Question 134678: 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. (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.
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! 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
E = 1.96*170.3784/sqrt(20)= 0.0514
CI: 346.5-0.0514 < mu < 346.5+0.0514
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(b) Why might normality be an issue here?
The population goes far beyond one phone directory.
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(c) What sample size would be needed to obtain an error of ±10 square millimeters with 99 percent confidence?
E = z*sigma/sqrt(n)
n = [z*sigma/E]^2
n = [1.96*170.3784/10]^2 = 1115.17
n rounded up = 1116
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(d) If this is not a reasonable requirement, suggest one that is.
I'll leave that to you.
Cheers,
Stan H.
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