STA 6327
STATISTICAL THEORY II
EXAM I - SOLUTIONS
PROBLEM 1
Note that
Fn
t P n X n t
X
n
t
P X n
n
t
1 P X n
n
t
1 P X n
n
t
1 FX n
n
n
t
1 ,
n
n
t
1 1 ,
n
if 0
t
0 t n . If we let 1 , then
n
n
t
t
Fn X t 1 1
1 e ,
n
n
n
if 0 t . T
STA 6326
CHAPTER 5 - SOLUTIONS
Problem 5.1
For a (simple) random sample of size n from this population let
1, if i th person is color-blind
Xi
0, otherwise
i 1, 2, , n . Consequently, X 1 , X 2 , , X n are iid Bernoulli(0.01). Hence,
P at least 1 person
STA 6327
CHAPTER 6 - SOLUTIONS
Problem 6.1
1
exp 2 x 2
2
2
If X ~ n 0, 2 then
1
f x | 2
2
1
2 1
1
exp 2 x
2
2
2
g x | 2 h x
where
g x | 2
2
1
exp 2 x
2
1
2
and
1
.
2
h x
Hence, by the factorization criteria T X X is sufficient for 2 .
Problem
STA 6327
CHAPTER 7 - SOLUTIONS
Problem 7.2
(a) Here,
x
1
L | x
x 1 exp i
i
i 1
n
n
1
1 1 n
n xi
i 1
1
exp
n
x
i 1
i
Hence,
n
ln L | x n ln n ln 1 ln xi
i 1
1
n
x
i
i 1
and
d
n 1
ln L | x
d
2
n
x
i
i 1
0
which has a unique solution
1
x.
Fur
STA 6327
CHAPTER 9 - SOLUTIONS
Problem 9.1
Note that
P L X U X
P L X U X
P L X P U X P L X U X .
Furthermore, since L X U X then L X U X . Hence,
1 P L X U X P L X L X 1 .
Therefore,
P L X U X 1
and consequently
P L X U X 1 1 1 2 1 1 1 2 .
Problem 9.2
STA 6327
STATISTICAL THEORY II
EXAM I - SOLUTIONS
PROBLEM 1
Note that
Fn
t P n X n t
X
n
t
P X n
n
t
1 P X n
n
t
1 P X n
n
t
1 FX n
n
n
t
1 ,
n
n
t
1 1 ,
n
if 0
t
0 t n . If we let 1 , then
n
n
t
t
Fn X t 1 1
1 e ,
n
n
n
if 0 t . T
STA 6326
CHAPTER 5 - SOLUTIONS
Problem 5.1
For a (simple) random sample of size n from this population let
1, if i th person is color-blind
Xi
0, otherwise
i 1, 2, , n . Consequently, X 1 , X 2 , , X n are iid Bernoulli(0.01). Hence,
P at least 1 person
STA 6327
CHAPTER 7 - SOLUTIONS
Problem 7.2
(a) Here,
x
1
L | x
x 1 exp i
i
i 1
n
n
1
1 1 n
n xi
i 1
1
exp
n
x
i 1
i
Hence,
n
ln L | x n ln n ln 1 ln xi
i 1
1
n
x
i
i 1
and
d
n 1
ln L | x
d
2
n
x
i
i 1
0
which has a unique solution
1
x.
Fur
STA 6327
CHAPTER 9 - SOLUTIONS
Problem 9.1
Note that
P L X U X
P L X U X
P L X P U X P L X U X .
Furthermore, since L X U X then L X U X . Hence,
1 P L X U X P L X L X 1 .
Therefore,
P L X U X 1
and consequently
P L X U X 1 1 1 2 1 1 1 2 .
Problem 9.2