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

What this dot point is asking

SCSA Unit 4 closes the course by linking the binomial and normal distributions to statistical inference. From a single random sample you estimate an unknown population proportion $p$ and quantify the uncertainty using a confidence interval. This dot point appears in the WACE written examination and rewards both correct calculation and precise interpretation.

The sampling distribution of the sample proportion

If $X$ is the number of successes in $n$ independent trials with success probability $p$, then $X \sim B(n, p)$ and the sample proportion is $\hat{p} = \dfrac{X}{n}$.

Because $\hat{p}$ is unbiased ($E(\hat{p}) = p$), it is the natural point estimate of $p$. The standard error shrinks as $n$ grows, so larger samples give more precise estimates.

Constructing the confidence interval

The true value of $p$ is unknown, so the standard error is estimated by replacing $p$ with $\hat{p}$:

The quantity $z \cdot \mathrm{SE}$ is the margin of error. A higher confidence level uses a larger $z$, which widens the interval; a larger sample size $n$ narrows it.

Interpreting the interval

A correct interpretation refers to the method, not to a single interval containing $p$ with a stated probability.

• Correct: we are $95\%$ confident that the population proportion lies between the endpoints, meaning that if many such samples were taken, about $95\%$ of the resulting intervals would contain the true $p$.
• The margin of error halves only when the sample size is roughly quadrupled, because the standard error has $\sqrt{n}$ in the denominator.

For example, in the poll above, we are $95\%$ confident the true support level is between about $55.2\%$ and $64.8\%$.