Topic 6.2 Constructing a Confidence Interval for a Population Proportion: identify the conditions, compute the point estimate, critical value, standard error, and margin of error, and construct and interpret a one-sample z-interval for a proportion.

What this topic is asking

The College Board (Topic 6.2) wants you to construct and interpret a one-sample z-interval for a population proportion $p$: check the conditions, compute the point estimate, the critical value $z^{*}$, the standard error, and the margin of error, then state the interval and interpret it in context.

The interval and its parts

The interval is the point estimate $\hat{p}$ extended by a margin of error in each direction. The standard error is the Topic 5.5 standard deviation $\sqrt{p(1-p)/n}$ with $\hat{p}$ substituted for the unknown $p$ (we do not know $p$, so we estimate the spread from the sample). The critical value $z^{*}$ sets how many standard errors wide the interval is, chosen so the procedure captures $p$ in $C\%$ of samples.

Checking the conditions

The large-counts check here uses $\hat{p}$, because in an interval there is no hypothesized value of $p$, only the estimate from data. (In a test, Topic 6.4, you use the claimed $p_0$ instead.) Checking conditions is not box-ticking; it is what earns the normal model and hence the $z^{*}$ multiplier.

What "confident" means

A $95\%$ confidence interval does not mean there is a $95\%$ probability that $p$ lies in this interval; $p$ is fixed, and a given interval either contains it or does not. The $95\%$ refers to the method: across many samples, $95\%$ of the intervals constructed this way capture the true $p$. The correct interpretation names the parameter and context: "We are $95\%$ confident that the true proportion of [context] is between [low] and [high]." A separate, commonly examined sentence interprets the confidence level itself, "in repeated sampling, about $95\%$ of intervals constructed this way would contain the true proportion." Keeping these two interpretations distinct is a frequent free-response discriminator.

Width, sample size, and precision

The margin of error $z^{*}\sqrt{\hat{p}(1-\hat{p})/n}$ shows three levers. A higher confidence level raises $z^{*}$ and widens the interval (more confidence costs precision). A larger sample raises $n$ and narrows it (precision improves with the square root of $n$, so quadrupling $n$ halves the margin of error). The value of $\hat{p}$ matters too: $\hat{p}(1-\hat{p})$ is largest at $\hat{p} = 0.5$, so proportions near a half give the widest intervals for a given $n$. These relationships are routinely tested, including "how large a sample is needed for a margin of error of at most $m$," solved by setting $z^{*}\sqrt{\hat{p}(1-\hat{p})/n} \le m$ and rearranging for $n$ (using $\hat{p} = 0.5$ when no estimate is available, for the most conservative size).

Try this

Q1. A sample of $n = 300$ gives $\hat{p} = 0.40$. Find the standard error. [1 point]

• Cue. $SE = \sqrt{\dfrac{0.40 \times 0.60}{300}} = \sqrt{0.0008} \approx 0.0283$.

Q2. Why does the large-counts check for an interval use $\hat{p}$ rather than a claimed $p_0$? [1 point]

• Cue. An interval has no hypothesized value; the only estimate of the proportion available is the observed $\hat{p}$.