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?
A) n = 239
B) n = 2149
C) n = 139
D) n = 1245

Question

Quizplus · Accepted Answer

To determine the sample size needed for estimating the mean with a specific confidence level and margin of error, the formula $n = \left(\frac{Z*\sigma}{E}\right)^2$ is used, where $Z$ is the Z-score corresponding to the confidence level, $\sigma$ is the standard deviation, and $E$ is the margin of error. Given a variance of 9, the standard deviation $\sigma$ is 3 (since $\sigma^2 = 9$). For a 99% confidence level, the Z-score is approximately 2.576. Plugging these values into the formula with $E = 0.5$ gives $n = \left(\frac{2.576*3}{0.5}\right)^2 \approx 239$.

Past Experience Indicates That the Variance in the Time It