<PRE><B>Question 461190</B>
A confidence interval estimate for the population mean is given to be 
(46.66, 54.74). 
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If the standard deviation is 11.822 and the sample size is 57, answer each of the following (show all work): 
(A) Determine the maximum error of the estimate, E.
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Fact: The width of the Confidence interval is ALWAYS 2*ME
2*ME = (54.74-46.66) = 8.08
ME = 4.04 
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(B) Determine the confidence level used for the given confidence interval
ME = zs/Sqrt(n)
4.04 = z*11.822/sqrt(57)
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z = 2.5
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Confidence level = 1 - 2*P(z &gt; 2.5) = 1 - 1.24% = 98.76%
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Cheers,
Stan H.
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