Solutions to EndofSection and Chapter Review Problems
53
CHAPTER 7
7.1
X
±
Z
⋅
σ
n
=
85
±
1.96
⋅
8
64
83.04
≤
μ
≤
86.96
7.2
X
±
Z
⋅
σ
n
=
125
±
2.58
⋅
24
36
114.68
≤
μ
≤
135.32
7.3
If all possible samples of the same size
n
are taken, 95% of them will include the true
population average monthly sales of the product within the interval developed. Thus we are
95 percent confident that this sample is one that does correctly estimate the true average
amount.
7.4
Since the results of only one sample are used to indicate whether something has gone wrong
in the production process, the manufacturer can never know with 100% certainty that the
specific interval obtained from the sample includes the true population mean. In order to
have every possible interval estimate of the true mean, the entire population (sample size
N
)
would have to be selected.
7.5
To the extent that the sampling distribution of sample means is approximately normal, it is
true that approximately 95% of all possible sample means taken from samples of that same
size will fall within 1.96 times the standard error away from the true population mean. But
the population mean is not known with certainty. Since the manufacturer estimated the mean
would fall between 10.99408 and 11.00192 inches based on a single sample, it is not
necessarily true that 95% of all sample means will fall within those same bounds.
7.6
Approximately 5% of the intervals will not include the true population mean somewhere in
the interval. Since the true population mean is not known, we do not know for certain
whether it is in the one interval we have developed, between 10.99408 and 11.00192 inches.
7.7
(a)
X
±
Z
⋅
σ
n
=
0.995
±
2.58
⋅
0.02
50
0.9877
≤
μ
≤
1.0023
(b)
Since the value of 1.0 is included in the interval, there is no reason to believe that the
average is different from 1.0 gallon.
(c)
No. Since
σ
is known and
n
= 50, from the central limit theorem, we may assume that
the sampling distribution of
X
is approximately normal.
(d)
An individual can with an observed value of 0.98 gallon of paint yields a
Z
value of
– 0.75. The confidence interval represents bounds on the estimate of the average of a
sample of 50 cans of paint, not an individual value.
(e)
The reduced confidence level narrows the width of the confidence interval.
(a)
X
±
Z
⋅
σ
n
=
0.995
±
1.96
⋅
0.02
50
0.9895
≤
μ
≤
1.0005
(b)
Since the value of 1.0 is still included in the interval, there is no reason to
believe that the average is different from 1.0 gallon.
53
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54
Chapter 7: Confidence Interval Estimation
7.8
(a)
(b)
No. The manufacturer cannot support a claim that the bulbs last an average of 400
hours. Based on the data from the sample, a mean of 400 hours would represent a
distance of 4 standard deviations above the sample mean of 350 hours.
(c)
No. Since
σ
is known and
n
= 64, from the central limit theorem, we may assume that
the sampling distribution of
X
is approximately normal.
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 Spring '08
 Harris
 Normal Distribution

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