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Solutions to Final Review Problems Part 2
15.
ˆ
p
= 34
/
100 = 0
.
34
(a) Hypotheses:
H
0
:
p
= 0
.
29
H
a
:
p
6
= 0
.
29
Test Statistic:
Z
=
ˆ
p

p
0
q
p
0
(1

p
0
)
n
=
0
.
34

0
.
29
q
(0
.
29)(0
.
71)
100
≈
1
.
10
Since this is a twotailed
Z
test, the Pvalue is the area under the
Z
curve
above and below the test statistic and it’s negative.
Pvalue = 2
·
P
(
Z >
1
.
10) = 2(0
.
1357) = 0
.
2714
(b) Pvalue
>
0
.
10 implies fail to reject
H
0
at 0
.
10 level of signiFcance.
(c) Anything above the Pvalue. ±or example
α
= 0
.
28.
16.
(a)
(
n

1)
S
2
χ
2
α/
2
,n

1
,
(
n

1)
S
2
χ
2
1

α/
2
,n

1
(10

1)24
2
23
.
587
,
(10

1)24
2
1
.
735
!
(219
.
7821
,
2987
.
896)
(b) Hypotheses:
H
0
:
σ
2
= 25
2
H
a
:
σ
2
<
25
2
Test Statistic:
W
=
(
n

1)
S
2
σ
2
0
=
(10

1)24
2
25
2
= 8
.
2944
Rejection Region: The rejection region is the lower tail (with area 0
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This note was uploaded on 01/23/2011 for the course APPM 5440 at Colorado.
 '08
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