How many samples for a 1% margin of error?

Question

You want to estimate an unknown proportion $p$ (say, the fill rate of an order type) with a 95% confidence interval of half-width at most 1 percentage point.

How many independent observations do you need? Why do opinion polls typically use only about 1,000 respondents, and why doesn't population size enter?

Accepted Answer

Setup. The estimator \hat{p} = k/n has standard error \sqrt{p(1-p)/n}, and the 95% CI is approximately \hat{p} \pm 1.96\sqrt{p(1-p)/n}. Requiring half-width E: 1.96\sqrt{\frac{p(1-p)}{n}} \leq E \implies n \geq \frac{1.96^2\,p(1-p)}{E^2}. Worst case. p(1-p) is maximized at p = 	frac12, where it equals 	frac14. Using this conservative value with E = 0.01: n \geq \frac{1.96^2 	imes 0.25}{0.01^2} = \frac{0.9604}{0.0001} = 9{,}604. So about 9,600 observations guarantee a ±1pp margin regardless of the true p. If you know p is near, say, 0.05, then p(1-p) \approx 0.0475 and n \approx 1{,}825 suf…

How many samples for a 1% margin of error?

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