Finding confidence interval for a binomial process using the normal distribution?

Question

See, when I was taught how to find confidence intervals, I always needed the sample variance to use a Student $t$ distribution to form the confidence interval. How does this work in the binomial case (when using the normal distribution to approximate) though? Do we just use $np(1-p)$ as our sample variance?

Answer 1

For reasonably large sample sizes, and proportions not too close to $0$ or $1$, you can use $\sqrt{n\hat{p}(1-\hat{p})}$, where $\hat{p}$ is the numerical sample proportion.

to push this somewhat lower, let $p$ be the unknown probability. Then if $X$ is the number of successses, we can say for example that with probability $0.95$, $X/n$ will lie between $p-1.96\frac{\sqrt{np(1-p)}}{\sqrt{n}}$ and $p+1.96\frac{\sqrt{np(1-p)}}{\sqrt{n}}$. This gives you (after some processing, or looking up the formula) a couple of quadratic inequalities that you can solve to get your bounds. Somewhat unpleasant, so one often in the old days crossed one's fingers and used the more simple-minded large sample approach.

Nowadays, there is software that will do the computations "exactly."