Korea University, Department of Statistics
STAT232 Mathematical Statistics – Assignment 7
Due date: 06/08/2022
1. You are told that a coin has probability of a Head either p = 34 or p = 14 . You must
test
H0 : p =
3
4
vs.
Ha : p =
1
4
You toss the coin 3 times and get Y Heads. You decide to reject H0 in favor of Ha if
Y = 0 or Y = 1. (Note that the count of Heads Y is the CSS for p.) What are the
two error probabilities ↵ and for this test? What is the power of this test at the
alternative Ha ?
Sol) Y ⇠ Bin(3, p)
⇣
3⌘
↵ = P (reject H0 |H0 is true) = Y = 0 or 1 | p =
4
✓
◆
✓
◆
✓
◆
1
⇣
⌘
⇣
⌘
⇣
⌘
⇣
⌘
X 3 3 y 1 3 y
3 3 0 1 3
3 ⇣ 3 ⌘1 ⇣ 1 ⌘2
=
=
+
y 4
4
0 4
4
1 4
4
y=0
=
5
32
⇣
1⌘
= P (not reject H0 |Ha is true) = P Y = 2 or 3 | p =
4
✓ ◆⇣ ⌘ ⇣ ⌘
✓ ◆⇣ ⌘ ⇣ ⌘
3 ✓ ◆⇣ ⌘ ⇣ ⌘
X
y
3
y
2
1
3 1
3
3 1
3
3 1 3 3 0
=
=
+
y 4
4
2 4
4
3 4
4
y=2
=
5
32
power = 1
=
27
32
2. In the setting of the previous problem, small values of Y are evidence against H0 in
favor of the alternative hypothesis Ha . What is the P -value when you observe Y = 2?
1
When you observe Y = 1?
Sol)
p-value when (Y = 2) = P (Y  2|H0 is true)
=1
=1
P (Y = 3|H0 is true)
✓ ◆⇣ ⌘ ⇣ ⌘
3 3 3 1 0 37
=
3 4
4
64
p-value when (Y = 1) = P (Y  1|H0 is true)
= P (Y = 0|H0 is true) + P (Y = 1|H0 is true)
✓ ◆⇣ ⌘ ⇣ ⌘
✓ ◆⇣ ⌘ ⇣ ⌘
3 3 0 1 3
3 3 1 1 2
=
+
0 4
4
1 4
4
5
=
32
3. You observe a random sample Y1 , Y2 , · · · , Y8 from a Poisson distribution with unknown
mean . You want to test
H0 :
vs.
= 0.5
You decide to reject H0 in favor of Ha if V
that V is the CSS for .)
Ha :
> 0.5
8, where V is the sum of Y1 to Y8 . (Note
(a) What is the level ↵ of this test? (Hint : What is the distribution of a sum of
independent Poisson variables?)
t
Sol) When Yi ⇠ P oisson( ), mgf MY (t) = e (e 1) .
P
t
Then, mgf of V (= 8i=1 Yi ) is MV (t) = e8 (e 1) and therefore V ⇠ P oisson(8 )
↵ = P (reject H0 |H0 is true) = P (V
=1
7
X
e 4 4v
v=0
v!
=1
2
8|8 = 4)
0.949 = 0.051
(b) What is the power of this test at = 0.75? At = 1.0? At
Poisson table in Appendix to give numerical answers.)
= 1.25? (Use the
Sol)
power = P (V
8
>
>
1
>
<
= 1
>
>
>
:1
8| ) = 1
7
X
e
v=0
v
v!
0.744 = 0.256,
= 0.75
0.453 = 0.547,
= 1.0
0.220 = 0.780,
= 1.25
4. In the setting of the previous problem, large values of V are evidence against H0 in
favor of Ha . What is the P -value if you observe V = 4? If you observe V = 10?
Sol) Under H0 , V ⇠ P oisson(4).
p-value when (V = 4) = P (V
=1
3
X
e 4 4v
=1
9
X
e 4 4v
=1
v=0
p-value when (V = 10) = P (V
=1
4| = 0.5)
v=0
v!
0.433 = 0.567
10| = 0.5)
v!
0.992 = 0.008
5. A certain size bag is designed to hold 25 pounds of potatoes. A farmer fills such bags
in the field. Assume that the weight Y of potatoes in a bag follows N (µ, 9). We would
like to test H0 : µ = 25 vs. Ha : µ < 25. Let Y1 , Y2 , Y3 , Y4 be a random sample of size
4 from this distribution, and let the rejection region for this test is {Ȳ  22.5}.
(a) What is the significance level of this test?
Sol)
↵ = P (reject H0 |H0 is true) = P (Ȳ  22.5|µ = 25)
✓
◆
✓ ◆
22.5 25
5
p
=
=
= 0.0478
3
3/ 4
where
(·) is the cdf of the standard normal distribution.
3
(b) What is the power of this test when µ = 23?
Sol)
power at (µ = 23) = P (Ȳ  22.5|µ = 23) = P
✓ ◆
1
=
= 0.3694
3
✓
22.5 23
p
Z
3/ 4
◆
where Z is a random variable following N (0, 1).
