Chapter 14 Solutions
14.1. (a)
The standard deviation of
x
is
σ/
√
1000
.
=
1
.
8974.
(b)
m
=
2
(
1
.
8974
)
.
=
3
.
8 (or “
±
3
.
8”).
(c)
The confidence intervals drawn may vary, of
course, but they should be 2
m
=
7
.
6 units wide.
(d)
95%.
281.9 283.8 285.7
278.1
276.2
274.3
280
14.2. (a)
51%
±
3%
=
48% to 54%.
(b)
“95% confidence” means that this interval was found
using a procedure that produces correct results (i.e., includes the true population proportion)
95% of the time.
14.3.
Shown below are sample output screens for (a) 10 and (b) 1000 SRSs. In 99.4% of all
repetitions of part (a), students should see between 5 and 10 hits (that is, at least 5 of the 10
SRSs capture the true mean
µ
). Out of 1000 80% confidence intervals, nearly all students
will observe between 76% and 84% capturing the mean.
14.4.
Search Table A for 0.0125 (half of the 2.5% that is
not
included in a 97.5% confidence interval). This area
corresponds to
z
∗
=
2
.
24. Software gives
z
∗
=
2
.
2414.
z
*
–
z
*
0
Standard
Normal curve
Probability
= 0.0125
Probability
= 0.0125
14.5. State:
What is the concentration of the active ingredient in this batch?
Formulate:
We will estimate the true concentration in this batch of the product—the mean
µ
of the distribution of all measurements—by giving a 95% confidence interval.
Solve:
We assume that the three measurements can be considered an SRS of all such
measurements. We have been told that repeated measurements follow a Normal distribution
with standard deviation 0.0068 g
/
l. The mean of the sample is
x
=
0
.
8403
+
0
.
8363
+
0
.
8447
3
=
0
.
8404
3 g
/
l
.
169
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170
Chapter 14
Confidence Intervals: The Basics
For 95% confidence, the critical value is
z
∗
=
1
.
960, so the 95% confidence interval for
µ
is
x
±
z
∗
σ
√
n
=
0
.
8404
3
±
1
.
960
0
.
0068
√
3
=
0
.
8404
3
±
0
.
00770
=
0
.
8327 to 0
.
8481 g
/
l
.
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 Fall '09
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 Normal Distribution, Standard Deviation, 2.5%, 54%, 51%

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