Coverage probability

Fix a method, a sample size, and a true parameter value, and the coverage probability becomes a computable number: the probability that the interval the method is about to produce contains that value. It is the rate the method achieves. The confidence level is the rate it advertises. When the two agree the method is exact, and when the first falls below the second the method is undercovering.

For the one-proportion interval $\hat{p} \pm z^*\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$ you can compute it exactly, because a sample of $n = 50$ has only 51 possible outcomes. Build the 95% interval from each count $x$ and ask whether it contains the true $p$. With $p = 0.60$ the counts 24 through 36 give intervals that capture it and the other 38 miss. Adding the binomial probabilities of those thirteen counts gives 0.9406, so the coverage probability is 94.1%, not 95%.

The claim to be careful with is that a 95% method covers 95% of the time. The 95% is an input you selected. The coverage probability is an output you have to compute, and for this interval the two rarely match. The gap is not always small: at $n = 20$ with $p = 0.90$ the same 95% formula captures $p$ only 87.6% of the time.

More data does not repair it smoothly. Because the count $x$ is discrete, coverage oscillates instead of climbing. At $p = 0.50$ the 95% interval covers 95.7% of the time when $n = 30$ and 91.9% when $n = 40$: ten more observations made it worse. The long trend is toward the nominal level, but no single increase in $n$ is guaranteed to improve on the one before it.

This is what the large counts condition is for. Requiring at least 10 successes and 10 failures rules out the collapse cases, where $\hat{p}$ lands so near 0 or 1 that the standard error vanishes. It does not make coverage exact and it does not promise a small shortfall: $n = 40$ with $p = 0.50$ clears the condition twice over and still covers 91.9%.