The sample-size cost of 99% versus 95% confidence

Question

You want to estimate a proportion to within $\pm 2$ percentage points, worst case over the true $p$ .

How many observations are needed at 95% confidence, and how many at 99%? Use $z_{95} = 1.96$ and $z_{99} = 2.576$ .

Accepted Answer

The worst-case sample size is n = z^2 \cdot 0.25 / E^2 (using p(1-p) = 0.25 at p = 	frac12), with E = 0.02. 95% confidence (z = 1.96): n = \frac{1.96^2 	imes 0.25}{0.02^2} = \frac{3.8416 	imes 0.25}{0.0004} = \frac{0.9604}{0.0004} = 2{,}401. 99% confidence (z = 2.576): n = \frac{2.576^2 	imes 0.25}{0.02^2} = \frac{6.6358 	imes 0.25}{0.0004} = \frac{1.6589}{0.0004} \approx 4{,}148. So going from 95% to 99% confidence, at the same \pm 2pp margin, jumps the requirement from about 2{,}400 to about 4{,}150, a factor of (2.576/1.96)^2 \approx 1.73, roughly 73\% more data. Takeaway. Sample size…

The sample-size cost of 99% versus 95% confidence

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