# Clopper-Pearson for non mathematicians

I was wondering if anyone can explain to me the intuition beyond the Clopper-Pearson CI for proportions.

As far as I know, every CI includes a variance in it. However, for proportions, even if my proportion is 0 or 1 (0% or 100%), the Clopper-Pearson CI can be calculated. I tried looking at the formulas, and I understand it has something with percentiles of the Binomial distribution and I understand that finding the CI involves iterations, but I wondered if anyone can explain the logic and rational in “simple words”, or with minimum math ?

You can get a Clopper–Pearson 95% (say) confidence interval for the parameter $\pi$ working directly with the binomial probability mass function. Suppose you observe $x$ successes out of $n$ trials. The p.m.f. is
Increase $\pi$ until the probability of $x$ or fewer successes falls to 2.5%: that’s your upper bound. Decrease $\pi$ until the probability of $x$ or more successes falls to to 2.5%: that’s your lower bound. (I suggest you actually try doing this if it’s not clear from reading about it.) What you’re doing here is finding the values of $\pi$ that when taken as a null hypothesis would lead to its (only just) being rejected by a two-tailed test at a significance level of 5%. In the long run, bounds calculated this way cover the true value of $\pi$, whatever it is, at least 95% of the time.