Sample size for a rare-event proportion at 90%

Question

You want to estimate a proportion to within a margin of error of $E = 0.03$ , with 90% confidence. A prior study suggests the rate is near $p = 0.1$ .

Using the prior estimate, what is the minimum sample size?

Accepted Answer

Setup. Use n = (z^{})^{2}\,p(1-p)/E^{2}. For 90% confidence, z^{} = 1.645, and with p = 0.1, p(1-p) = 0.09. Compute. n = \frac{(1.645)^{2}(0.09)}{(0.03)^{2}} = \frac{2.706025 	imes 0.09}{0.0009} = \frac{0.24354}{0.0009} = 270.6. Round up: n = 271. Interpret. Two factors keep this sample small: the modest 90% confidence and the low rate p = 0.1, which makes p(1-p) far below the worst-case 0.25. Takeaway. For rare events with a trusted prior, the required sample can be much smaller than the conservative estimate. Just make sure the assumed rate is defensible before relying on it.

Sample size for a rare-event proportion at 90%

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