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

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24 Sep 2015

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187

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.

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.

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 =

8−10

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.