Question 127096
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.
The mean of the sample (or sample mean) is x-bar = 346.5
The sample standard deviation is s = 170.38
t* for 19 degrees of freedom on a two-tailed test with alpha = 5% is 2.093
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E=t*[s/sqrt(20)]= 2.093[170.38/sqrt(20)]= 0.0549
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95% C.I. is (346.5-0.0549,346.5+0.0549)
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(b) Why might normality be an issue here? 
The C.I. is a statement about the population. The sample of 20 might not
have been randomly selected from 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? 
Formula: E = t*[s/sqrt(n)]
Solve for "n":
n =[t*s/E]^2
n = [1.96*170.38/10]^2 = 1116 when rounded up.
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(d) If this is not a reasonable requirement, suggest one that is.
Either raise the allowed error or lower the confidence requirement.
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Cheers,
Stan H.