Statistical Sciences 1024A/B Chapter 18: Chapter+18

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Chapter 18 Solutions
18.1: The standard error of the mean is
s/n=63.9/ 1000
= 2.0207 minutes.
18.2: The standard error of the mean is
s/n=15/ 10
= 4.743 beats per minute.
18.3: (a)
t*
= 2.132. (b)
t*
= 2.479.
18.4: Here, df = 30 – 1 = 29. (a)
t
*
= 2.045. (b)
t*
= 0.683.
18.5: (a) df = 12 – 1 = 11, so
t
*
= 2.201. (b) df = 18 – 1 = 17, so
t*
= 2.898.
(c) df = 6 – 1 = 5, so
t*
= 2.015.
18.6: (a) The stemplot provided shows a slight skew to the right, but not so strong that it would
invalidate the t procedures. (b) With
x
= 18.66, s = 10.2768, and
t*
= 2.093 (df = 19), the 95%
confidence interval for μ is 18.66 ± 2.093
10.2768
20
= 18.66 ± 4.8096 = 13.8504 to 23.4696.
18.7: STATE: What is the mean percent μ of nitrogen in ancient air? PLAN: We will estimate μ
with a 90% confidence interval. SOLVE: We are told to view the observations as an SRS. A
stemplot shows some left-skewness; however, for such a small sample, the data are not
unreasonably skewed. There are no outliers. With
x
= 59.5889% and s = 6.2553% nitrogen, and
t
*
= 1.860 (df = 8), the 90% confidence interval for μ is 59.5889 ± 1.860
6.2553
9
= 59.5889 ±
3.8783 = 55.71% to 63.47%. CONCLUDE: We are 90% confident that the mean percent of
nitrogen in ancient air is between 55.71% and 63.47%.
18.8: (a) df = 20 – 1 = 19. (b) t = 1.84 is bracketed by
t
*
= 1.729 (with right-tail probability
0.05) and
t
*
= 2.093 (with right-tail probability 0.025). Hence, since this is a one-sided
significance test, 0.025 < P < 0.05. (c) This test is significant at the 5% level since the P < 0.05.
It is not significant at the 1% level since the P > 0.01. (d) From software, P = 0.0407.
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188 Chapter 18 Inference about a Population Mean
18.9: (a) df = 15 – 1 = 14. (b) t = 2.12 is bracketed by
t
*
= 1.761 (with two-tail probability 0.10)
and
t
*
= 2.145 (with two-tail probability 0.05). Hence, since this is a two-sided significance test,
0.05 < P < 0.10. (c) This test is significant at the 10% level since the P < 0.10. It is not
significant at the 5% level since the P > 0.05. (d) From software, P = 0.0524.
18.10: STATE: Is there evidence that the percent of nitrogen in ancient air was different from the
present 78.1%? PLAN: We test
H
0
: μ = 78.1% vs.
H
a
: μ
78.1%. We use a two-sided
alternative because, prior to seeing the data, we had no reason to believe that the percent of
nitrogen in ancient air would be higher or lower. SOLVE: We addressed the conditions for
inference in Exercise 18.7. In that solution, we found
x
= 59.5889% and s = 6.2553% nitrogen,
so t =
59.5889 78.1
6.2553 9
= 8.88. For df = 8, this is beyond anything shown in Table C, so P <
0.001 (software gives P = 0.00002). CONCLUDE: We have very strong evidence (P < 0.001)
that Cretaceous air contained less nitrogen than modern air.
18.11: PLAN: Take μ to be the mean difference (monkey call minus pure tone) in firing rate. We
test
H
0
: μ = 0 vs.
H
a
: μ > 0, using a one-sided alternative because the researchers suspect a
stronger response to the monkey calls. SOLVE: We must assume that the monkeys can be
regarded as an SRS. For each monkey, we compute the call minus pure tone differences; a
stemplot of these differences (provided) shows no outliers or deviations from Normality. The
mean and standard deviation are
x
= 70.378 and s = 88.447 spikes/second, so t =
70.378 0
88.447 37
=
4.84 with df = 36. This has a very small P-value: P < 0.0001. CONCLUDE: We have very
strong evidence that macaque neural response to monkey calls is stronger than the response to
pure tones.
18.12: Using the information in Exercise 18.11, note that df = 36 is not contained in the table, so
use df = 30 (round down in order to be conservative). Here, with
t*
= 2.750, the 99% confidence
interval is given by 70.378 ± 2.750
88.447
37
= 70.378 ± 39.987 = 30.391 to 110.365
spikes/second.
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Solutions 189
18.13: The stemplot suggests that the distribution of nitrogen contents is heavily skewed.
Although t procedures are robust, they should not be used if the population being sampled from
is this heavily skewed. In this case, t procedures are not reliable.
18.14: The provided stemplot of carbon-13 ratios suggests no strong skew, so use of t
procedures is appropriate, assuming the sample is random. With
x
= –2.8825, s = 1.0360, df =
24 – 1 = 23, and
t
*
= 1.714, a 90% confidence interval for the mean carbon-13 ratio is given by –
2.8825 ± 1.714
1.0360
24
= –2.8825 ± 0.3625 = –3.2450 to –2.5200.
18.15: (b) We virtually never know the value of
σ
.
18.16: (b) t =
810
4 16
= –2.
18.17: (c) df = 25 – 1 = 24.
18.18: (c) P < 0.01.
18.19: (a) 2.718. Here, df = 11.
18.20: (b) t < –3.497 or t > 3.497.
18.21: (c) 72.7 to 97.3. The interval is computed as 85 ± 3.250
12
10
.
18.22: (a) There is a clear outlier in the data. The t procedures are robust against mild skew, and
they are used when
σ
is unknown.
18.23: (b) If you sample 64 unmarried male students, and then sample 64 unmarried female
students, no matching is present.
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