THE UNIVERSITY OF HONG KONG

DEPARTMENT OF STATISTICS AND ACTUARIAL SCIENCE

STAT1302/2602 Probability and Statistics II (1st semester, 2016-2017)

Suggested solutions for Assignment 3

Q1. Let X1 , X2 , . . . , Xn be an independent random sample from N (µ, σ 2 ) with known mean µ. Describe
how you would construct a confidence interval for the unknown variancePσ 2 . (5 marks)
n 2 2
(X i − X̄) nS
Solution: Method 1: To construct the confidence interval for σ 2 , we use i=1 2 = 2 ∼ χ2n−1
σ σ
which is a pivotal quantity involving σ 2 . Let α be the significance level, then

nS 2
 
2 2
1 − α = P χ1−α/2,n−1 6 2 6 χα/2,n−1
σ
!
nS 2 2 nS 2
= P 6σ 6 2 ,
χ2α/2,n−1 χ1−α/2,n−1

where χ2α,n satisfies P (T > χ2α,n ) = α for T ∼ χ2n . So, a 1 − α confidence interval of σ 2 is given by
!
nS 2 nS 2
, .
χ2α/2,n−1 χ21−α/2,n−1

Method 2: Since Pn
i=1 (Xi − µ)2
∼ χ2 (n),
σ2
a 1 − α confidence interval of σ 2 is given by
Pn Pn !
2 2
i=1 (Xi − µ) i=1 (Xi − µ)
, .
χ2α/2,n χ21−α/2,n

Q2. Let X̄ and S 2 be the mean and the variance of a random sample of size n = 17 from a distribution
N (µ, σ 2 ). Find the constant c such that
 
4(X̄ − µ)
P −c < < c = 0.95.
S

(5 marks)
Solution: We have
X̄ − µ 4(X̄ − µ)
T = √ = ∼ tn−1 ,
S/ n − 1 S

1
and  
4(X̄ − µ)
P −t0.025,17−1 < < t0.025,17−1 = 0.95.
S
Thus
c = t0.025,17−1 = 2.1199.

Q3. Let X1 , X2 , · · · , Xn be an independent random sample from N (µ, σ 2 ), where µ and σ 2 ae unknown.
Suppose that √ √ 
nS nS
1−α=P √ 6σ6 √ .
b a
√ √ 
nS nS
Hence, √ , √ is the 1 − α confidence interval of σ, and it has the length
b a
√ √
nS nS
k= √ − √ .
a b
Show that the values of a and b for σ of minimum length satisfy the following equations:

(i) G(b) − G(a) = 1 − α;

(ii) an/2 e−a/2 − bn/2 e−b/2 = 0,

where G(·) is the c.d.f. of χ2n−1 . (10 marks)

Solution: (i) First we have
nS 2
∼ χ2n−1 ,
σ2
then
√ √ 
nS nS
1−α = P √ 6σ6 √
b a
2
 
nS
= P a6 2 6b
σ
= G(b) − G(a).
√ √
nS nS √ 1 1
(ii) The length of the interval is k = √ − √ = nS( √ − √ ), denote
a b a b
1 1
f (a, b) = √ − √ .
a b
We need to minimize f (a, b) with the constrain G(b) − G(a) = (1 − α). Set
1 1
L(a, b, λ) = √ − √ − λ [G(b) − G(a) − (1 − α)] .
a b

2
Then we have
1 3
L0a (a, b, λ) = − a− 2 + λg(a),
2
1 3
L0b (a, b, λ) = b− 2 − λg(b),
2
where g(·) is the p.d.f. of χ2n−1 . For X ∼ χ2k , the p.d.f is given by
1 k x
g(x) = k k

x 2 −1 e− 2 .
2 Γ2
2

It follows that
1 2g(a) 2g(b)
= −3 = −3 .
λ a 2 b 2
Substitute g(a) and g(b) into the above equations, we have

an/2 e−a/2 − bn/2 e−b/2 = 0.

Q4. Let X be N (µ, 100).

(i) To test H0 : µ = 230 versus H1 : µ > 230, what is the rejection region specified by the generalized
likelihood ratio test? (5 marks)
(ii) Is this test uniformly most powerful? (5 marks)
(iii) If a random sample of n = 16 yielded x̄ = 232.6, is H0 accepted at a significance level of
α = 0.10? (5 marks)
Solution: (i) First we have

Ω = {µ: µ > 230} ,

Ω0 = {µ: µ = 230} ,

Ω1 = {µ: µ > 230} .

The likelihood function of the sample is

 n " n
#
1 1 X
L(µ) = √ exp − (Xi − µ)2 .
10 2π 200 i=1

On Ω0 , the likelihood function is

 n " n
#
1 1 X
L(Ω0 ) = √ exp − (Xi − 230)2 .
10 2π 200 i=1

3
On Ω, the maximum value of L(µ) is L(µ̂), where

X̄, if X̄ > 230,
µ̃ =
230, if X̄ < 230.

Therefore,   n
1
 1 Pn 2

 √ exp − 200 i=1 (X i − X̄) , if X̄ > 230,
L(Ω) =  10 2π n
1
 1
P n 2

 √
10 2π
exp − 200 i=1 (Xi − 230) , if X̄ < 230.

