SOLUTION: 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 me

Question 124963: 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(48568) (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.
I used the T-Interval function on a TI calculator to get
(237.50,455.50) as the 99% C.I. for the population mean.
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b) Why might normality be an issue here? The method for selecting
the newspapers in the sample may not have been random.
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(c) What sample size would be needed to obtain an error of ±10 square millimeters with 99 percent confidence?
E = t*sigma/sqrt(n)
n = [t*sigma/E]^2
n = [2.575*170.378/10]^2
n = 43.87^2
n = 1925 when rounded up.
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(d) If this is not a reasonable requirement, suggest one that is.
Reduce the confidence requirement or enlarge the error limit.
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Cheers,
Stan H.