Interval for a proportion
Section of Confidence intervals, part of Statistical inference
Interval for a proportion
| Type | Section |
| Taught in | Confidence intervals |
| Part | Statistical inference |
Let X\sim\operatorname{Bin}(n,p) and \widehat p=X/n. An exact interval is awkward because the binomial law is discrete. For large n the central limit theorem gives
\frac{\widehat p-p}{\sqrt{p(1-p)/n}}\approx\operatorname{N}(0,1),
so \mathbb{P}(-z_{1-\alpha/2}<(\widehat p-p)/\sqrt{p(1-p)/n}<z_{1-\alpha/2})\approx1-\alpha. Solving for p leaves p inside the standard error. For large n, \widehat p is close to p, and replacing the unknown standard error by \sqrt{\widehat p(1-\widehat p)/n} produces an interval whose ends depend only on the data.
Definition 1 (Wald interval) For X\sim\operatorname{Bin}(n,p) with \widehat p=X/n, the Wald interval with approximate confidence 1-\alpha is
\widehat p\pm z_{1-\alpha/2}\sqrt{\frac{\widehat p(1-\widehat p)}{n}}.
It is an approximation. It behaves badly when n is small or \widehat p is near 0 or 1, because the estimated standard error is then a poor stand-in and the normal shape has not set in. At \widehat p=0 it collapses to a single point. A common rule is to use it only when n\widehat p and n(1-\widehat p) are both at least 5. It also models the sample as independent draws: a poll with refusals or untruthful answers has errors no interval formula can see.
For two independent samples the same approximation applies to \widehat p_1-\widehat p_2, whose variance is the sum of the two variances. The interval is
\widehat p_1-\widehat p_2\pm z_{1-\alpha/2}\sqrt{\frac{\widehat p_1(1-\widehat p_1)}{n_1}+\frac{\widehat p_2(1-\widehat p_2)}{n_2}}.
Example 1 (Nightmares in two groups) Of 160 men, 55 report frequent nightmares; of 192 women, 60 do.
Question. Give an approximate 95% confidence interval for the difference p_1-p_2 in the population proportions, and interpret it.
Solution. \widehat p_1=55/160=0.344 and \widehat p_2=60/192=0.313, a difference of 0.031. The standard error is
\sqrt{\frac{0.344\times0.656}{160}+\frac{0.313\times0.688}{192}}=0.0503,
so the interval is 0.031\pm1.96\times0.0503=(-0.067,\ 0.130). It contains 0. The data are consistent with no difference between men and women, and also with a difference of several percentage points either way.
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See also
Confidence interval · Pivot · One-sample t pivot · Pooled variance · Two-sample t pivot · Variance-ratio F pivot · Wald interval · Coverage