A rare event fires 2 times in 200 days. Give the 95% Wald interval for its daily rate, explain why the actual coverage of that interval is far below 95% here, and compute the Wilson interval as the replacement.

A rare event fires 2 times in 200 days. Give the 95% Wald interval for its daily rate, explain why the actual coverage of that interval is far below 95% here, and compute the Wilson interval as the replacement.

Approach: Compute the normal-approximation interval, then invert the score test instead of the Wald test and compare the two on the same data.

The Wald interval is 0 to 0.024 and the Wilson interval is 0.0027 to 0.0357. The Wald interval is p_hat plus or minus 1.96*sqrt(p_hat*(1 - p_hat)/n) = 0.01 plus or minus 0.0138, so its lower end is negative and gets truncated at 0. It fails for two reasons at once: it centres on p_hat, and it plugs p_hat into its own standard error, so an unusually low count produces an unusually narrow interval, and the actual coverage of that normal approximation at p near 0.01 with n = 200 is well under 90%. The Wilson interval inverts the score test, solving |p_hat - p| = 1.96*sqrt(p*(1 - p)/n) for p, which uses the null value inside the standard error. Its centre is shifted to (p_hat + z^2/(2n))/(1 + z^2/n) = 0.0192 and it never leaves the unit interval, giving 0.0027 to 0.0357 here, close to the exact Clopper-Pearson result.

Follow-up: With zero events in 200 days, what is the upper end of a one-sided 95% interval for the rate, and what rule of thumb does it produce?

Key concepts: Wald interval, Wilson interval, coverage, normal approximation.