Large Counts condition

Large Counts is the shape check for proportion procedures: at least 10 expected successes and at least 10 expected failures before you may treat the sampling distribution of $\hat{p}$ (p-hat, the sample proportion) as normal. It exists because $\hat{p}$ is a binomial count divided by $n$, and a binomial is symmetric only when $p = 0.5$, piling up against a wall as $p$ slides toward 0 or 1. Large Counts is the textbook name; the AP course calls the same check the normality condition.

Which value of $p$ goes in depends on the procedure. A significance test has a hypothesized value, so verify $np_0 \ge 10$ and $n(1-p_0) \ge 10$ using $p_0$ (p-naught). A confidence interval assumes nothing about $p$, so verify $n\hat{p} \ge 10$ and $n(1-\hat{p}) \ge 10$, which is just counting the successes and failures you observed. Testing $p_0 = 0.25$ with $n = 60$ passes, since $np_0 = 15$ and $n(1-p_0) = 45$. If those same 60 observations contained only 9 successes, the interval version fails on $n\hat{p} = 9$. One data set, two procedures, two different answers, and that is not a contradiction.

The misreading is "$n = 60$ is more than 30, so the normal model is fine." The 30 belongs to sample means and never appears in a proportion problem. How large $n$ must be depends entirely on $p$: at $p = 0.05$ the condition demands $n \ge 200$, which is why the rule counts outcomes instead of counting observations.

Passing is not the same as exact. With $n = 50$ and $p = 0.60$ both counts clear the bar comfortably, yet the exact coverage of the nominal 95 percent z-interval is 94.1 percent. Break the condition at $n = 20$ and $p = 0.90$, where $n(1-p) = 2$, and coverage falls to 87.6 percent. The coverage simulator reports the exact coverage at any setting. The bar of 10 is a convention rather than a theorem, and some textbooks use 5 or 15, but AP standardized on 10.