Statistical Sciences 1024A/B Chapter Notes - Chapter 16: Null Hypothesis, Standard Deviation, Binomial Distribution

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Chapter 16 Solutions
16.1: The most important reason is (c); this is a convenience sample consisting of the first 20
students on a list. This is not an SRS. Anything we learn from this sample will not extend to the
larger population. The other two reasons are valid, but less important issues. Reason (a) — the
size of the sample and large margin of error would make the interval less informative, even if
the sample were representative of the population. Reason (b) – nonresponse — is a potential
problem with every survey, but there is no particular reason to believe it is more likely in this
situation.
16.2: (a) The 95% confidence interval is
x
±
z*
σ
n
= 1.92
±
1.96
1.83
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 telephone numbers were represented in the sample, and the low
response rate (10.9% =
5,029 45,956
) means that even that group may not be well represented
by this sample.
16.3: Any number of things could go wrong with this convenience sample. The day after
Thanksgiving is widely regarded (rightly or wrongly) as a day on which retailers offer great
deals — and the kinds of shoppers found that day probably don’t represent shoppers generally.
Also, the sample isn’t random.
16.4: (a) The 95% confidence interval is 159
±
1.96
35
501
= 159
±
3.06 = 155.94 pounds to
162.06 pounds. (b) The sample is an SRS, and it is certainly large enough so that the central
limit theorem applies — we know that
x
has a sampling distribution that is approximately
Normal. However, people are often inclined to misrepresent (lie or underestimate) their weights.
Gallup did not weigh these women — they reported their own weights. If their responses are
honest, then the confidence interval above can be relied upon. However, this may well not be the
case.
16.5: No. The confidence interval does not describe the range of future values of
x
. Instead, if
we repeated the experiment over and over, each time computing a 95% confidence interval for
the population mean
µ
, then 95% of these confidence intervals would capture the true population
mean.
16.6: With z* = 1.96 and
σ
= 75, the margin of error is z*
σ
n
=
. (a) If n = 100, the margin
of error is 1.47. (b) If n = 400, the margin of error is 0.735. If n = 1600, the margin of error is
0.3675. (c) As the sample size increases, the margin of error decreases. Notice that when the
sample size quadruples, the margin of error is halved.
16.7: The margin of error only addresses chance variation in the random selection of a sample.
Hence, the answer is (c). Sources of bias described in (a) and (b) are not accounted for in the
margin of error, and are difficult to assess.
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Solutions 173
16.8: Computing by hand, we get results almost identical to those obtained using the Applet.
The standard deviation of
x
is
σ
n
=
116
50
= 16.405. The test statistic is
x 515
16.405
. (a) If
x
=
541, the test statistic is z = 1.58. The P-value is P =
P Z 1.58
( )
= 0.057. Hence, this is (barely)
not significant at the 5% level. (b) If
x
= 542, the test statistic is z = 1.65. The P-value is P =
P Z 1.65
( )
= 0.049. Hence, this is (barely) significant at the 5% level. Notice that there’s no
practical difference between
x
= 541 and
x
= 542, yet the decision we would make changes,
strictly applying the 5% significance level.
16.9: The full applet output for n = 5 is below on the left. Next to this are the Normal curves
drawn for n = 15 and n = 40. The reported P-values agree with the “hand-computed” values z =
4.8 5
0.5 n
, and P =
P Z z
( )
, given in the table.
16.10: The confidence intervals are given in the table to the right.
In each case, the interval is 4.8
±
1.96
0.5
n
.
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