Standard Error of the Mean vs Standard Error of a Proportion

Across all samples of size $n$ drawn from one population, the sample mean $\bar{x}$ (x-bar) varies around the population mean $\mu$ (mu) with standard deviation $\frac{\sigma}{\sqrt{n}}$ (sigma over the square root of n), provided the observations are independent. You rarely know $\sigma$, so you estimate it from the sample and report $SE_{\bar{x}} = \frac{s}{\sqrt{n}}$. Either version describes how far a mean lands from $\mu$, never how far one observation lands from the mean.

Five commute times, in minutes: 31, 28, 35, 40, 26. The mean is 32 and the sample standard deviation is $s = 5.6125$, so $SE_{\bar{x}} = \frac{5.6125}{\sqrt{5}} = 2.5100$. Individual commutes scatter about 5.6 minutes wide. The average of five of them typically sits about 2.5 minutes from the true mean commute.

The sentence to catch is "about 68% of commutes fall within one standard error of 32 minutes." That reading is about $s$, not the standard error: if commutes were roughly normal the 68% span would be 26.4 to 37.6, not 29.5 to 34.5. Using the standard error where the sample standard deviation belongs describes the data as $\sqrt{n}$ times more consistent than it is, and it is the usual reason a student's interval comes out absurdly tight.

The square root is where intuition fails. Doubling the sample does not halve the standard error; quadrupling it does. Twenty commutes with the same $s = 5.6125$ give $SE_{\bar{x}} = 1.2550$, exactly half of 2.5100. The standard error is also silent on accuracy: if all five commutes came from the same route on the same morning, collecting more of them only tightens the estimate around the wrong center. It measures precision, and bias is invisible to it.

This is the denominator of every one-sample procedure for a mean in Unit 4 of the AP course, from the sampling distributions in topic 4.1 through the interval in 4.2 and the test in 4.5.

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.