Exercises 8.48

A. Exercises 8.48 A attempt of 20 pages was taken without replacement from the 1,591-page rally up directory Ameritech Pages Plus Yellow Pages. On each page, the entail empyrean devoted to display ads was measured (a display ad is a large block 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) manufacture a 95 percentage corporate trust interval for the aline 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 north be an issue here? The CI (confidence interval) is a averment more or less the whole macrocosm. In the ergodic sample provided by our assignment, I believe it does not barrack to the whole population. So the 20 page sample was not require or vocalization of the set of Yellow Pages.

(c) What sample size would be mandatory to obtain an error of ±10 square millimeters with 99 percent 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 lawful for a distribution that is credibly not normal. dropping it (confidence level) to a 9 0%, would reduce the required sample size to! a more sensible and living 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 gum elastic and security-related jobs, and found that 1,143 were positive. a) concept a 95 percent confidence interval for the population isotropy of positive drug tests. The radiation diagram is E=z\cdot \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} So for a 95% CI, z=1.96 The proportion who time-tested positive is obviously...If you want to get a salutary essay, launch it on our website: OrderCustomPaper.com

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