SOLUTION: Probability is not my high point. I have tried to come up with an answer, but I need assistance, please. Here is the question: A sample of 20 pages was taken without replacement

Question 148331This question is from textbook
: Probability is not my high point. I have tried to come up with an answer, but I need assistance, please. Here is the question:
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.
Thank you! This question is from textbook

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
---------
x-bar = 346.5 ; s = 170.378
(a) Construct a 95 percent confidence interval for the true mean.
I'm assuming you want a Z-Interval:
E = 1.96[170.378/sqrt(20] = 74.6715
95% CI: (x-bar-E, x-bar+E) = (346.5-74.6715 , 346.5+74.6715)

(b) Why might normality be an issue here?
These 20 samples gathered in the way described is probably not a random
sample.
----------------
(c) What sample size would be needed to obtain an error of �10 square millimeters with 99 percent confidence?
E = z
n = [2.326s/E]^2
n = [2.326*170.378/10]^2 = 1571 when rounded up.
---------------------
(d) If this is not a reasonable requirement, suggest one that is.
You would get a smaller "n" if you increase the size of E or
decrease the size of z* by requiring less confidence.
------------------
Cheers,
Stan H.

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