A. Exercises 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.

(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 normality be an issue here?

The CI (confidence interval) is a statement about the whole population. In the random sample provided by our assignment, I believe it does not correspond to the whole population. So the 20 page sample was not required or representative of the set of Yellow Pages.

(c) What sample size would be needed 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 one that is.

A 99% confidence level is not logical 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 realistic answer.

B. Exercise 8.62

In 1992, the FAA conducted 86,991 pre-employment drug tests on job applicants who were to be engaged in safety and security-related jobs, and found that 1,143 were positive.

a) Construct a 95 percent confidence interval for the population proportion of positive drug tests.

The formula is E=z\cdot \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} So for a 95% CI, z=1.96… The proportion who tested positive is obviously...