Question 749377
A local retail company wants to estimate the mean amount spent. Their budget limits the number of surveys to 292. What is their maximum error of the estimated mean amount spent for a 99% level of confidence and an estimated standard deviation of $9? 
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ME = z*s/sqrt(n)
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ME = 2.3263*9/sqrt(292) = 1.2253
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Cheers,
Stan H.
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