AP Statistics · Topic 3.7 · Unit 3

AP Stats 3.7: Carrying Out a Proportion Test

By Jude Wallis · Published July 28, 2026

The one-sample z-statistic for a proportion is (p-hat minus p0) divided by sqrt(p0(1-p0)/n). Find the p-value from the standard normal distribution. If the p-value is at most alpha, reject the null; otherwise fail to reject, and state the conclusion in context.

AP Statistics: Unit 3 (topics 3.7). CED topic 3.7 (Carrying Out a Test for a Population Proportion), skills 3.E and 4.G.

What topic 3.7 covers

Topic 3.7 completes the significance test you set up in topic 3.5. You calculate the test statistic and p-value, compare the p-value to the significance level, and write a conclusion in context. The general form of a test statistic is $\dfrac{\text{statistic} - \text{parameter}}{\text{standard error}}$ , which here becomes a z-statistic. This four-step rhythm, hypotheses then conditions then calculation then conclusion, is the same one you will reuse for every test in the course.

The test statistic and p-value

For a one-sample z-test for a proportion the test statistic is

z = \dfrac{\hat{p} - p_0}{\sqrt{\dfrac{p_0(1-p_0)}{n}}}.

The standard error in the denominator uses the hypothesized $p_0$ , because the whole calculation assumes the null is true. When the null holds, this $z$ has a standard normal distribution, so you find the p-value from the standard normal table or technology using the direction of $H_a$ from topic 3.6. The z-statistic tells you how many standard errors the observed $\hat{p}$ sits from the hypothesized value.

Making and stating the decision

The significance level $\alpha$ is the preset probability of rejecting a true null. Compare the p-value to $\alpha$ : if the p-value is less than or equal to $\alpha$ , reject $H_0$ ; if it is greater, fail to reject $H_0$ . Rejecting means there is convincing evidence for the alternative; failing to reject means there is not. A test can never prove the null is true, so state the conclusion in context using non-definitive language and reference the parameter and population.

Writing a conclusion that earns credit

A complete conclusion links the numbers to the decision to the context. State the comparison explicitly, for example "because the p-value of 0.079 is greater than $\alpha = 0.05$ , we fail to reject the null." Then translate that into the setting: there is or is not convincing evidence for the alternative claim about the population.

Avoid definitive words like "prove" or "accept," and always tie the result back to the response variable and population named in the question. A conclusion that reports only the decision, with no p-value comparison and no context, leaves points on the table even when the arithmetic is correct. Matching the z and p-value from your calculator output to your own by-hand values is a quick way to catch data-entry mistakes.

One-sample z-test for a proportion

A shopper tests $H_0\colon p = 0.25$ against $H_a\colon p > 0.25$ using a random sample of 150 candies, of which 45 are red. Use $\alpha = 0.05$ .

Sample proportion: $\hat{p} = \dfrac{45}{150} = 0.30$ .
Standard error: $\sqrt{\dfrac{0.25 \times 0.75}{150}} = \sqrt{\dfrac{0.1875}{150}} = \sqrt{0.00125} \approx 0.03536$ .
Test statistic: $z = \dfrac{0.30 - 0.25}{0.03536} = \dfrac{0.05}{0.03536} \approx 1.41$ .
p-value ( $H_a$ uses $>$ ): $P(Z \ge 1.41) = 1 - 0.9207 = 0.0793$ .
Compare: $0.0793 > 0.05$ .

Because the p-value 0.0793 is greater than 0.05, fail to reject $H_0$ . There is not convincing evidence that more than 25% of the candy mix is red.

Frequently asked questions

Why does the test use p0 in the standard error but the interval uses p-hat?

A hypothesis test assumes the null is true, so it uses the hypothesized $p_0$ to build the null distribution. A confidence interval has no hypothesized value, so it estimates the standard error from the observed $\hat{p}$ .