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Past experience indicates that the variance in the time it takes for a \"fast lu

ID: 3380167 • Letter: P

Question

Past experience indicates that the variance in the time it takes for a "fast lube" operation to actually complete the lube and oil change for customers is 9.00 minutes. The manager wishes to estimate the mean time with 99 percent confidence and a margin of error of ± 0.50 minutes. Given this, what must the sample size be? Past experience indicates that the variance in the time it takes for a "fast lube" operation to actually complete the lube and oil change for customers is 9.00 minutes. The manager wishes to estimate the mean time with 99 percent confidence and a margin of error of ± 0.50 minutes. Given this, what must the sample size be? Past experience indicates that the variance in the time it takes for a "fast lube" operation to actually complete the lube and oil change for customers is 9.00 minutes. The manager wishes to estimate the mean time with 99 percent confidence and a margin of error of ± 0.50 minutes. Given this, what must the sample size be?

Explanation / Answer

Compute Sample Size
n = (Z a/2 * S.D / ME ) ^2
Z/2 at 0.01% LOS is = 2.58 ( From Standard Normal Table )
Standard Deviation ( S.D) = 3
ME =0.5
n = ( 2.58*3/0.5) ^2
= (7.74/0.5 ) ^2
= 239.63 ~ 240