Question 127096:  Please help me.  I need desperate help.  Thank you
 
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! 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 
---------------- 
E=t= 2.093[170.38/sqrt(20)]= 0.0549 
----------- 
95% C.I. is (346.5-0.0549,346.5+0.0549) 
---------------- 
(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. 
----------------------- 
c) What sample size would be needed to obtain an error of ±10 square millimeters 
with 99 percent confidence?  
Formula: E = t 
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. 
 
 
 
 
 
 
 
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