A neighborhood bakery believes that 25% of its customers order a gluten-free item. In a random sample of 160 customers, 32 order a gluten-free item. At α = 0.05, is there convincing evidence that the true proportion of gluten-free orders differs from 0.25?

z ≈ -1.46, two-sided p-value ≈ 0.144. Fail to reject H₀ at α = 0.05; no convincing evidence the gluten-free order rate differs from 25%.

Proportion z-test practice problems

By Jude Wallis · Published July 28, 2026

This set has 8 problems on one-proportion and two-proportion z-tests: writing hypotheses, checking conditions, computing the test statistic with a pooled standard error for two-sample tests, finding the p-value, and stating a conclusion in context. Solve each on paper, then open the steps.

AP Statistics: Unit 3 (topics 3.5 Setting Up a Test for a Population Proportion, 3.6 p-Values, 3.7 Carrying Out a Test for a Population Proportion, 3.12 Setting Up a Test for the Difference Between Two Population Proportions, 3.13 Carrying Out a Test for the Difference Between Two Population Proportions). These problems cover the one-proportion z-test (Unit 3 topics 3.5 to 3.7) and the two-proportion z-test with a pooled standard error (topics 3.12 to 3.13) in the Fall 2026 AP Statistics course. Unit 3 is 15 to 25% of the multiple-choice section.

What these problems build

These 8 problems build the full workflow of a significance test for proportions: stating the null and alternative hypotheses, checking the random, 10%, and large-counts conditions, computing the standardized test statistic, finding a p-value from the standard normal curve, and writing a conclusion in context. The first problems use a one-proportion z-test, which compares a single sample proportion $\hat{p}$ (read 'p-hat', the successes divided by the sample size $n$ ) to a claimed value $p_0$ . The later problems use a two-proportion z-test, which compares two groups and builds its standard error from the pooled proportion $\hat{p}_c$ , the combined success rate found by assuming the two proportions are equal. Throughout, $z$ is the test statistic and $\alpha$ (alpha) is the significance level you compare the p-value against.

Work each problem with the four-step State, Plan, Do, Conclude structure before opening the solution. For the setup, conditions, and formulas, read one-proportion vs two-proportion z-test, and check your arithmetic with the proportion z-test calculator. More sets are on the practice page.

Problem 1

A neighborhood bakery believes that 25% of its customers order a gluten-free item. In a random sample of 160 customers, 32 order a gluten-free item. At $\alpha = 0.05$ , is there convincing evidence that the true proportion of gluten-free orders differs from 0.25?

Show the worked solution

State. Let $p$ be the true proportion of the bakery's customers who order a gluten-free item. $H_0: p = 0.25$ versus $H_a: p \ne 0.25$ , with $\alpha = 0.05$ .
Plan. One-proportion z-test. Random: the 160 customers are a random sample. 10%: 160 is less than 10% of all the bakery's customers. Large counts under $H_0$ : $np_0 = 160(0.25) = 40$ and $n(1 - p_0) = 160(0.75) = 120$ , both at least 10.
Find the sample proportion. $\hat{p} = \frac{32}{160} = 0.20$ .
Find the standard error from $p_0$ . $SE = \sqrt{\frac{0.25(0.75)}{160}} = \sqrt{\frac{0.1875}{160}} = \sqrt{0.001171875} = 0.03423$ .
Find the test statistic. $z = \frac{0.20 - 0.25}{0.03423} = \frac{-0.05}{0.03423} = -1.46$ .
Find the p-value. Two-sided, so p-value $= 2 \times P(Z < -1.46) = 2 \times 0.0721 = 0.1442$ .
Conclude. Since $0.1442 > 0.05$ , fail to reject $H_0$ . There is not convincing evidence that the true gluten-free order rate differs from 25%.

$z \approx -1.46$ , two-sided p-value $\approx 0.144$ . Fail to reject $H_0$ at $\alpha = 0.05$ ; no convincing evidence the gluten-free order rate differs from 25%.

Problem 2

A dental office has historically had a 15% appointment no-show rate. After it starts sending text-message reminders, a random sample of 200 appointments has 22 no-shows. At $\alpha = 0.05$ , is there convincing evidence that the reminders lowered the no-show rate below 0.15?

