Standard error of a proportion

The standard error of a proportion, $SE_{\hat{p}} = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$, estimates the standard deviation of the sampling distribution of $\hat{p}$ (p-hat, the sample proportion). Read it as the typical distance between a sample proportion and the true $p$ across repeated random samples of size $n$. Unlike a mean, a proportion needs no separate measure of spread: a success-or-failure variable has its variance determined by its own mean, so $\hat{p}$ and $n$ are the only inputs.

With $\hat{p} = 0.40$ from a sample of 400, $SE_{\hat{p}} = \sqrt{0.40(0.60)/400} = \sqrt{0.0006} = 0.024495$. Take 1600 instead and it halves to 0.012247, because the sample size enters through $\sqrt{n}$.

The misreading is "the standard error is 0.0245, so the true proportion is within 0.0245 of 0.40." One standard error each side is roughly the middle 68% of the normal model, not the middle 95%. The 95% reach is $1.96 \times 0.024495 = 0.0480$, giving 0.352 to 0.448. The standard error is the unit of distance; a margin of error is a critical value's worth of those units, and quoting the first as if it were the second shrinks the stated reach by a factor of 1.96.

Because $\hat{p}(1-\hat{p})$ peaks at 0.5, the standard error is largest there and falls off slowly on either side: at $n = 400$ it is 0.025000 at $\hat{p} = 0.50$, 0.022913 at 0.30, and 0.015000 at 0.10. It reaches exactly 0 only when $\hat{p}$ is 0 or 1, where the formula claims perfect certainty from a sample that has simply never seen the other outcome. That collapse is one reason the interval asks for at least 10 successes and 10 failures. The formula also assumes the draws are independent, which is why the sample should stay under 10% of the population.

The sampling distribution this quantity estimates is Unit 3, topic 3.2, Sampling Distributions for Sample Proportions, and every proportion interval and test on the course is built on it.