This is my first time taking statistics, please help me with the problem step by step.
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.
2 answers
26% t(9)/h*4
Although I am not going to answer your questions, I want you to know that the area of the display ad on any page is not a "mean area."
Find the mean first = sum of scores/number of scores
Subtract each of the scores from the mean and square each difference. Find the sum of these squares. Divide that by the number of scores to get variance.
Standard deviation (SD) = square root of variance
95% conf. interval = mean ± 1.96 SD
Is there a minimum requirement for the size of the ad on any page? If so, the distribution is unlikely to be normal.
I hope this helps a little.
Also, I don't see any connection of Lolli's post with your question..
Find the mean first = sum of scores/number of scores
Subtract each of the scores from the mean and square each difference. Find the sum of these squares. Divide that by the number of scores to get variance.
Standard deviation (SD) = square root of variance
95% conf. interval = mean ± 1.96 SD
Is there a minimum requirement for the size of the ad on any page? If so, the distribution is unlikely to be normal.
I hope this helps a little.
Also, I don't see any connection of Lolli's post with your question..