Download Statistical Theory (STAT 467/667) Homework 8 Solution

Statistical Theory (STAT 467/667)
Homework 8 Solution - Spring 2015
Department of Mathematics and Statistics, University of Nevada, Reno
Problem 6.4.8
Will n = 45 be a sufficiently large sample size to test H0 : µ = 10 versus H1 : µ 6= 10 at the α = 0.05 level of
significance if the experimenter wants the type II error probability to be no greater than 0.20 when µ = 12? Assume
that σ = 4.
Let β be the probability of type II error when µ = 12 assuming that σ = 4 at the α = 0.05 level of significance, then
β
=
=
=
=
P (accept H0 | µ = 12)
4
σ
σ
4
P µ0 − zα/2 √ ≤ x̄ ≤ µ0 + zα/2 √ | µ = 12 = P 10 − 1.96 √ ≤ x̄ ≤ 10 + 1.96 √ | µ = 12
n
n
45
45
8.831 − 12
11.169 − 12
√
√
P (8.831 ≤ x̄ ≤ 11.169 | µ = 12) = P
≤Z≤
4/ 45
4/ 45
P (−5.315 ≤ Z ≤ −1.394) ≈ 0.0823 < 0.20.
So, n = 45 is a sufficiently large sample so that type II error probability is no greater than 0.20 when µ = 12 at the
α = 0.05 level of significance assuming σ = 4.
Problem 6.4.11
9
= 0.064. Similarly,
In this context, α is the proportion of incorrect decisions made on innocent suspects, that is
140
15
β is the proportion of incorrect decisions made on guilty suspects, which is
= 0.107 in this case. A Type I
140
error (convicting an innocent defendant) would be considered more serious than a Type II error (acquitting a guilty
defendant).
Problem 6.4.12.
An urn contains 10 chips. An unknown number of the chips are white; the others are red. We wish to test
H0 : {exactly half the chips are white } versus H1 : {more than half of the chips are white }. We will draw, without
replacement, three chips and reject H0 if two or more are white. Find α. Also, find β when the urn is (a) 60% white
and (b) 70% white.
1
2
Let’s denote X to be the number of white chips in the sample and E be the event that exactly half of the chips are
white. Then (noting that X has a hypergeometric distribution)
α
=
=
P (reject H0 | H0 is true) = P (X ≥ 2 | E)
     
5
5
5
5
     
2
1
3
0
P (X = 2 | E) + P (X = 3 | E) =   +   = 0.5.
10
10
 
 
3
3
(a) Let E1 be the event that the urn contains 6 white and 4 red chips.
β
=
=
P (accept H0 | E1 )
     
6
4
6
4
     
0
3
1
2
1
P (X ≤ 1 | E1 ) =   +   = .
3
10
10
 
 
3
3
(b) Let E2 be the event that the urn contains 7 white and 3 red chips.
β
= P (accept H0 | E2 )
     
7
3
7
3
     
0
3
1
2
11
= P (X ≤ 1 | E2 ) =   +   =
≈ 0.183.
60
10
10
 
 
3
3
Problem 6.4.18.
An experimenter takes a sample of size 1 from a Poisson probability model with pdf pX (k) = e−λ λk /k!, k = 0, 1, 2...,
and wishes to test
H0 : λ = 6
vs.
H1 : λ < 6
by rejecting H0 if k ≤ 2
(a) α = P (reject H0 | H0 is true) = P (X ≤ 2 | λ = 6) =
P2
(b) β = P (accept H0 | λ = 4) = 1 − P (X ≤ 2 | λ = 4) = 1 −
k=0
P2
e−6 6k /k! = 0.0619688
k=0
e−4 4k /k! = 0.7618967
3
Problem 6.4.21.
Problem 6.5.2.
Let Y1 , Y2 , ..., Y10 be a random sample from an exponential pdf with unknown parameter λ. I assume that the exponential distribution we deal with is parametrized so that its mean is 1/λ. Find the form of the GLRT for H0 : λ = λ0
versus H1 : λ 6= λ0 . What integral would have to be evaluated to determine the critical value if α were equal to 0.05?
Start by writing down the parameter spaces: the full parameter space Ω = Ω0 ∪ Ω1 = {λ : λ > 0}, Ω0 = {λ0 }, and
Ω1 = {λ : λ 6= λ0 , λ > 0}. The likelihood function is
L(λ) =
10
Y
λe−λyi = λ10 e−λ
P10
i=1
yi
.
i=1
We find that the MLE (maximum likelihood estimate) for λ is λ̂ =
10
P10
i=1
yi
. Then, the generalized likelihood ratio Λ
is
P10
maxΩ0 L(λ)
L(λ0 )
λ10 e−λ0 i=1 yi
=0
Λ=
=
=
10
maxΩ L(λ)
L(λ̂)
P 10
e−10
10
i=1
eλ0
10
10
yi
e
−λ0
P10
i=1 yi
10
X
!10
yi
.
i=1
Now, by Definition 6.5.2 the generalized likelihood ratio test (GLRT) rejects H0 whenever 0 < Λ ≤ λ∗ where λ∗ is
chosen so that
P (0 < Λ ≤ λ∗ | H0 is true) = α,
for significance level α (0.05 in this case). To compute λ∗ , we note that fΛ (t | H0 ) is the PDF of Λ when H0 is true,
and then we evaluate the following integral
Z
λ∗
fΛ (t | H0 ) dt = 0.05
0