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APPM 4/5570
Solutions to Problems from Section 8.3
37.
ˆ
p
= 82
/
150
Step One: State the hypotheses:
H
0
:
p
= 0
.
40
H
a
:
p
6
= 0
.
40
Step Two: The test statistic:
Z
=
ˆ
p

p
0
q
p
0(1

p
0
)
/n
=
(82
/
150)

0
.
40
q
(0
.
40)(0
.
60)
/
150
≈
3
.
67
Step Three: The rejection region:
The rejection region is twotailed. For
α
= 0
.
01, it is given by the region on the
z
curve above 2
.
575 and below

2
.
575.
Step Four: The conclusion:
The test statistic is in the rejection region, so we reject
H
0
in favor of
H
a
at
0
.
01 level of signi±cance. These data suggest that the actual percentage of type
A blood donation di²ers from 40%.
This conclusion would not change for
α
= 0
.
05 since the new rejection region
would then be above 1
.
96 and below

1
.
96.
The test statistic is still in the
rejection region.
39.
n
= 1000
H
0
:
p
= 0
.
02
H
a
:
p <
0
.
02
a.
ˆ
p
= 15
/
1000 = 0
.
015 The test statistic is
Z
=
ˆ
p

p
0
q
p
0
(1

p
0
)
/n
=
0
.
015

0
.
02
q
(0
.
02)(0
.
98)
/
1000
≈ 
1
.
13
.
The rejection region is given by the region on the
z
curve below

1
.
645.
The test statistic does not fall in the rejection region. There is not signif
icant evidence, given by this data set, to reject
H
0
at 0
.
05 level of signif
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