ch17 - Chapter 17 Solutions 17.1. s / n = 21 . 88 / 20 . =...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Chapter 17 Solutions 17.1. s / n = 21 . 88 / 20 . = 4 . 8925 minutes. 17.2. The mean is x = 251 ms. Because s / n = s / 12 = 45 ms, we have s . = 155 . 88 ms. 17.3. (a) t = 2 . 015 for df = 5 and = . 05. (b) t = 2 . 518 for df = 21 and = . 01. 17.4. Use df = 24. (a) t = 2 . 064 for = . 025. (b) t = . 685 for = . 25. 17.5. (a) Use df = 9 and = . 025: t = 2 . 262. (b) Use df = 19 and = . 005: t = 2 . 861. (c) Use df = 6 and = . 05: t = 1 . 943. 17.6. (a) The stemplot shows a slight skew to the right, but not so strong that it would invalidate the t procedures. (b) With x = 18 . 66 and s . = 10 . 2768, and t = 2 . 093 (df = 19), the 95% confidence interval for is 18 . 66 2 . 093 10 . 2768 20 = 18 . 66 4 . 8096 = 13.8504 to 23.4696. 67888 1 0004 1 67 2 012 2 57 3 00 3 78 17.7. State: What is the mean percent of nitrogen in ancient air? Plan: We will estimate 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 unrea- sonably 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 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 be- tween 55.71% and 63.47%. 4 9 5 1 5 5 4 5 5 6 6 33 6 445 17.8. (a) df = 14. (b) t = 1 . 82 is between 1.761 and 2.145, for which the one-sided P-values are 0.05 and 0.025, respectively. (Software reports that P = . 0451.) (c) t = 1 . 82 is significant at = . 05 but not at = . 01. 17.9. (a) df = 24. (b) t = 1 . 12 is between 1.059 and 1.318, so 0 . 20 < P < . 30. (Software reports that P = . 2738.) (c) t = 1 . 12 is not significant at either = . 10 or = . 05. 17.10. State: Is there evidence that the percent of nitrogen in ancient air was different from the present 78.1%? Plan: We test H : = 78 . 1% vs. H a : 6= 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 17.7. 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, 195 196 Chapter 17 Inference about a Population Mean this is beyond anything shown in Table C, so P < . 001 (software gives P = . 00002). Conclude: We have very strong evidence ( P < . 001) that Cretaceous air contained less nitrogen than modern air....
View Full Document

This note was uploaded on 11/23/2010 for the course STAT 2325151 taught by Professor T during the Spring '10 term at Waters College.

Page1 / 9

ch17 - Chapter 17 Solutions 17.1. s / n = 21 . 88 / 20 . =...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online