<PRE><B>Question 1108906</B>
Scores on a certain test are normally distributed with a variance of 22. A researcher wishes to estimate the mean score achieved by all adults on the test. 
Find the sample size needed to assure with 95% confidence that the sample mean will not differ from the population mean by more than 2 units. 
--------
Since ME = z*s/sqrt(n), n = [z*s/ME]^2
---------------------------
Ans: n = [1.96*22/2]^2 = 21.56 ~ 22 when rounded up
------------
Cheers,
Stan H.
-----------</PRE>