Show the worked solution

State. Let $p$ be the true no-show proportion after reminders. $H_0: p = 0.15$ versus $H_a: p < 0.15$ , with $\alpha = 0.05$ .
Plan. One-proportion z-test. Random: the 200 appointments are a random sample. 10%: 200 is less than 10% of all the office's appointments. Large counts under $H_0$ : $np_0 = 200(0.15) = 30$ and $n(1 - p_0) = 200(0.85) = 170$ , both at least 10.
Find the sample proportion. $\hat{p} = \frac{22}{200} = 0.11$ .
Find the standard error from $p_0$ . $SE = \sqrt{\frac{0.15(0.85)}{200}} = \sqrt{\frac{0.1275}{200}} = \sqrt{0.0006375} = 0.02525$ .
Find the test statistic. $z = \frac{0.11 - 0.15}{0.02525} = \frac{-0.04}{0.02525} = -1.58$ .
Find the p-value. Left-tailed, so p-value $= P(Z < -1.58) = 0.0571$ .
Conclude. Since $0.0571 > 0.05$ , fail to reject $H_0$ . Even though the sample rate fell to 11%, there is not convincing evidence that the reminders lowered the true no-show rate.

$z \approx -1.58$ , p-value $\approx 0.057$ . Fail to reject $H_0$ at $\alpha = 0.05$ ; the drop to 11% is not strong enough to conclude the reminders lowered the no-show rate.

Problem 3

Nationally, 70% of first-year students at a university return for a second year. After a new peer-mentorship program, a random sample of 250 first-year students shows 190 who return. At $\alpha = 0.05$ , is there convincing evidence that the return rate under the program is greater than 0.70?

Show the worked solution

State. Let $p$ be the true second-year return proportion under the mentorship program. $H_0: p = 0.70$ versus $H_a: p > 0.70$ , with $\alpha = 0.05$ .
Plan. One-proportion z-test. Random: the 250 students are a random sample. 10%: 250 is less than 10% of all first-year students. Large counts under $H_0$ : $np_0 = 250(0.70) = 175$ and $n(1 - p_0) = 250(0.30) = 75$ , both at least 10.
Find the sample proportion. $\hat{p} = \frac{190}{250} = 0.76$ .
Find the standard error from $p_0$ . $SE = \sqrt{\frac{0.70(0.30)}{250}} = \sqrt{\frac{0.21}{250}} = \sqrt{0.00084} = 0.02898$ .
Find the test statistic. $z = \frac{0.76 - 0.70}{0.02898} = \frac{0.06}{0.02898} = 2.07$ .
Find the p-value. Right-tailed, so p-value $= P(Z > 2.07) = 1 - 0.9808 = 0.0192$ .
Conclude. Since $0.0192 < 0.05$ , reject $H_0$ . There is convincing evidence that the true return rate under the program is greater than 70%.

$z \approx 2.07$ , p-value $\approx 0.019$ . Reject $H_0$ at $\alpha = 0.05$ ; convincing evidence the second-year return rate exceeds 70%.

Problem 4

An online store runs an experiment on its checkout button color. Of 300 visitors randomly shown a green button, 66 complete a purchase; of 300 visitors randomly shown an orange button, 90 complete a purchase. At $\alpha = 0.05$ , is there convincing evidence that the two button colors lead to different purchase rates?

Show the worked solution

State. Let $p_1$ and $p_2$ be the true purchase proportions for the green and orange buttons. $H_0: p_1 = p_2$ versus $H_a: p_1 \ne p_2$ , with $\alpha = 0.05$ .
Plan. Two-proportion z-test. Random: visitors were randomly assigned a button color, so the groups are independent. Large counts: use the pooled proportion $\hat{p}_c = \frac{66 + 90}{300 + 300} = \frac{156}{600} = 0.26$ ; the expected counts $300(0.26) = 78$ and $300(0.74) = 222$ for each group are all at least 10.
Find the sample proportions. $\hat{p}_1 = \frac{66}{300} = 0.22$ and $\hat{p}_2 = \frac{90}{300} = 0.30$ .
Find the pooled standard error. $SE = \sqrt{0.26(0.74)\left(\frac{1}{300} + \frac{1}{300}\right)} = \sqrt{0.1924 \times 0.006667} = \sqrt{0.0012827} = 0.03581$ .
Find the test statistic. $z = \frac{0.22 - 0.30}{0.03581} = \frac{-0.08}{0.03581} = -2.23$ .
Find the p-value. Two-sided, so p-value $= 2 \times P(Z < -2.23) = 2 \times 0.0129 = 0.0258$ .
Conclude. Since $0.0258 < 0.05$ , reject $H_0$ . There is convincing evidence that the two button colors lead to different purchase rates.

