THE UNIVERSITY OF HONG KONG
DEPARTMENT OF STATISTICS AND ACTUARIAL SCIENCE
STAT 1302 PROBABILITY AND STATISTICS II
(201011)
Assignment 4 (Sketch Solution)
Section A
Test statistics, pvalues & critical regions
A1.
(a)
(i) The likelihood is
‘
(
θ
) =
θ

2
e

S/θ
.
Likelihood ratio is LR=
‘
(2)
/‘
(1) = 2

2
e
S/
2
.
Thus
{
LR
> k
0
}
is equivalent to
{
S > k
}
=
C
k
.
(ii) Type I error probability of the test is
P
(
S > k

θ
= 1) =
R
∞
k
s e

s
ds
= (1 +
k
)
e

k
.
(iii) Type II error probability of the test is
P
(
S
≤
k

θ
= 2) =
R
k
0
s e

s/
2
/
4
ds
= 1

e

k/
2
(
k
+
2)
/
2.
(iv) Substitute
k
= 4
.
744 into the type I error probability to get approximately 0
.
05.
(v) From the data we have
S
= 5
.
3
>
4
.
744. We reject
H
0
at 5% level.
Pvalue =
P
(
S >
5
.
3

θ
= 1) =
R
∞
5
.
3
s e

s
ds
= 0
.
031447. Pvalue
<
0
.
05
⇒
above conclusion.
(b)
(i) The power function of the test is
w
(
θ
) =
P
(
S > k

θ
) =
R
∞
k
f
(
s
)
ds
=
e

k/θ
(
k
+
θ
)
/θ
. To
check it is increasing, we may examine the derivative of its logarithm.
(ii) The size of the test is sup
θ
≤
1
w
(
θ
) =
w
(1) =
e

k
(
k
+ 1).
(iii)
pv
(
s
) = sup
θ
≤
1
P
(
S > s

θ
) =
e

s
(
s
+ 1) (from (i) and (ii)). Note that
pv
0
(
s
) =

s e

s
<
0,
so that
pv
(
s
) is decreasing in
s
.
A2.
(a) Type I error probability =
P
(
U >
1

θ
= 0) = 0.
(b) Type II error probability =
P
(
U
≤
1

θ
= 1) = 1
/
2.
(c) The size of the test is type I error probability = 0.
(d) The power function of the test is
w
(
θ
) =
P
(
U >
1

θ
) =
0
,
θ <
0
,
θ/
2
,
0
≤
θ
≤
2
,
1
,
θ >
2
.
(e) The likelihood is
‘
(
θ
) = 2

1
1
{
θ

1
≤
U
≤
θ
+ 1
}
. The likelihood ratio is
1
{
0
≤
U
≤
2
}
/
1
{
1
≤
U
≤
1
}
=
0
,

1
< U <
0
,
1
,
0
≤
U
≤
1
,
∞
,
1
< U <
2
.
A3.
(a) The power function is
w
1
(
θ
) =
P
(Binomial(10
, θ
)
>
8) = 10
θ
9
(1

θ
) +
θ
10
.
1
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(b) The power function
w
1
is increasing on (0
,
1). Thus the size of the test is
w
(1
/
2) = 11
/
2
10
.
(c) The power function
w
1
is increasing on (0
,
1). Thus the size is
w
(2
/
3).
(d)
(i) The power function is
w
2
(
θ
) =
θ
10
+ (1

θ
)
10
. The size is sup
{
w
2
(
θ
) : 1
/
3
≤
θ
≤
2
/
3
}
=
w
2
(1
/
3).
(ii)
T
=

¯
X

0
.
5

,
k
= 0
.
4.
(iii)
T
is observed to be
t
= 0, so that
pv
(0)
=
sup
H
0
'
1

P
(
¯
X
= 0
.
5

θ
)
“
=
1

inf
H
0
P
(Binomial(10
, θ
) = 5)
=
1

inf
H
0
10
5
¶
θ
5
(1

θ
)
5
=
1

10
5
¶
1
3
¶
5
2
3
¶
5
=
5665
6561
≈
0
.
863435
.
Section B
Hypotheses & likelihood ratios
B1. Let
X
be the no. of marked fish among a sample of 2,000 caught from the lake and
N
be the total
no. of fish in the lake.
(a) (i)
H
0
:
N
≤
21000 vs
H
1
:
N >
21000.
(ii) Note that the likelihood is
‘
(
N
) =
(
1000
100
)(
N

1000
1900
)
/
(
N
2000
)
. Consider
‘
(
N
+ 1)
/‘
(
N
) = 1

100(
N

19999)(
N
+ 1)

1
(
N

2899)

1
>
1
,
2900
≤
N
≤
19998
,
= 1
,
N
= 19999
,
<
1
,
N
≥
20000
.
Thus
‘
(
N
) increases as
N
increases up to 19,999, stays constant at
N
= 19999 and 20000, and
then decreases as
N
increases beyond 20,000.
Under
H
1
,
‘
(
N
) is maximized at
N
=
ˆ
N
1
= 21000. Under
H
0
,
‘
(
N
) is maximized at
N
=
ˆ
N
0
=
20000 (or 19999). The likelihood ratio is
‘
(
ˆ
N
1
)
/‘
(
ˆ
N
0
) = 0
.
872511.
(b) (i)
H
0
:
N >
21000 vs
H
1
:
N
≤
21000.
(ii) Under
H
0
,
‘
(
N
) is maximized at
N
=
ˆ
N
0
= 21000.
Under
H
1
,
‘
(
N
) is maximized at
N
=
ˆ
N
1
= 20000 (or 19999). The likelihood ratio is
‘
(
ˆ
N
1
)
/‘
(
ˆ
N
0
) = 1
.
146118.
B2. Let
X
s
, X
m
, X
n
be nos. substantially, mildly and not improved by the ordinary treatment respectively.
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 Spring '11
 Dr.Yun
 H0, Pearson's chisquare test, Poisson, Likelihoodratio test, likelihood ratio

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