Question 145004: A sample of 20 pages was taken without replacement from 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 multi colored 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 tru 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 from a project by MBA student Daniel R. Dalach.)
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! 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.
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)
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(b) Why might normality be an issue here
The CI is a statement about the whole population. The random
sample is probably not representative of the whole population.
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(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)
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(d) If this is not a reasonable requirement, suggest one that is.
Increase E or decrease the confidence level; both will have the
effect of lowering "n".
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Cheers,
Stan H.
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