Originally Posted by

**Aala** 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.

Add all numbers together: Mean 6943 sq mm

Divide total by 20: Standard deviation 347.15

(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.) DisplayAds

8.64 Biting an unpopped kernel of popcorn hurts! As an experiment, a self-confessed connoisseur of

cheap popcorn carefully counted 773 kernels and put them in a popper. After popping, the unpopped

kernels were counted. There were 86. (a) Construct a 90 percent confidence interval for the proportion

of all kernels that would not pop. (b) Check the normality assumption. (c) Try the Very Quick

Rule. Does it work well here? Why, or why not? (d) Why might this sample not be typical?