$\hat{p}_c = 0.26$ , $z \approx -2.23$ , two-sided p-value $\approx 0.026$ . Reject $H_0$ at $\alpha = 0.05$ ; convincing evidence the button colors differ in purchase rate.

Problem 5

A public health worker compares flu-vaccination rates at two high schools. An independent random sample of 200 students at School A includes 128 who are vaccinated; an independent random sample of 200 students at School B includes 108 who are vaccinated. At $\alpha = 0.05$ , is there convincing evidence that School A has a higher vaccination rate than School B?

Show the worked solution

State. Let $p_1$ and $p_2$ be the true vaccination proportions at School A and School B. $H_0: p_1 = p_2$ versus $H_a: p_1 > p_2$ , with $\alpha = 0.05$ .
Plan. Two-proportion z-test. Random: two independent random samples, each less than 10% of its school. Large counts: pooled proportion $\hat{p}_c = \frac{128 + 108}{200 + 200} = \frac{236}{400} = 0.59$ ; expected counts $200(0.59) = 118$ and $200(0.41) = 82$ for each group are all at least 10.
Find the sample proportions. $\hat{p}_1 = \frac{128}{200} = 0.64$ and $\hat{p}_2 = \frac{108}{200} = 0.54$ .
Find the pooled standard error. $SE = \sqrt{0.59(0.41)\left(\frac{1}{200} + \frac{1}{200}\right)} = \sqrt{0.2419 \times 0.01} = \sqrt{0.002419} = 0.04918$ .
Find the test statistic. $z = \frac{0.64 - 0.54}{0.04918} = \frac{0.10}{0.04918} = 2.03$ .
Find the p-value. Right-tailed, so p-value $= P(Z > 2.03) = 1 - 0.9788 = 0.0212$ .
Conclude. Since $0.0212 < 0.05$ , reject $H_0$ . There is convincing evidence that School A has a higher vaccination rate than School B.

$\hat{p}_c = 0.59$ , $z \approx 2.03$ , p-value $\approx 0.021$ . Reject $H_0$ at $\alpha = 0.05$ ; convincing evidence School A's vaccination rate is higher than School B's.

Problem 6

A seed company advertises that 90% of its wildflower seeds germinate. A gardener plants a random sample of 120 of these seeds and 102 germinate. (a) State the hypotheses. (b) Check the conditions. (c) Find the test statistic and p-value. (d) At $\alpha = 0.05$ , is there convincing evidence that the true germination rate is less than 90%?

Show the worked solution

(a) Let $p$ be the true germination proportion for the company's seeds. $H_0: p = 0.90$ versus $H_a: p < 0.90$ , with $\alpha = 0.05$ .
(b) One-proportion z-test. Random: the 120 seeds are a random sample. 10%: 120 is less than 10% of all seeds the company produces. Large counts under $H_0$ : $np_0 = 120(0.90) = 108$ and $n(1 - p_0) = 120(0.10) = 12$ , both at least 10.
(c) Sample proportion and standard error. $\hat{p} = \frac{102}{120} = 0.85$ , and $SE = \sqrt{\frac{0.90(0.10)}{120}} = \sqrt{\frac{0.09}{120}} = \sqrt{0.00075} = 0.02739$ .
(c) Test statistic. $z = \frac{0.85 - 0.90}{0.02739} = \frac{-0.05}{0.02739} = -1.83$ .
(c) p-value. Left-tailed, so p-value $= P(Z < -1.83) = 0.0336$ .
(d) Conclude. Since $0.0336 < 0.05$ , reject $H_0$ . There is convincing evidence that the true germination rate is less than the advertised 90%.

(a) $H_0: p = 0.90$ , $H_a: p < 0.90$ ; (c) $z \approx -1.83$ , p-value $\approx 0.034$ ; (d) reject $H_0$ , convincing evidence the germination rate is below 90%.

Problem 7

A gym tests two welcome emails to see which gets more new members to book a first class. Of 350 members randomly sent Email A, 84 book a class; of 400 members randomly sent Email B, 120 book a class. (a) At $\alpha = 0.05$ , is there convincing evidence that the two emails lead to different booking rates? (b) Describe what a Type I error would mean here, and say whether the decision in part (a) could have produced one.

