Finite Population Corrections for Sample Proportions

When a sample of size $n$ is drawn without replacement from a finite population of size $N,$ the standard error of the sample proportion $\hat{p}$ shrinks compared to sampling with replacement. The adjustment is made by multiplying by the finite population correction (FPC) factor — the same factor used for sample means:

$$SE(\hat{p}) = \sqrt{\dfrac{p(1-p)}{n}} \cdot \underbrace{\sqrt{\dfrac{N-n}{N-1}}}_{\textrm{FPC}}$$

Here $p$ is the population proportion, $N$ is the population size, and $n$ is the sample size.

For example, with $N=400,$ $n=100,$ and $p=0.5{:}$

$$\begin{align*} SE(\hat{p}) &= \sqrt{\dfrac{(0.5)(0.5)}{100}}\cdot\sqrt{\dfrac{400-100}{400-1}}\\[4pt] &= \sqrt{0.0025}\cdot\sqrt{\dfrac{300}{399}}\\[4pt] &= 0.0500 \cdot 0.8671\\[4pt] &\approx 0.0434. \end{align*}$$

Without the FPC, the standard error would be $0.0500$ — the FPC shrinks it by about $13\%.$