409
Fall STAT 2009
Examples for 11/02/2009
Example 1:
1
2
3
4
5
p0(x )
0.25
0.15
0.20
0.05
0.35
p1(x )
0.20
0.20
0.20
0.20
0.20
H 0 : X has p.m.f. p 0 ( x )
vs.
H 1 : X has p.m.f. p 1 ( x ).
1
(x)
2
3
4
5
1.25
0.75
1
0.25
1.75
Reject H 0 if x = 4
= 0.05,
Power = 0.20.
Reject H 0 if x = 4 or 2
= 0.20,
Power = 0.40.
= 0.10,
Power =
Reject H 0 if x = 4
Reject H 0 with probability
1
if x = 2
3
8
.
30
Example 2:
1
2
3
4
5
p0(x )
0.25
0.15
0.20
0.05
0.35
p1(x )
0.05
0.30
0.40
0.20
0.05
H 0 : X has p.m.f. p 0 ( x )
vs.
H 1 : X has p.m.f. p 1 ( x ).
1
(x)
2
3
4
5
5
0.5
0.5
0.25
7
= 0.05,
Power = 0.20.
1
if x = 2
3
= 0.10,
Power = 0.30.
1
if x = 3
4
= 0.10,
Power = 0.30.
= 0.10,
Power = 0.30.
Reject H 0 if x = 4
Reject H 0 if x = 4
Reject 0 H with probability
Reject H 0 if x = 4
Reject H 0 with probability
Reject H 0 if x = 4
1
if x = 2
6
1
Reject H 0 with probability
if x = 3
8
Reject H 0 with probability
H0 : 0
vs.
0 1 = .
H1 : 1
max L( )
Reject H 0
=
if
0
max L( )
max L( )
k
0
* =
1
max
0 1
L( )
k
since * = min ( , 1 )
H0 : = 0
H1: 0
vs.
Reject H 0
if
=
L( 0 )
k
L
()
Example 3:
Let X 1 , X 2 , , X n be a random sample of size
distribution with mean 1 / . That is,
n
from an Exponential
f X ( x ) = e x ,
H0 : = 0
L( ) =
vs.
n
e
H1: 0
xi
i =1
n
= n exp x i .
i =1
=
L( 0 )
=
L
()
x > 0.
n
1
=
.
X
n
0 exp 0 x i
()
{
}
i =1
n
n
= e n 0 x exp n 0 x .
n
1 n
1
exp x i
x
x i =1
k
x exp{ 0 x } c
Reject H 0
if
x exp{ 0 x } c.