Chapter 16 Solutions
16.1. (a)
Yes:
(
1
.
96
)(
$1125
/
√
958
)
.
=
$71, and $8740
±
$71
=
$8669 to $8811.
(b)
No:
Because the numbers are based on voluntary response rather than an SRS, the confidence
interval methods of this chapter cannot be used; the interval does not apply to the whole
population.
16.2. (a)
x
±
1
.
96
σ/
√
880
=
1
.
92
±
0
.
1209
=
1
.
80 to 2.04 motorists.
(b)
The large sample size
means that, because of the central limit theorem, the sampling distribution of
x
is roughly
Normal even if the distribution of responses is not.
(c)
Only people with listed phone
numbers were represented in the sample, and the low response rate (10
.
9%
.
=
5029
45
,
956
) means
that even that group may not be well represented by this sample.
16.3. (a)
Not included: This is a ﬂaw in the sampling design.
(b)
Not included: This error
arises from the sampling process.
(c)
Included: This random error is the only error
addressed by confidence interval methods.
16.4. (a)
The central limit theorem tells us that the sampling distribution of
x
is approximately
Normal because we have a reasonably large sample.
(b)
The 99% confidence interval is
x
±
2
.
576
σ/
√
250
=
$237
±
10
.
59
=
$226
.
4 to $247.6.
(c)
The sample may be badly biased:
At best, this sample represents only the opinions of those who shop in malls similar to the
one where the sample was chosen.
16.5. (a)
z
=
544
−
518
114
/
√
50
.
=
1
.
61—not significant at the 5% level (
P
=
0
.
0537).
(b)
z
=
545
−
518
114
/
√
50
.
=
1
.
67—significant at 5% level (
P
=
0
.
0475),
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16.6.
The full applet output for
n
=
5 is below on the left; on the right are the Normal curves
drawn for
n
=
15 and
n
=
40. The reported
P
values (0.1867, 0.0606, and 0.0057) agree
with the “handcomputed” values given in the solution to the next exercise.
16.7.
The
z
statistics and the
P
values are given in the table on
the right. In each case, the computation is
z
=
4
.
8
−
5
0
.
5
/
√
n
and
P
=
P
(
Z
<
z
).
n
z
P
(a)
5
−
0.89
0.1867
(b)
15
−
1.55
0.0606
(c)
40
−
2.53
0.0057
16.8.
The confidence intervals are given in the table on the right.
In each case, the interval is
4
.
8
±
1
.
960
0
.
5
√
n
.
n
=
5
4.362 to 5.238
n
=
15
4.547 to 5.053
n
=
40
4.645 to 4.955
16.9. (a)
“Statistically insignificant” means that the differences observed were no more than
might have been expected to occur by chance even if SES had no effect on LSAT results.
(b)
If the results are based on a small sample, then even if the null hypothesis were not true,
the test might not be sensitive enough to detect the effect. Knowing the effects were small
tells us that the test was not insignificant merely because of a small sample size.
16.10. (a)
No: In a sample of size 500, we expect to see about 5 people who have a
P
value
of 0.01 or less [5
=
(
500
)(
0
.
01
)
]. These four
might
have ESP, or they may simply be among
the “lucky” ones we expect to see.
(b)
The researcher should repeat the procedure on these
four to see whether they again perform well.
16.11.
Recall that effect size
=
true mean response
−
hypothesized response
standard deviation of response
. The hypothesized
response is 115, and the standard deviation is 30.
(a)
The effect size is
130
−
115
30
=
0
.
5, so
from the table in the text, we need a sample of size
n
=
35 to have 90% power against
the alternative
µ
=
130.
(b)
The effect size is
139
−
115
30
=
0
.
8, so we need a sample of size
n
=
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