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Chapter 8, fare 8.48
A sample of 20 pages was interpreted without replacement from the 1,591-page phone directory Ameritech Pages Plus Yellow Pages. On to each one page, the mean area devoted to ostentation ads was measured (a display ad is a large block of multicolored illustrations, maps, and text). The information (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) prepare a 95 percent confidence interval for the current mean. (b) wherefore might normality be an issue present? (c) What sample size would be needed to obtain an fallacy 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 - Mean
s=170.378 Standard deviation
T-critical care for 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)
= 426.240 upper confidence limit
= 366.760 lower confidence limit
B) A CI statement is about the whole population. This random sample is in all probability not
the whole population.




C) n = (t * s / E)
n = (2.576*170.378/10)^2 = 1926.067
n = 1927
D) Increase E or decrease the confidence level

Chapter 8, Exercise 8.64
Biting an unpopped kernel of popcorn hurts! As an experiment, a self-confessed(prenominal) connoisseur of cheap popcorn care luxurianty 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 draw well here? Why, or why not? (d) Why might this sample not be typical?
A) E = z * (p-hat/sqrt (n))
E =...If you want to get a full essay, order it on our website: Ordercustompaper.com



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