(c) Suppose we observe y1 = 22, y2 = 24.5, y3 = 23, y4 = 26.5. What is the P -value
associated with this sample?
Sol) Ȳ = 24, n = 4
p-value = P (Ȳ  24|µ = 25)
✓
◆
24 25
p
=P Z
3/ 4
✓ ◆
2
=
= 0.2525
3
6. You have one observation Y , which has one of the pdf’s
3
f0 (y) = y 1/2
2
4
f1 (y) = y 1/3
3
0y1
0y1
Find the most powerful rejection region of level ↵(0 < ↵ < 1 given) for testing
H0 : f0 is true
vs.
4
Ha : f1 is true
Sol) Using Neyman-Pearson Lemma,
1
3 2
y
L(✓0 )
f0 (y)
9 1
=
= 2 1 = y 6 , where 0  y  1
4 3
L(✓a )
f1 (y)
8
3y
⇢
⇢
1
9 1
8
6
Most Powerful Rejection Region =
y < k = y6 < k
8
9
(
✓ ◆6 )
✓ ◆6
8
8
6
⇤
⇤
= y<
k
= {y < k }, k =
k6
9
9
Z k⇤
3 1
3 3 3 ⇤
↵ = P (reject H0 |H0 is true) =
y 2 dy = / y 2 |k0
2
2 2
0
3
= (k ⇤ ) 2
2
k⇤ = ↵ 3
2
) MPRR = {y < ↵ 3 }
7. You have one observation Y , which has one of the discrete pdf’s
y
0
1
2
3
4
5
6
f0 (y)
0.1
0.1
0.1
0.1
0.2
0.1
0.3
f1 (y)
0.3
0.1
0.1
0.2
0.1
0.1
0.1
You want to test
H0 : f0 is true
vs.
Ha : f1 is true
(a) Here is a test: reject H0 if Y = 0, 1, 2, 3, or 5. What are the probabilities of the
two types of error for this test? What is the power of the test at the alternative?
Sol)
↵ = P (reject H0 |H0 is true) = 0.1 + 0.1 + 0.1 + 0.1 + 0.1 = 0.5
= P (not reject H0 |Ha is true) = 0.1 + 0.1 = 0.2
power = 1
= 0.8
5
(b) What is the most powerful test for H0 against Ha of level ↵ = 0.1? What is the
most powerful test of level ↵ = 0.2?
Sol)
RR
f0 /f1  13
f0 /f1  12
f0 /f1  1
f0 /f1  2
f0 /f1  3
Y
0
0, 3
0, 1, 2, 3, 5
0, 1, 2, 3, 4, 5
0, 1, 2, 3, 4, 5, 6
↵
0.1
0.2
0.5
0.7
1
MP test is the test that reject H0 if Y = 0 when ↵ = 0.1.
MP test is the test that reject H0 if Y = 0 or 3 when ↵ = 0.2.
8. Let Y1 , · · · , Yn be a random sample from an exponential distribution with mean ✓. We
would like to test H0 : ✓ = 3 against Ha : ✓ = 5 based on this random sample.
(a) Find the form of the most powerful rejection region.
Sol)
1 yi
fYi (yi ) = e ✓
✓
✓ ◆n
n
Y
1 Pn
1
L(✓) =
fYi (yi ) =
e ✓ i=1 yi
✓
i=1
✓ ◆n P
✓ ◆n
n
1
1
L(✓0 )
✓a
5
)
i=1 yi ( ✓0
✓a =
=
e
e
L(✓a )
✓0
3
n
X
RR : {
yi k ⇤ }
2 Pn
i=1 yi
15
i=1
n
X
Since Yi ⇠ exponential(✓) = Gamma(1, ✓),
Under H0 ,
2
Pn
i=1 Yi
✓0
i=1
n
=
2X
Yi ⇠
3
2
Yi ⇠ Gamma(n, ✓)
(2n)
i=1
n
n
X
2X
) M P RR = {
Yi
3
2
2n,↵ } = {
i=1
6
i=1
Yi
3
2
2
2n,↵ }
(b) Suppose n = 12. Find the MP rejection region of level 0.1.
Sol)
n
M P RR = {
2X
Yi
3
2
24,0.1 } = {
i=1
n
X
Yi
49.79437}
i=1
(c) Is the rejection region in (b) the uniformly most powerful rejection region of level
0.1 for testing H0 : ✓ = 3 vs. Ha : ✓ > 3?
Sol)
Neither the test statistic nor the rejection region for this ↵-level test depends on the
particular value assigned to ✓a . Thus, for any value of ✓a greater than ✓0 , we obtain
exactly the same rejection region.
) It is the uniformly most powerful rejection region.
(d) [not to be turned in] Find the likelihood ratio test of level ↵ for testing H0 : ✓ = 3
vs. Ha : ✓ 6= 3.
