= Wilson score interval
{c}
{title2=$\frac{\widehat p+z^2/(2n)\pm z\sqrt{\widehat p(1-\widehat p)/n+z^2/(4n^2)}}{1+z^2/n}$}
= Wilson interval
{c}
{synonym}
For $K\sim\operatorname{Bin}(n,p)$, $n>0$, put $\widehat p=K/n$ and let $z$ be the $1-\alpha/2$ <quantile> of the <standard normal distribution>. Inverting the approximate <score test> retains precisely the parameters satisfying
$$
\frac{n(\widehat p-p)^2}{p(1-p)}\le z^2.
$$
Multiplication and collection of terms gives the <quadratic inequality>
$$
(n+z^2)p^2-(2n\widehat p+z^2)p+n\widehat p^2\le0.
$$
Its two roots are the displayed endpoints. The <Wilson score interval> stays within $[0,1]$ and remains nontrivial when $K=0$ or $K=n$, unlike the unmodified <Wald confidence interval>. Its <coverage probability> is approximate and varies with $p$; it is not a uniformly exact 95% guarantee. Separate pointwise intervals for several groups are not a simultaneous confidence statement. The method is named after <Edwin Bidwell Wilson>.<Binomial Proportion Confidence Intervals by NIST>[§7.2.4.1: inversion of the score statistic and endpoint formulas]{ref}
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