Confidence Intervals for One Proportion

A confidence interval for a population proportion $p$ is built around the point estimate $\hat{p}=x/n,$ where $x$ is the number of successes in a sample of size $n.$ When $n$ is large, $\hat{p}$ is approximately normal with mean $p$ and standard deviation $\sqrt{p(1-p)/n}.$ Since $p$ is unknown, we estimate the standard deviation using $\hat{p}$ itself, giving the standard error

$$SE = \sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}.$$

The $(1-\alpha)$ confidence interval for $p$ is

$$\hat{p} \;\pm\; z_{\alpha/2}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}},$$

where $z_{\alpha/2}$ is the standard normal critical value. The most common choices are:

$$\begin{array}{c|c} \text{Confidence level} & z_{\alpha/2} \\ \hline 90\% & 1.645 \\ 95\% & 1.960 \\ 99\% & 2.576 \end{array}$$

For example, suppose a survey of $n=100$ voters finds that $x=64$ support a candidate. Then $\hat{p}=0.64,$ and

$$SE = \sqrt{\dfrac{0.64\cdot 0.36}{100}} = \sqrt{0.002304} \approx 0.048.$$

The 95% confidence interval is

$$0.64 \pm 1.96\cdot 0.048 \;\approx\; 0.64 \pm 0.094 \;=\; (0.546,\ 0.734).$$