Confidence Intervals for One Proportion: Finite Population Corrections

When we sample with replacement, or when the population is effectively infinite, a $(1-\alpha)$ confidence interval for a population proportion $p$ is

$$\hat{p} \pm z^* \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}.$$

When we sample without replacement from a finite population of size $N$, the observations are no longer independent and the standard error above is too large. We fix this by multiplying the standard error by the finite population correction (FPC) factor

$$\sqrt{\frac{N-n}{N-1}}.$$

The corrected confidence interval becomes

$$\hat{p} \pm z^* \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \cdot \sqrt{\frac{N-n}{N-1}}.$$

For example, suppose $N = 100,$ $n = 25,$ and $\hat{p} = 0.4.$ Then

$$\begin{align*} \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} &= \sqrt{\frac{0.4 \cdot 0.6}{25}} = \sqrt{0.0096} \approx 0.0980, \\[5pt] \sqrt{\frac{N-n}{N-1}} &= \sqrt{\frac{75}{99}} \approx 0.8704, \\[5pt] \text{SE} &\approx 0.0980 \cdot 0.8704 \approx 0.0853. \end{align*}$$

Using $z^* = 1.96$ for a $95\%$ interval, the margin of error is $1.96 \cdot 0.0853 \approx 0.167,$ giving the interval $(0.233,\, 0.567).$