AP Statistics · Topic 3.3 · Unit 3

AP Stats 3.3: Confidence Interval for a Proportion

By Jude Wallis · Published July 28, 2026

A one-sample z-interval for a population proportion is p-hat plus or minus z* times sqrt(p-hat(1-p-hat)/n). Check randomization, the 10% condition, and at least 10 observed successes and 10 observed failures before you build it.

AP Statistics: Unit 3 (topics 3.3). CED topic 3.3 (Constructing a Confidence Interval for a Population Proportion), skills 2.C, 3.E, 4.E.

What topic 3.3 covers

Topic 3.3 shows how to turn a single point estimate into an interval estimate for a population proportion. The correct procedure is the one-sample z-interval for a population proportion. The interval takes the form point estimate plus or minus a margin of error, and it estimates the true proportion of a named response variable in a named population. Because you built only a point estimate in topic 3.1, this topic is where a range of plausible values first appears.

The formula and its pieces

The interval is

\hat{p} \pm z^{*}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}

Here $\hat{p}$ is the sample proportion, $n$ is the sample size, and $z^{*}$ is the critical value, the boundary that encloses the middle $C\%$ of the standard normal distribution. Common critical values are 1.645 for 90%, 1.96 for 95%, and 2.576 for 99%. The standard error $SE = \sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}$ estimates the standard deviation of the sampling distribution, and the margin of error is $z^{*} \times SE$ , half the total width of the interval.

Conditions to check first

A one-sample z-interval requires three conditions:

Randomization: the data come from a random sample.
10% condition: when sampling without replacement, $n \le 0.10N$ .
Normality: the observed successes $n\hat{p}$ and observed failures $n(1-\hat{p})$ are both at least 10.

Notice that the interval uses $\hat{p}$ in the normality check because $p$ is unknown, which differs from the test in topic 3.5 that uses the hypothesized $p_0$ . State each condition with the specific numbers from the problem rather than a vague reference to normality, since exam readers expect the actual counts.

Choosing a sample size

You can rearrange the margin of error formula to find the sample size needed for a target margin of error $M$ :

n = \left(\dfrac{z^{*}}{M}\right)^{2}\hat{p}(1-\hat{p}).

When no prior estimate of $\hat{p}$ is available, use $\hat{p} = 0.5$ , which makes $\hat{p}(1-\hat{p})$ as large as possible and gives a safe upper bound on $n$ . Always round the result up to the next whole number, because rounding down would leave the margin of error slightly too large. Keep enough decimal places in the standard error while you work, since rounding it too early can shift the margin of error and move both endpoints of the interval.

95% confidence interval for a proportion

In a random sample of 400 households, 240 own a pet. Construct a 95% confidence interval for the population proportion that owns a pet.

Point estimate: $\hat{p} = \dfrac{240}{400} = 0.60$ .
Conditions: successes $240$ and failures $160$ are both at least 10 (assume random sample and 10% condition met).
Standard error: $SE = \sqrt{\dfrac{0.60 \times 0.40}{400}} = \sqrt{\dfrac{0.24}{400}} = \sqrt{0.0006} \approx 0.02449$ .
Critical value for 95%: $z^{*} = 1.96$ .
Margin of error: $1.96 \times 0.02449 \approx 0.0480$ .
Interval: $0.60 \pm 0.0480$ .

The 95% confidence interval is about $(0.552, 0.648)$ . You are 95% confident this interval captures the true proportion of households that own a pet.

Frequently asked questions

Why use p-hat in the standard error instead of p?

For a confidence interval the population proportion $p$ is unknown, so you estimate the standard deviation of the sampling distribution using the sample proportion $\hat{p}$ . That estimate is why it is called a standard error rather than a standard deviation.