Homework #6 Solutions
9.1
(a)
With no prior estimate of p,
062
.
663
)
5
.
0
)(
5
.
0
(
05
.
0
575
.
2
ˆ
ˆ
2
*
*
2
2
=
=
=
q
p
E
z
n
α
or 664
Therefore, 664 subscribers should be surveyed.
(b)
With a prior estimate of p=0.30,
972
.
556
)
7
.
0
)(
3
.
0
(
05
.
0
575
.
2
ˆ
ˆ
2
*
*
2
2
=
=
=
q
p
E
z
n
α
or 557
Therefore, 557 subscribers should be surveyed.
(c)
If a 40% nonresponse rate is anticipated, we need to end up with 557 surveys, after
nonresponse. So 0.6N=557 or N=928 surveys must be mailed. Such a nonresponse rate
may cause a bias in the results if the nonresponse rate is related to the proportion of
interest, i.e. subscribers with income above $100,000 may be less likely to respond to
the survey.
9.2
(a)
5
.
0
:
0
=
p
H
vs
5
.
0
:
1
≠
p
H
. The alternative should be twosided because there is
no prior knowledge of whether heads should be more or less likely than tails.
(b)
The test statistic is
34
.
1
10000
)
5
.
0
)(
5
.
0
(
5
.
0
5067
.
0
ˆ
0
0
0
=

=

=
n
q
p
p
p
z
The Pvalue is
180
.
0
0901
.
0
2
))
34
.
1
(
1
(
2
=
×
=
Φ

=
P
Since
05
.
0
=
α
P
, do not reject
0
H
and conclude that heads and tails are equally likely.
(c)
A 95% CI is given by
]
5165
.
0
,
4969
.
0
[
10000
)
4933
.
0
)(
5067
.
0
(
96
.
1
5067
.
0
ˆ
ˆ
96
.
1
ˆ
=
±
=
±
n
q
p
p
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9.4
(a)
With a prior estimate of p=0.01,
96
.
9507
)
99
.
0
)(
01
.
0
(
002
.
0
96
.
1
ˆ
ˆ
2
*
*
2
2
=
=
=
q
p
E
z
n
α
or 9508
Therefore, 9508 parts should be sampled.
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 Fall '10
 WeiZhu
 Statistics, Normal Distribution, Trigraph, Type I and type II errors, p1

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