Sample size for mean call duration at 99%

Question

A call center wants to estimate the mean call duration to within a margin of error of $E = 15$ seconds, with 99% confidence. The standard deviation of call length is $\sigma = 90$ seconds.

What is the minimum number of calls to sample?

Accepted Answer

Setup. Use n = (z^{}\sigma/E)^{2}. For 99% confidence, z^{} = 2.576. Compute. n = \left(\frac{2.576 	imes 90}{15}ight)^{2} = \left(\frac{231.84}{15}ight)^{2} = (15.456)^{2} = 238.89. Round up: n = 239. Interpret. Demanding 99% instead of 95% confidence raises the critical value from 1.96 to 2.576, inflating the required sample. Takeaway. Higher confidence enters n through z^{} squared, so pushing from 95% to 99% costs roughly a (2.576/1.96)^{2} \approx 1.73 times larger sample.

Sample size for mean call duration at 99%

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