College Board AP 2022 (style)-style practice: Section II (free response). A study tests whether a reminder text raises appointment attendance. Of 250 patients who received a text, 200 attended; of 250 who did not, 170 attended. Test at α = 0.05 whether the text group has a higher attendance proportion. Compute the pooled proportion, the test statistic and P-value, and state a conclusion in context (justify in context).

A 4-point complete two-proportion z-test. (1) (1 point) Let p_1, p_2 be the true attendance proportions for the text and no-text groups. H_0: p_1 = p_2 versus H_a: p_1 > p_2. Random/independent and large counts (using p_c) hold. (2) (1 point) p_1 = 200/250 = 0.80, p_2 = 170/250 = 0.68. p_c = 200 + 170500 = 0.74. (3) (1 point) SE = 0.74(0.26)(1250 + 1250) = sqrt(0.1924 × 0.008) = sqrt(0.0015392) = 0.03923. z = 0.80 - 0.680.03923 = 0.120.03923 ≈ 3.06. P-value = P(Z > 3.06) ≈ 0.0011. (4) (1 point) Since 0.0011 < 0.05, reject H_0. There is convincing evidence that the text group has a higher true attendance proportion than the no-text group. Markers reward the pooled proportion, the pooled-SE z statistic, the one-sided P-value, and a contextual conclusion.

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How do you compute the test statistic and P-value and conclude a test comparing two proportions?

Topic 6.11 Carrying Out a Test for the Difference of Two Population Proportions: compute the two-sample z test statistic using the pooled standard error, find the P-value, and state a conclusion in context.

A focused answer to AP Statistics Topic 6.11, on computing the two-sample z statistic with the pooled standard error, finding the P-value, and stating a conclusion in context, with a full worked two-proportion test.

Generated by Claude Opus 4.811 min answerUpdated 2026-06-04

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this topic is asking
The pooled test statistic
From z to the P-value and decision
Test and interval, pooled and unpooled
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What this topic is asking

The College Board (Topic 6.11) wants you to carry out and conclude a two-proportion test: compute the z statistic using the pooled standard error, find the P-value in the direction of $H_a$ , compare to $\alpha$ , and state a conclusion in context.

The pooled test statistic

Because $H_0$ assumes a common proportion, the standard error uses $\hat{p}_c$ in place of both sample proportions. This is the test's signature, and the contrast with the interval's unpooled standard error (each $\hat{p}_i$ separate) is the most heavily tested distinction in this part of the unit. The numerator measures the observed gap between groups; dividing by the pooled standard error standardizes it to a z-score.

From z to the P-value and decision

Match the tail to the alternative exactly, and double for a two-sided test. The conclusion sentence states the decision, ties it to the $P$ -versus- $\alpha$ comparison, and interprets in context ("there is convincing evidence that the proportion of ... is higher for group 1 than group 2"). Never write "accept $H_0$ "; the alternatives are "reject" or "fail to reject."

Test and interval, pooled and unpooled

The two-proportion test (pooled SE) and the two-proportion interval (unpooled SE) answer related questions, "is there a difference?" and "how big is the difference?", but use different standard errors. They usually agree about whether $0$ is plausible, yet because the standard errors differ they can occasionally disagree in borderline cases. On the exam, choose the pooled SE whenever you are running a test of $H_0: p_1 = p_2$ , and the unpooled SE whenever you are building an interval for $p_1 - p_2$ . Selecting the right standard error for the task is itself a graded skill.

Complete two-proportion z-test

A company tests whether a redesigned page increases sign-ups. Of $300$ visitors to the new page, $96$ signed up; of $300$ to the old page, $72$ signed up. Test at $\alpha = 0.05$ whether the new page has a higher sign-up proportion.

step 1 Hypotheses and conditions

Let $p_1, p_2$ be the true sign-up proportions for the new and old pages. $H_0: p_1 = p_2$ versus $H_a: p_1 > p_2$ . Random/independent assumed; large counts checked with $\hat{p}_c$ below.

step 2 Sample and pooled proportions

$\hat{p}_1 = \dfrac{96}{300} = 0.32$ , $\hat{p}_2 = \dfrac{72}{300} = 0.24$ .

\hat{p}_c = \frac{96 + 72}{300 + 300} = \frac{168}{600} = 0.28.