Sol)
L(✓) =
n
Y
fYi (yi ) =
i=1
ln L(✓) =
n ln ✓
@
ln L(✓) =
@✓
then, ✓ˆ =
1
n
n
Y
1
✓
i=1
n
X
1
✓
n
1
+ 2
✓
✓
n
X
e
yi
✓
1
e
✓n
1 Pn
i=1 yi
✓
2nȳ
=
✓ˆ3
n
< 0.
✓ˆ2
=
yi
i=1
n
X
let
yi = 0
i=1
Yi = Ȳ
i=1
2
@
n
To be more rigorous, check @✓
ˆ = ˆ2
2 ln L(✓)|✓=✓
✓
7
Pn
1
y
L(✓0 )
✓0 n e ✓0 i=1 i
=
=
1 Pn
ˆ
L(✓)
✓ˆ n e ✓ˆ i=1 yi
Pn
1
1
ȳ
ȳ
)
nȳ( ȳ3ȳ3 )
i=1 yi ( 3
ȳ = ( )n e
= ( )n e
3
3
ȳ n n (ȳ 3)
ȳ n n ȳ set
n
=( ) e 3
= e ( ) e 3 = g(ȳ)
3
3
ȳ
n
nt
g(t) = ct e , where t = and c = en
3
let
g 0 (t) = cntn 1 e nt + nctn e nt = 0
ȳ
then t = = 1 provides the maximum value of g(t) since g 00 (t)|t=1 < 0.
3
n
n
X
X
Therefore, g(ȳ) < k where
Yi  c1 or
Yi c2
Under H0 ,
2
i=1
Pn
i=1 Yi
✓0
=
n
2X
3
i=1
i=1
2
Yi ⇠
(2n).
Then the likelihood ratio test of level ↵ is the test which rejects H0
P
P
when ni=1 Yi  32 22n,1 ↵/2 or ni=1 Yi 32 22n,↵/2 .
9. Let Y1 , · · · , Yn be a random sample from N (0, 2 ). We would like to test H0 :
against Ha : 2 > 02 based on this random sample.
(a) Find the unifomly most powerful rejection region of level ↵.
sol)
L(
L( 02 )
=
L( 2 )
✓
2
2
0
2
2
) = (2⇡
◆n/2
exp
X
yi2
n
2
)
 ✓
exp
✓ P
Yi2
2 2
◆
◆X
Yi2
c/ 2 )} where
X
2
2
2
0
2 2
0
>
2
0
Yi2 / 2 ⇠
2
for
2
P 2
Therefore the likelihood ratio is decreasing in
Yi .
The uniformly most powerful rejection region of level ↵ is
C = {(y1 , · · · , yn :
= {(y1 , · · · , yn :
X
then P
c)}
yi2 / 2
⇣X
Yi2 / 02
U M P RR = {
8
X
2
n,↵ |H0
Yi2
⌘
=↵
2 2
0 n,↵ }
(n)
2
=
2
0
(b) [not to be turned in] Find the likelihood ratio test of level ↵ for testing H0 :
2
2 6= 2
0 vs. Ha :
0
sol)
L(
2
) = (2⇡
2
)
n
2
exp
✓ P
Yi2
◆
2 2
P 2
n
Yi
ln L( 2 ) =
ln(2⇡ 2 )
2
2
P 22
@
n
Yi
ln L( 2 ) =
+
=0
2
2
@
2
2( 2 )2
1X 2
then ˆ 2 =
Yi
n
P 2
@2
n
Yi
2
Since
ln L( ) =
| 2 2
2
2
2
2
2
@( )
2( )
( )3 =ˆ
P 2

1
n
Yi
n 1
= 2 2
| 2 =ˆ 2 =
< 0,
2
( ) 2
2 (ˆ 2 )2
ˆ 2 is the MLE of
L( 02 )
=
=
L(ˆ 2 )
✓
ˆ2
2
0
◆n/2
exp
2
✓
.
n
2
✓
ˆ2
2
0
◆
n
+
2
◆
=g
✓
ˆ2
2
0
◆
⇣ n ⌘
⇣n⌘
ˆ2
n/2
g(t) = ct exp
t
where t = 2 and c = exp
2
2
0
⇣ n ⌘
⇣ n ⌘
n
n
n
n
let
g 0 (t) = c t 2 1 exp
t
c t 2 exp
t =0
2
2
2
2
⇣
⌘
⇣
⇣
⌘ n ⌘
n
n
n
n n
n
g 00 (t) = c exp
t
t 2 nt 2 1 +
1 t2 2
2
2
2
2
Then t =
Since
and
X
ˆ2
00
2 = 1 provides the maximum value of g(t) since g (t)|t=1 < 0.
0
L( 02 )
is increasing where ˆ 2 / 02 < 1 and decreasing otherwise
L(ˆ 2 )
Yi2 / 02 ⇠
2
(n) under H0 , then the rejection region of level ↵ is
X
X
2
RR = {
Yi2 / 02  2n,1 ↵/2 or
Yi2 / 02
n,↵/2 }
X
X
2 2
={
Yi2  02 2n,1 ↵/2 or
Yi2
0 n,↵/2 }
9
2 =