Exercises 8.48

A. Exercises 8.48 A exemplar of 20 pages was taken without replacement from the 1,591-page impose up directory Ameritech Pages Plus Yellow Pages. On each page, the tight 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) hit a 95 percentage say-so interval for the full-strength 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 atomic number 7 be an issue here? The CI (confidence interval) is a report more or less the whole universe. In the stochastic sample provided by our assignment, I believe it does not touch to the whole population. So the 20 page sample was not needful 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 probably not normal. dropping it (confidence level) to a 90%, would reduce the required sample size to a more sensible and veridical 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) render a 95 percent confidence interval for the population isotropy of posi tive 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 well-tried positive is obviously...If you want to get a salutary essay, frame it on our website: OrderCustomPaper.com

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