Use the distribution of sample proportions to construct and interpret confidence intervals for a population proportion.

What this dot point is asking

When you cannot survey a whole population, you estimate a proportion from a sample. The sample proportion $\hat{p} = \dfrac{\text{number of successes}}{n}$ is a point estimate of the true population proportion $p$. Because a different sample would give a slightly different $\hat{p}$, we describe the variability and report an interval.

Constructing a confidence interval

A confidence interval takes the point estimate and adds and subtracts a margin of error.

Since $p$ is unknown, we use $\hat{p}$ inside the standard error, which is why the formula contains $\hat{p}(1-\hat{p})$ rather than $p(1-p)$.

The trade-offs

Higher confidence uses a larger $z$, which widens the interval. A larger sample size $n$ shrinks the standard error and narrows the interval. So precision (a narrow interval) and confidence pull in opposite directions, and you buy precision by increasing $n$.

Choosing a sample size for a target margin

A common design question reverses the formula: given a required margin of error $E$, find the sample size $n$ needed. Rearranging $E = z\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}$ gives

When no prior estimate of $\hat{p}$ is available, use $\hat{p} = 0.5$, because $\hat{p}(1-\hat{p})$ is largest there and so gives the safest (largest) sample size. Always round $n$ up to the next whole number, since a fractional person cannot be surveyed and rounding down would leave the margin slightly too wide. For example, a $95\%$ interval with margin $0.03$ and $\hat{p} = 0.5$ needs $n = \dfrac{1.96^{2}\times 0.25}{0.03^{2}} \approx 1067.1$, so survey $1068$ people.

Finding the confidence level from a given interval

Some questions hand you a completed interval and ask which confidence level produced it. Take half the interval width as the margin $E$, divide by the standard error to recover $z$, then identify the confidence level from the standard $z$-values ($1.645$ for $90\%$, $1.96$ for $95\%$, $2.576$ for $99\%$). This reverses the construction and is a favourite TASC twist on the standard interval question.

Summary

Compute $\hat{p}$, then the standard error $\sqrt{\hat{p}(1-\hat{p})/n}$, then multiply by the $z$ for your confidence level to get the margin of error, and add and subtract it from $\hat{p}$. Remember that increasing confidence widens the interval while increasing the sample size narrows it, and always interpret the interval as a statement about the long-run reliability of the method.