: Probability histogram for the sum of a pair of dice
10 Section 2 Mathematical Expectation
2-12 P(X≥ 4 |X≥1) =
P(X≥4)
P(X≥1) =
1 −P(X≤3)
1 −P(X= 0)
=
1 −[1− 1 /2 + 1 / 2 − 1 /3 + 1 / 3 − 1 /4 + 1 / 4 − 1 /5]
1 −[1− 1 /2]
=
2
5
.
2-14P(X≥1) = 1−P(X= 0) = 1 −
3
0
17
5
20
5
= 1 − 91
228
=
137
228
= 0. 60.
2-16 (a)P(2, 1 , 6 ,10) means that 2 is in position 1 so 1 cannot be selected. Thus
P(2, 1 , 6 ,10) =
1
0
1
1
8
5
10
6
= 56
210
= 4
15
;
(b)P(i, r, k, n) =
i− 1 r− 1
1
1
n−i k−r
n k
.
2 Mathematical Expectation
2-
2-4 1 =
6
x=
f(x) = 9 10
+c
1
1
+ 1
2
+ 1
3
+ 1
4
+ 1
5
+ 1
6
c = 2 49
;
E(Payment) = 2 49
1 · 1
2
+ 2 · 1
3
+ 3 · 1
4
+ 4 · 1
5
+ 5 · 1
6
= 71
490
units.
2-6Note that
∞
x=
6
π 2 x 2
= 6
π 2
∞
x=
1
x 2
= 6
π 2
π 2 6
= 1, so this is a pdf
E(X) =
∞
x=
x 6 π 2 x 2
= 6
π 2
∞
x=
1
x
= + ∞
and it is well known that the sum of this harmonic series is notfinite.
2-8E(|X−c|) =
1
7
x∈S
|x−c|, where S={ 1 , 2 , 3 , 5 , 15 , 25 , 50 }.
Whenc= 5,
E(|X− 5 |) =
1
7
[(5−1) + (5−2) + (5−3) + (5−5) + (15−5) + (25−5) + (50−5)].2 2 2 2 2 2
2 2
E X ( )= 0 )( 2 941490
27 27
E X ( )= 0 )( 2 90 )( 1 4 (0) 1 )(1 4 (2) 980
27 27 27 27 27 27
E X ( 3 9) E X ( ) 3 E X ( ) 9+ 80 0 9 323
27 27
+0( 1 )+( 0 )+( 1 )+(2)=
27 27 27
+ + + + =
0 X + = 0 = 0+=
Chapter 2 Discrete Distributions 11
Ifcis either increased or decreased by 1, this expectation is increased by 1/7. Thus c= 5, the median, minimizes this expectation whileb=E(X) = μ, the mean, minimizes E[(X−b) 2 ]. You could also leth(c) = E(|X−c|) and show that h(c) = 0 when c= 5.
2-
2-12 (a)The average class size is
(16)(25) + (3)(100) + (1)(300)
20
= 50;
(b)
f(x) =
§
̈
©
4 , x= 25 , 3 , x= 100, 3 , x= 300,
(c)E(X) = 25(0) + 100(0) + 300(0) = 130.
2 Special Mathematical Expectations
2-2 (a) μ = E(X)
=
3
x=
x 3! x! (3 −x)!
1
4
x 3 4
3 −x
= 3
1
4
2
k=
2!
k! (2 −k)!
1
4
k 3 4
2 −k
= 3
1
4
1
4 +
3
4
2
=
3
4 ;
E[X(X−1)] =
3
x=
x(x−1) 3! x! (3 −x)!
1
4
x 3 4
3 −x
= 2(3)
1
4
2
3
4
+ 6
1
4
3
= 6
1
4
2
= 2
1
4
3
4
;
σ 2 = E[X(X−1)] +E(X)−μ 2
= (2)
3
4
1
4
+
3
4
−
3
4
2
= (2)
3
4
1
4
+
3
4
1
4
= 3
1
4
3
4
;
21 5
(15)
36 36 2
15 21 5
(15)
36 36 2
(20) 63055
36 36 6
15
+
=
0
( 15)
+0( 15)= 0
+0( 15)= 0
Chapter 2 Discrete Distributions 13
2-10We haveN=N 1 +N 2. Thus
E[X(X−1)] =
n
x=
x(x−1)f(x)
=
n
x=
x(x−1)
N 1!
x! ( N 1 −x)!·
N 2!
(n−x)! (N 2 −n+x)! N n
= N 1 (N 1 −1)
n
x=
(N 1 −2)!
(x−2)! (N 1 −x)!
· N 2!
(n−x)! (N 2 −n+x)! N n
.
In the summation, letk=x−2, and in the denominator, note that N n
= N!
n!(N−n)!
=N(N−1)
n(n−1)
N− 2
n− 2
.
Thus
E[X(X−1)] =
N 1 (N 1 −1)
N(N−1)
n(n−1)
n− 2
k=
N 1 − 2
k
N 2
n− 2 −k
N− 2
n− 2
= N 1 (N 1 −1)(n)(n−1) N(N−1)
.
2-
2-14P(X≥100) =P(X > 99) = (0) 99 = 0. 3697.
2-16 (a)f(x) = (1 /2)x− 1 , x = 2 , 3 , 4 ,… ;
XÅ F ( X )
0 0.
1 0.
2 0.
3 0.
4 0.
5 0.
6 0.
Mean = 1. Variance = 2. Standard deviation = 1. 0 ( 8Å > 3) = 0 ( 8Å = 4) + 0 ( 8Å = 5) + 0 ( 8Å = 6) = 0. 0 ( 8Å < 3) = 1 ñ 0 ( 8Å > 3) ñ 0 ( 8Å = 3) = 1 ñ 0 ñ 0 = 0.
14 Section 2 The Binomial Distribution
(b) M(t) = E[etx] =
∞
x=
etx(1/2)x− 1
= 2
∞
x=
(et/2)x
=
2(et/2) 2 1 −et/ 2 =
e 2 t 2 −et, t
k +j|X > k ) = P(X > k +j) P(X > k )
= q
k+j qk
=qj=P(X > j ).
2 The Binomial Distribution
2-
2-4 (a) X is b(Å ); (b) Errata: The value of 0 renders this part unsolvable. This will be revised and corrected.
2-6 (a) X FOLLOWSÅ BÅ); Y ÅÅÅ− X FOLLOWSÅBÅ);
(e) % X ÅÅ ÅÅÅÅÅÅ6ARÅ X ÅÅ ÅÅÅÅÅÅ3TATNDARDÅDEVIATIONÅÅ
2-8 (a) 0 8 ÅȴÅ ÅÅÅ Å 0 8 ÅȳÅ ÅÅ; (b) 0 8 ÅÅ ÅÅ ÅÅÅ 2-10 (a) X is b(8, 0 .90); (b) (i) 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; (iii) P(X≥6) = P(8−X≤2) = 0. 9619.
Errata: Å0ARTSÅB ÅC ÅANDÅD ÅCANNOTÅBEÅSOLVEDÅASÅTHEÅPROBABILITYÅVALUESÅ OFÅÅANDÅÅDOÅNOTÅEXISTÅINÅTHEÅGIVENÅTABLESÅINÅthe !PPENDIX
2 2 2
( 0 5) 117 ;
18 18
( 0 5) 11 +(5) 7 1.
8 18
( 05 +1) 11 +(5+1) 23.
8 18
f , f (5)
μ
σ
= =
= =
= 7
=
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