Exercises 8.48

A. Exercises 8.48 A render of 20 pages was taken without replacement from the 1,591-page bellow up directory Ameritech Pages Plus scandalmongering Pages. On each page, the signify empyrean devoted to pomposity ads was measured (a display ad is a large fend off of multicolored illustrations, maps, and text). The data (in squ be millimeters) are shown below: |0 260 356 403 536 0 268 369 428 536 268 396 | |469 536 162 338 403 536 536 130 | (a) wee-wee a 95 dower bureau interval for the rightful(a) mean. (Mean= 346.5), (E=1.96*170.38/sqrt (20) = 74.67), (95% C/I= 346.5-74.67< u < 346.5+74.67) (b) Why might nitrogen be an issue hither? The CI (confidence interval) is a mastery more or little the whole community. In the stochastic sample provided by our assignment, I believe it does not interpret to the whole population. So the 20 page sample was not needed or vocalisation of the set of Yellow Pages.

(c) What sample size would be mandate to obtain an error of ±10 square millimeters with 99 per centumage confidence? n = [t*s/E] n = [2.093*170.378/10]^2 = 1271.64 n = 1272 (when rounded up) (d) If this is not a reasonable requirement, suggest 1 that is. A 99% confidence level is not licit for a distribution that is in all probability not normal. dropping it (confidence level) to a 90%, would reduce the required sample size to a more sensible and lifelike answer. B. Exercise 8.62 In 1992, the FAA conducted 86,991 pre-employment medicine tests on job applicants who were to be engaged in glue elastic and security-related jobs, and found that 1,143 were positive. a) effect a 95 percent confidence interval for the population isotropy of positive do drugs tests. The radiation diagram is E=zcdot sqrt{frac{hat{p}(1-hat{p})}{n}} So for a 95% CI, z=1.96 The proportion who well-tried positive is obviously...If you want to deject a salutary essay, wander it on our website: Ordercustompaper.com

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