Estimating Sample Sizes for Proportions: Finite Population Correction

When the population size $N$ is finite, the sample size required to estimate a proportion $p$ at confidence level $1-\alpha$ with margin of error $E$ is

[ n = \dfrac{n_0}{1 + \dfrac{n_0 - 1}{N}}, ]

where $n_0$ is the unadjusted sample size for an infinite population:

[ n_0 = \dfrac{z_{\alpha/2}^{,2},\hat{p}(1-\hat{p})}{E^2}. ]

Here $z_{\alpha/2}$ is the critical value for the desired confidence level, and $\hat{p}$ is a planning estimate of the proportion (from a pilot study or prior information). We always round $n$ up to the next integer, since sample sizes must be whole numbers.

Quick example: Suppose we want a $95\%$ confidence interval ($z_{0.025}=1.96$) with margin of error $E=0.05,$ planning estimate $\hat{p}=0.2,$ and population $N=500.$ Then

$$\begin{align*} n_0 &= \dfrac{(1.96)^2(0.2)(0.8)}{(0.05)^2} = \dfrac{0.6147}{0.0025} \approx 245.86, \\[5pt] n &= \dfrac{245.86}{1 + \dfrac{244.86}{500}} = \dfrac{245.86}{1.4897} \approx 165.04. \end{align*}$$

Rounding up, $n = 166.$