Chapter 8 Exercises 8.48 and 8.64

Chapter 8 Exercises 8.48 and 8.64

8.48) A sample of 20 pages was taken without replacement from the 1,591-page phone directoryAmeritech 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 (insquare 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.)

A)

x-bar = 346.5

s=170.378

t-critical value for 95% CI with df =19= 2.093

E= 2.093 * 170.378/sqrt (20)=2.0931.96 * 38.0976 = 79.74

95% CI: (346.5 – 79.74, 346.5 + 79.74)

B) A CI statement is about the whole population. This random sample is probably not the whole population.

C)

n = (t * s / E)

n = (2.093*170.378/10)^2 = 1271.64

n = 1272

D) Increase E or decrease the confidence level

8.48) 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?

A)

E = z * (p-hat/sqrt(n))

E = 1.645 * sqrt[(86/773)(687/773)/773) = 0.01131

90% C.I. = ((86/773)-0.01131 , (86/773)+0.01131)

B)

pn= (86/773)773>5; nq=(687/773)773>5

C)

D) Since the C.I. is a statement about the whole population of popcorn

kernnels, the sample here might not be representative.