Wilson Confidence Interval (Proportion)

Background

The Wilson score interval is the proper confidence interval for a proportion. It corrects the canonical Wald interval's pathologies at small $n$ and near $p = 0$ or $p = 1$, where Wald can give nonsensical bounds outside $[0, 1]$.

Problem statement

Implement wilson_ci(successes, n, alpha=0.05, z=None):

$$\hat{p} = s/n,\qquad \text{centre} = \frac{\hat{p} + \frac{z^2}{2n}}{1 + \frac{z^2}{n}},\qquad \text{half-width} = \frac{z}{1 + z^2/n}\sqrt{\frac{\hat{p}(1-\hat{p})}{n} + \frac{z^2}{4n^2}}$$

Return (lo, hi) clipped to $[0, 1]$.

Input

• successes - int >= 0.
• n - int >= 1, total trials, $n \ge$ successes.
• alpha - significance (default $0.05$ → 95% CI).
• z - optional override of the z-score (default: 1.96 for alpha=0.05).

Output

• (lo, hi) - tuple of plain Python floats.

Examples

Input: successes=8, n=10, alpha=0.05
Output: roughly (0.49, 0.94)

Constraints

• ValueError on bad inputs (n < 1, successes < 0, successes > n).
• Output bounded in $[0, 1]$.

Notes

• vs Wald. Wald = $\hat p \pm z\sqrt{\hat p(1-\hat p)/n}$. At $s = 0, n = 10$ Wald gives $[0, 0]$; Wilson gives a non-degenerate upper bound.