How would you compute a 95% confidence interval for a proportion?

• A. Use a binomial proportion CI: p ± z*sqrt[p(1-p)/n].
• B. Use a t-distribution around the mean.
• C. Use a chi-square to compute the variance of the proportion.
• D. CI cannot be computed for a proportion.

Answer: (A)

The key idea is constructing a confidence interval for a population proportion using the normal approximation to the binomial (the Wald interval). You estimate the proportion with p̂ = x/n. The standard error of p̂ is sqrt[p̂(1 − p̂)/n], so a 95% interval is p̂ ± z0.975 × sqrt[p̂(1 − p̂)/n], where z0.975 is about 1.96. This directly matches the formula p ± z × sqrt[p(1−p)/n] when p is interpreted as the estimated proportion p̂.

Why this works: for large samples the distribution of p̂ is approximately normal, so the center is p̂ and the spread is governed by the standard error above. The 95% cutoff uses the 1.96 multiplier from the standard normal to capture 95% of the distribution.

The other approaches aren’t appropriate here: the t-distribution is used for means with unknown variance, not for a binomial proportion; a chi-square is used for variance-related inferences in different contexts; and you can compute a CI for a proportion (so saying it cannot be done is false).

Note that for small samples or extreme proportions, the standard Wald interval can be inaccurate, and alternatives like the Wilson score interval or exact (Clopper–Pearson) interval may be preferred.