Sampling distribution of the sample proportion

Fix a population proportion $p$ and a sample size $n$. Each sample yields a count of successes $X$ and a sample proportion $\hat{p} = X/n$ (p-hat), and the collection of those values across all samples is this sampling distribution. Its center is $\mu_{\hat{p}} = p$ and its spread is $\sigma_{\hat{p}} = \sqrt{\frac{p(1-p)}{n}}$. When the draws are independent, $X$ is binomial, so $\hat{p}$ is a binomial count divided by $n$ and its shape is only ever approximately normal.

With $p = 0.6$ and $n = 150$, the spread is $\sqrt{\frac{0.6(0.4)}{150}} = \sqrt{0.0016} = 0.04$, so a sample proportion of 0.68 sits two standard deviations above center. The Large Counts condition passes here, since $np = 90$ and $n(1-p) = 60$. Shrink to $n = 10$ and $p = 0.3$ and the same formulas still give the center and spread correctly, 0.3 and 0.1449, but $\hat{p}$ can only take one of the eleven values 0, 0.1, and so on up to 1.0, and $np = 3$ fails the check. Center and spread hold at every $n$. The normal shape does not.

Now the misreading: "a bigger population needs a bigger sample." There is no $N$ in the formula. A random sample of 1,000 pins down a proportion just as precisely in a town of 15,000 as in a country of 300 million, which is why national polls report a margin of error near 3 points on samples that size: $1.96\sqrt{0.25/1000} = 0.031$.

The formula uses the parameter $p$, which you usually do not know. Put a sample value in its place and the result is called the standard error. It also assumes independent draws, which sampling without replacement only approximates, so the 10% condition comes with it. Topic 3.2 is Sampling Distributions for Sample Proportions.