Then   1 Pn Pn

L(Ω0 ) 2 2
exp − 200 i=1 (X i − 230) − i=1 (X i − X̄) , if X̄ > 230,
Λ= =
L(Ω) 1, if X̄ < 230.
The rejection region is {Λ 6 k} for some constant k ∈ (0, 1). Then {Λ 6 k} ⊆ X̄ > 230 is equivalent to

n(X̄ − 230)2
 
exp − 6 k,
200

or
X̄ − 230 > k 0 .

Let √ √ 0
0

n(X̄ − 230) nk
α = P X̄ − 230 > k =P > ,
10 10
√
nk0
so we should set 10
= zα . Hence, the generalized likelihood ratio test has the rejection region
√ 
n(X̄ − 230)
> zα
10

at the significance level α.

(ii) The most powerful test for H0 : µ = 230 against H1 : µ = µ1 > 230 having significance level
α∗ 6 α is the one having the rejection region
√ 
n(X̄ − 230)
> zα .
10

Since this rejection region does not depend on the value of µ1 , we know immediately that the uniformly
most powerful test with significance level for testing H0 : µ = 230 against H1 : µ > 230 has the same
n√ o
n(X̄−230)
rejection region 10
> zα . Thus, we can conclude that this test is most powerful test.
(iii) Critical region method: When α = 0.10, the rejection region is
(√ )
16(X̄ − 230) 
C= > z0.10 = 1.282 = X̄ > 233.203 .
10

4
The observed x̄ = 232.6 ∈
/ C, thus H0 cannot be rejected at α = 0.10.
p-value method:
√ √ !
16(X̄ − 230) 16(232.6 − 230)
p − value = P > = P (Z > 1.04) = 0.14917 > 0.10,
10 10

thus, H0 cannot be rejected at α = 0.10.

Q5. To test H0 : µ = 335 versus H1 : µ < 335, under normal assumptions, a random sample of size 17
yielded x̄ = 324.8 and s = 40. Is H0 accepted at an α = 0.10 significance level? (5 marks)
Solution: Critical region method: The rejection region is

X̄ − µ
C = {X̄ : √ < −tα,n−1 }.
S/ n − 1
x̄ − µ 324.8 − 335
The realized test statistic = √ = √ = −1.02 > −t0.10,16 = −1.337. Thus x̄ = 324.8 ∈
/
s/ n − 1 40/ 17 − 1
C, we cannot reject H0 at α = 0.10 significance level.

Q6. Let X1 , · · · , Xn be an independent random sample from N (µ, σ 2 ) with unknown mean µ and
σ 2 . Show the details to find the generalized likelihood ratio test for hypotheses H0 : µ > µ0 versus
H1 : µ < µ0 . (10 marks)
Solution: First, we have

Ω = {(µ, σ): − ∞ < µ < ∞, σ > 0} ,

Ω0 = {(µ, σ): µ > µ0 , σ > 0} ,

Ω1 = {(µ, σ): µ < µ0 , σ > 0} .

The likelihood function of the sample is

 n "n
#
1 1 X
L(µ, σ) = √ exp − 2 (Xi − µ)2 .
σ 2π 2σ i=1

On Ω, the maximum value of L(µ, σ) is L(µ̂, σ̂), where

n
1X2
µ̂ = X̄ and σ̂ = (Xi − X̄)2 .
n i=1

Therefore,  n
1  n
L(Ω) = L(µ̂, σ̂) = √ exp − .
σ̂ 2π 2

5
On Ω0 , the maximum value of L(µ, σ) is L(µ̃, σ̃), where

X̄, if X̄ > µ0 ,
µ̃ =
µ0 , if X̄ < µ0 ,

∂`(µ, σ) ∂`(µ, σ)
since > 0 when µ < X̄ and < 0 when µ > X̄. And
∂µ ∂µ
n
1X
σ̃ 2 = (Xi − µ̃)2 .
n i=1

Therefore,  n
1  n
L(Ω0 ) = L(µ̃, σ̃) = √ exp − .
σ̃ 2π 2
Then the generalized likelihood ratio statistic is

 2 n/2  1, if X̄ > µ0 ,
L(Ω0 ) σ̂  Pn 2
n/2
Λ= = = (X − X̄)
L(Ω) σ̃ 2  Pni=1 i 2
, if X̄ < µ0 .
i=1 (Xi − µ0 )

The rejection region is {Λ > k} for some constant k ∈ (0, 1). Then {λ > k} ⊆ X̄ < µ0 } and Λ 6 k is
equivalent to
Pn Pn
(Xi − X̄)2 i=1 (Xi − X̄)
2
1
i=1
Pn 2
= Pn = n(X̄−µ0 )2
6 k 2/n ,
i=1 (Xi − µ0 )
2 2
i=1 (Xi − X̄) + n(X̄ − µ0 ) 1+ Pn 2
i=1 (Xi −X̄)

that is
(X̄ − µ0 )2 n(X̄ − µ0 )2
= P n > k −2/n − 1,
S2 i=1 (X i − X̄) 2

or
X̄ − µ0
√ 6 c,
S/ n − 1
p
where c = − (n − 1)(k −2/n − 1). To determine the value of c, set the significance level is α, we have
 
X̄ − µ0
P √ 6 c = α,
S/ n − 1

thus we should let c = −tα,n−1 . Hence, the generalized likelihood ratio test has the rejection region
 
X̄ − µ0
√ 6 −tα,n−1
S/ n − 1

at the significance level α.