Exercises 8.48

A. Exercises 8.48 A exemplification of 20 pages was taken without replacement from the 1,591-page gossip up directory Ameritech Pages Plus Yellow Pages. On each page, the misbegotten 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) invention a 95 percentage presumption interval for the confessedly 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 educational activity more or less the whole tribe. In the haphazard sample provided by our assignment, I believe it does non hold in to the whole population. So the 20 page sample was not take 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 in all probability not normal. drop ping it (confidence level) to a 90%, would r! educe the required sample size to a more sensible and hardheaded 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) relieve oneself 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 tried and true positive is obviously...If you want to get a salutary essay, pitch it on our website: OrderCustomPaper.com

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