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How do you construct a confidence interval for the difference between two population proportions?

Topic 6.8 Confidence Intervals for the Difference of Two Proportions: check the conditions and construct a two-sample z-interval for the difference between two population proportions, using the unpooled standard error.

A focused answer to AP Statistics Topic 6.8, on building a two-sample z-interval for the difference of two population proportions - checking conditions for both samples and using the unpooled standard error - with a full worked interval.

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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Quick answer

A two-sample z-interval for $p_1 - p_2$ is

(\hat{p}_1 - \hat{p}_2) \pm z^{*}\sqrt{\frac{\hat{p}_1(1-\hat{p}_1)}{n_1} + \frac{\hat{p}_2(1-\hat{p}_2)}{n_2}},

using the unpooled standard error (add the two sample variances, then square-root). Check, for each sample:

Random and the two samples independent;
Large counts: $n_1\hat{p}_1, n_1(1-\hat{p}_1), n_2\hat{p}_2, n_2(1-\hat{p}_2)$ all $\ge 10$ ;
$10\%$ : each sample at most $10\%$ of its population.

If the interval excludes $0$ , there is evidence of a real difference.

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What this topic is asking
The interval and the unpooled standard error
Checking the conditions for two samples
Interpreting and using the interval
Try this

What this topic is asking

The College Board (Topic 6.8) wants you to construct a two-sample z-interval for the difference of two population proportions $p_1 - p_2$ : check the conditions for both samples and use the unpooled standard error (each sample's own $\hat{p}$ ).

The interval and the unpooled standard error

The point estimate is the observed difference $\hat{p}_1 - \hat{p}_2$ . The standard error follows the Topic 5.6 rule: variances of independent quantities add, even for a difference. Each sample contributes $\hat{p}_i(1-\hat{p}_i)/n_i$ ; you sum these and take the square root. This is unpooled because an interval has no common hypothesized proportion, so each sample is estimated separately. (A two-proportion test, Topic 6.10, pools the proportions; the interval does not. Keeping these apart is essential.)

Checking the conditions for two samples

Everything from the one-sample interval is doubled. There are four large-counts checks (successes and failures in each sample), and an extra independence requirement between samples. The large-counts check uses the observed $\hat{p}_1$ and $\hat{p}_2$ (it is an interval, so there is no null value to use).

Interpreting and using the interval

Interpret as usual: "We are $C\%$ confident the difference $p_1 - p_2$ is between [low] and [high]." The decisive feature is whether the interval contains $0$ . If the interval is entirely positive, $p_1 > p_2$ is supported; entirely negative, $p_1 < p_2$ ; and if it straddles $0$ , then $0$ is a plausible difference, so the data do not establish that the two proportions differ. This zero-check is the two-sample analogue of the one-sample plausible-value test and previews the two-proportion significance test. The order of subtraction matters for the sign: state clearly which group is group $1$ so a positive interval is interpreted correctly.

Two-sample z-interval for a difference in proportions

A study compares two teaching methods. Of $120$ students taught by method A, $84$ passed; of $130$ students taught by method B, $78$ passed. The two groups are independent random samples. Construct and interpret a $95\%$ confidence interval for $p_A - p_B$ .

step 1 Point estimates and conditions

$\hat{p}_A = \dfrac{84}{120} = 0.70$ , $\hat{p}_B = \dfrac{78}{130} = 0.60$ . Difference $= 0.70 - 0.60 = 0.10$ . Random and independent: stated. Large counts: $84, 36, 78, 52$ all $\ge 10$ . $10\%$ : each sample plausibly under $10\%$ of its population. Conditions met.

step 2 Unpooled standard error

SE = \sqrt{\frac{0.70(0.30)}{120} + \frac{0.60(0.40)}{130}} = \sqrt{\frac{0.21}{120} + \frac{0.24}{130}} = \sqrt{0.00175 + 0.001846} = \sqrt{0.003596} \approx 0.05997.

step 3 Critical value and margin of error

For $95\%$ , $z^{*} = 1.96$ . Margin of error $= 1.96 \times 0.05997 \approx 0.1175$ .

step 4 Construct the interval

0.10 \pm 0.1175 = (-0.018,\ 0.218).

step 5 Interpret in context

We are $95\%$ confident the difference in pass proportions (method A minus method B) is between $-0.018$ and $0.218$ . Because the interval contains $0$ , it is plausible that the two methods have the same pass proportion; the data do not provide convincing evidence that method A and method B differ in pass rate.

Try this

Q1. Write the standard error for a two-proportion interval and name why it is "unpooled." [2 points]

Cue. $SE = \sqrt{\dfrac{\hat{p}_1(1-\hat{p}_1)}{n_1} + \dfrac{\hat{p}_2(1-\hat{p}_2)}{n_2}}$ ; unpooled because each sample uses its own $\hat{p}_i$ (no common proportion is assumed).

Q2. A $95\%$ interval for $p_1 - p_2$ is $(0.04, 0.19)$ . Is there evidence of a difference? [1 point]

Cue. Yes; the interval excludes $0$ , so $0$ is not plausible and there is convincing evidence the proportions differ.

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). For a two-sample interval for

p_1 - p_2

, the standard error is (A)

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

(B)

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

(C)

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

(D)

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

Show worked answer →

The correct answer is (B).

A confidence interval for $p_1 - p_2$ uses the unpooled standard error: each sample contributes its own $\hat{p}_i(1-\hat{p}_i)/n_i$ , and the two are added (variances add for independent samples), then square-rooted.

(A) is the pooled standard error, used only in a two-proportion test, not an interval. (C) subtracts the variances (wrong). (D) is not a valid standard error.

AP 2022 (style)4 marksSection II (free response). In a random sample of

150

men,

90

own a bicycle; in an independent random sample of

200

women,

96

own a bicycle. (a) Check the conditions for a two-sample z-interval for

p_1 - p_2

, the difference in proportions (men minus women). (b) Construct a

95\%

confidence interval. (c) Justify in context whether there is convincing evidence of a difference in bicycle ownership.

Show worked answer →

A 4-point two-proportion interval.

(a) (1 point) Random and independent: two independent random samples. Large counts: men $90$ and $60$ , women $96$ and $104$ , all $\ge 10$ . $10\%$ : both samples plausibly under $10\%$ of their populations.
(b) (2 points) $\hat{p}_1 = 90/150 = 0.60$ , $\hat{p}_2 = 96/200 = 0.48$ . Difference $= 0.12$ . $SE = \sqrt{\dfrac{0.60(0.40)}{150} + \dfrac{0.48(0.52)}{200}} = \sqrt{0.0016 + 0.001248} = \sqrt{0.002848} = 0.05337$ . Interval $= 0.12 \pm 1.96(0.05337) = 0.12 \pm 0.1046 = (0.015,\ 0.225)$ .
(c) (1 point) The entire interval is above $0$ , so $0$ is not a plausible value for $p_1 - p_2$ ; there is convincing evidence that the proportion owning a bicycle differs (men higher than women).

Markers reward checking conditions for both samples, the unpooled standard error, the $z^{*} = 1.96$ interval, and judging the difference against $0$ .

Related dot points

Sources & how we know this

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