Sample size for a proportion at 99% confidence

Question

You want to estimate a proportion to within a margin of error of E = 0.04, with 99% confidence, and you have no prior estimate. What is the minimum sample size?

Accepted Answer

Setup. Use n = (z^{})^{2}\,p(1-p)/E^{2} with the conservative p = 0.5. For 99% confidence, z^{} = 2.576. Compute. n = \frac{(2.576)^{2}(0.25)}{(0.04)^{2}} = \frac{6.635776 	imes 0.25}{0.0016} = \frac{1.658944}{0.0016} = 1036.8. Round up: n = 1037. Interpret. The higher 99% confidence enters through (z^{})^{2}, so it inflates the sample relative to a 95% design with the same margin. Takeaway. Both the confidence level and the margin enter the proportion sample-size formula through squares, so raising confidence or tightening the margin both raise the count sharply.

Sample size for a proportion at 99% confidence

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