Show the worked solution

(a) State. Let $p_1$ and $p_2$ be the true booking proportions for Email A and Email B. $H_0: p_1 = p_2$ versus $H_a: p_1 \ne p_2$ , with $\alpha = 0.05$ .
(a) Plan. Two-proportion z-test. Random: members randomly assigned an email, so groups are independent. Large counts: pooled $\hat{p}_c = \frac{84 + 120}{350 + 400} = \frac{204}{750} = 0.272$ ; expected counts $350(0.272) = 95.2$ , $350(0.728) = 254.8$ , $400(0.272) = 108.8$ , $400(0.728) = 291.2$ are all at least 10.
(a) Sample proportions. $\hat{p}_1 = \frac{84}{350} = 0.24$ and $\hat{p}_2 = \frac{120}{400} = 0.30$ .
(a) Pooled standard error. $SE = \sqrt{0.272(0.728)\left(\frac{1}{350} + \frac{1}{400}\right)} = \sqrt{0.198016 \times 0.0053571} = \sqrt{0.0010608} = 0.03257$ .
(a) Test statistic and p-value. $z = \frac{0.24 - 0.30}{0.03257} = \frac{-0.06}{0.03257} = -1.84$ , so the two-sided p-value $= 2 \times P(Z < -1.84) = 2 \times 0.0329 = 0.0658$ .
(a) Conclude. Since $0.0658 > 0.05$ , fail to reject $H_0$ . There is not convincing evidence that the two emails lead to different booking rates.
(b) A Type I error would mean concluding the two emails have different booking rates when in truth they are equal. A Type I error can happen only when you reject $H_0$ ; because part (a) failed to reject $H_0$ , this decision could not have produced a Type I error, though it could be a Type II error if a real difference exists.

(a) $\hat{p}_c = 0.272$ , $z \approx -1.84$ , p-value $\approx 0.066$ ; fail to reject $H_0$ , no convincing evidence the emails differ. (b) A Type I error means calling the emails different when they are equal; failing to reject cannot produce one.

Problem 8

A clinic compares text-message and mailed reminders for getting patients to a follow-up visit. Of 240 patients randomly assigned text reminders, 156 attend; of 240 patients randomly assigned mailed reminders, 132 attend. (a) At $\alpha = 0.05$ , is there convincing evidence that text reminders give a higher attendance rate? (b) Would your conclusion change at $\alpha = 0.01$ ? (c) Interpret the p-value in context.

Show the worked solution

(a) State. Let $p_1$ and $p_2$ be the true attendance proportions for text and mailed reminders. $H_0: p_1 = p_2$ versus $H_a: p_1 > p_2$ , with $\alpha = 0.05$ .
(a) Plan. Two-proportion z-test. Random: patients randomly assigned a reminder type, so groups are independent. Large counts: pooled $\hat{p}_c = \frac{156 + 132}{240 + 240} = \frac{288}{480} = 0.60$ ; expected counts $240(0.60) = 144$ and $240(0.40) = 96$ for each group are all at least 10.
(a) Sample proportions and standard error. $\hat{p}_1 = \frac{156}{240} = 0.65$ , $\hat{p}_2 = \frac{132}{240} = 0.55$ , and $SE = \sqrt{0.60(0.40)\left(\frac{1}{240} + \frac{1}{240}\right)} = \sqrt{0.24 \times 0.0083333} = \sqrt{0.002} = 0.04472$ .
(a) Test statistic and p-value. $z = \frac{0.65 - 0.55}{0.04472} = \frac{0.10}{0.04472} = 2.24$ , so the right-tailed p-value $= P(Z > 2.24) = 1 - 0.9875 = 0.0125$ .
(a) Conclude. Since $0.0125 < 0.05$ , reject $H_0$ . There is convincing evidence that text reminders give a higher attendance rate than mailed reminders.
(b) At $\alpha = 0.01$ , the p-value $0.0125 > 0.01$ , so you fail to reject $H_0$ . The same data no longer clear the stricter cutoff, so the conclusion changes to no convincing evidence of a higher rate.
(c) Interpret the p-value. If the two reminder methods truly produced equal attendance rates, there would be about a 1.25% chance of seeing the text group's attendance proportion exceed the mail group's by at least this much from the random assignment alone.

(a) $\hat{p}_c = 0.60$ , $z \approx 2.24$ , p-value $\approx 0.0125$ ; reject $H_0$ , text reminders show a higher attendance rate. (b) At $\alpha = 0.01$ you fail to reject, so the conclusion changes. (c) About a 1.25% chance of a gap this large if the methods were equally effective.