individual exerc - 8.48 A sample of 20 pages was taken...

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8.48 A sample of 20 pages was taken without replacement from the 1,591-page phone directory Ameritech Pages Plus Yellow Pages. On each page, the mean area devoted to display ads was measured (a display ad is a large block of multicolored illustrations, maps, and text). The data (in square 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) Construct a 95 percent confidence interval for the true mean. (b) Why might normality be an issue here? (c) What sample size would be needed to obtain an error of . }10 square millimeters with 99 percent confidence? (d) If this is not a reasonable requirement, suggest one that is. (Data are from a project by MBA student Daniel R. Dalach.) ANS: 0,260,356,403,536,0,268,369,428,536,268,396,469,536,162,338,403,536,536,130 First, let's use the data to get the mean and standard deviation: mean = 346.5 standard dev = 170.378 Since our sample is small and we do not know the population stdev, we'll need to use a t value, with df = 19. t (95% interval) = 2.0930 z (99% interval) for the later part = 2.5758 The confidence interval goes from: mean - t*sdev/sqrt(N) to mean + t*sdev/sqrt(N) 346.5 - 2.0930*170.378/sqrt(20) to 346.5 + 2.0930*170.378/sqrt(20)
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