Large counts using

\hat{p}_c

300(0.28) = 84

and

300(0.72) = 216

in each group, all

\ge 10

step 3 Pooled standard error and z

SE = \sqrt{0.28(0.72)\left(\frac{1}{300} + \frac{1}{300}\right)} = \sqrt{0.2016 \times 0.006667} = \sqrt{0.001344} \approx 0.03666.

z = \frac{0.32 - 0.24}{0.03666} = \frac{0.08}{0.03666} \approx 2.18.

step 4 P-value

One-sided upper: $\text{P-value} = P(Z > 2.18) \approx 0.0146$ .

step 5 Conclusion in context

Since P-value $0.0146 < 0.05 = \alpha$ , reject $H_0$ . There is convincing evidence that the new page has a higher true sign-up proportion than the old page.

Try this

Q1. Write the test statistic and identify the standard error type. [2 points]

Cue. $z = \dfrac{\hat{p}_1 - \hat{p}_2}{\sqrt{\hat{p}_c(1-\hat{p}_c)(1/n_1 + 1/n_2)}}$ ; it uses the pooled standard error.

Q2. A two-sided two-proportion test gives $z = -1.90$ . Find the P-value. [1 point]

Cue. $\text{P-value} = 2 \cdot P(Z > 1.90) = 2(0.0287) \approx 0.0574$ .

Exam-style practice questions

Practice questions written in the style of College Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AP 2018 (style)1 marksSection I (multiple choice). In a two-proportion test, the standard error used in the z statistic is (A)

\sqrt{\dfrac{\hat{p}_1(1-\hat{p}_1)}{n_1} + \dfrac{\hat{p}_2(1-\hat{p}_2)}{n_2}}

(B)

\sqrt{\hat{p}_c(1-\hat{p}_c)\left(\dfrac{1}{n_1} + \dfrac{1}{n_2}\right)}

(C)

\sqrt{\hat{p}_c(1-\hat{p}_c)(n_1 + n_2)}

(D)

\dfrac{\hat{p}_1 - \hat{p}_2}{n_1 + n_2}

Show worked answer →

The correct answer is (B).

A two-proportion test uses the pooled standard error $\sqrt{\hat{p}_c(1-\hat{p}_c)\left(\dfrac{1}{n_1} + \dfrac{1}{n_2}\right)}$ , with the combined proportion $\hat{p}_c$ , because $H_0$ assumes a single common proportion.

(A) is the unpooled standard error, used for the confidence interval, not the test. (C) and (D) are not valid standard errors.

AP 2022 (style)4 marksSection II (free response). A study tests whether a reminder text raises appointment attendance. Of

250

patients who received a text,

200

attended; of

250

who did not,

170

attended. Test at

\alpha = 0.05

whether the text group has a higher attendance proportion. Compute the pooled proportion, the test statistic and P-value, and state a conclusion in context (justify in context).

Show worked answer →

A 4-point complete two-proportion z-test.

(1) (1 point) Let $p_1, p_2$ be the true attendance proportions for the text and no-text groups. $H_0: p_1 = p_2$ versus $H_a: p_1 > p_2$ . Random/independent and large counts (using $\hat{p}_c$ ) hold.
(2) (1 point) $\hat{p}_1 = 200/250 = 0.80$ , $\hat{p}_2 = 170/250 = 0.68$ . $\hat{p}_c = \dfrac{200 + 170}{500} = 0.74$ .
(3) (1 point) $SE = \sqrt{0.74(0.26)\left(\dfrac{1}{250} + \dfrac{1}{250}\right)} = \sqrt{0.1924 \times 0.008} = \sqrt{0.0015392} = 0.03923$ . $z = \dfrac{0.80 - 0.68}{0.03923} = \dfrac{0.12}{0.03923} \approx 3.06$ . P-value $= P(Z > 3.06) \approx 0.0011$ .
(4) (1 point) Since $0.0011 < 0.05$ , reject $H_0$ . There is convincing evidence that the text group has a higher true attendance proportion than the no-text group.

Markers reward the pooled proportion, the pooled-SE z statistic, the one-sided P-value, and a contextual conclusion.

Related dot points

Sources & how we know this

AP Statistics Course and Exam Description — College Board (2020)