JOSEPH W. M CKEAN - EIGHTH EDITION Robert V. Hogg Joseph W. McKean ...

SOLUTIONS

MANUAL

JOSEPH W. M CKEAN

INTRODUCTION TO

MATHEMATICAL STATISTICS

EIGHTH EDITION

Robert V. Hogg Joseph W. McKean Allen T. Craig 1 / 4

Contents

• Probability and Distributions 1
• Multivariate Distributions 11
• Some Special Distributions 19
• Some Elementary Statistical Inferences 31
• Consistency and Limiting Distributions 53
• Maximum Likelihood Methods 59
• Sufficiency 75
• Optimal Tests of Hypotheses 89
• Inferences about Normal Models 97

10 Nonparametric and Robust Statistics 113 11 Bayesian Statistics 131 iii

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Chapter 1 Probability and Distributions

1.2.1 Part (c):C1∩C2={(x, y):1

1.2.3C

1∩C2={mary,mray}.

1.2.6 lim

k→∞Ck={x:0

1.2.7 Part (b): lim

k→∞Ck=φ, because no point is in all the setsCk,k=1,2,3,...

1.2.9 Becausef(x) = 0 when 1≤x<10,

Q(C 3)= φ 10

f(x)dx= φ 1

6x(1−x)dx=1.

1.2.11 Part (c): Draw the regionCcarefully, noting thatx<2/3 because 3x/2<1.

Thus Q(C)= φ 2/3

Δ φ 3x/2 x/2 dy π dx= φ 2/3

xdx=2/9.

1.2.14 Note that

25 =Q(C)=Q(C

1)+Q(C 2)−Q(C 1∩C2)=19+16−Q(C1∩C2).

Hence,Q(C

1∩C2) = 10.

1.2.15 By studying a Venn diagram with 3 intersecting sets, it should be true that

11≥8+6+5−3−2−1=13.

It is not, and the accuracy of the report should be questioned.

1.3.3 P(C)= 1 2 + 1 4 + 1 8

+···=

1/2

1−(1/2)

=1.1

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Probability and Distributions2 1.3.6 P(C)= φ ∞

−∞

e −|x| dx= φ

−∞

e x dx+ φ ∞

e −x dx=2ξ=1.We must multiply by 1/2.

1.3.8 P(C c 1 ∪C c 2

)=P[(C 1∩C2)

c

]=P(C)=1,

becauseC 1∩C2=φandφ c =C.

1.3.11 The probability that he does not win a prize is

ξ 990 5 σ / ξ 1000 5 σ .

1.3.13 Part (a): We must have 3 even or one even, 2 odd to have an even sum.

Hence, the answer is ψ 10 3 ρψ 10

ρ ψ 20 3 ρ+ ψ 10 1 ρψ 10 2 ρ ψ 20 3 ρ.

1.3.14 There are 5 mutual exclusive ways this can happen: two “ones”, two “twos”,

two “threes”, two “reds”, two “blues.” The sum of the corresponding proba- bilities is ψ 2 2 ρψ 6

ρ + ψ 2 2 ρψ 6

ρ + ψ 2 2 ρψ 6

ρ + ψ 5 2 ρψ 3

ρ + ψ 3 2 ρψ 5

ρ ψ 8 2 ρ.

1.3.15

(a) 1− ψ 48 5 ρψ 2

ρ ψ 50 5 ρ (b) 1− ψ 48 n ρψ 2

ρ ψ 50 n ρ≥ 1 2 ,Solve for n.

1.3.20 Choose an integern

0>max{a −1 ,(1−a) −1 }. Then{a}=∩ ∞ n=n

ψ a− 1 n ,a+ 1 n ρ .Hence by (1.3.10), P({a}) = lim n→∞ P θξ a− 1 n ,a+ 1 n σχ = 2 n =0.

1.3.21 Choosen

0such that 0

• Since{a}=∩

∞ n=n

An,wehave P({a})=P ψ ∩ ∞ n=n

An ρ = lim n→∞ P(A n) = lim n→∞ 2 n =0..

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