When do I pool and when do I not pool the proportions?

Pool for the two-proportion z-test (hypothesis test), because H_0 assumes the proportions are equal. Do NOT pool for the two-proportion z-interval, because a confidence interval does not assume equality — it uses each sample's own p.

How do I find the counts if the problem only gives proportions?

Multiply each sample proportion by its sample size: x_i = n_i p_i, then round to the nearest whole number since counts of successes must be integers. Use these counts in the pooled proportion formula.

How is my conclusion graded on the AP exam?

You must connect the p-value to , state reject or fail to reject H_0, and describe the result in the context of the two groups. A conclusion without context, or one that says you 'accept' or 'prove' the null, loses credit.

AP-STATS-6.10-6.11

U6.10 Two-Proportion z-Test

Q: What conditions do I need to check for a two-proportion z-test?

Independence (two independent random samples or two groups from random assignment), the 10% condition if sampling without replacement, and the large-counts condition using the pooled proportion: n_1p_c, n_1(1-p_c), n_2p_c, and n_2(1-p_c) must all be at least 10.

Master the two-proportion z-test for AP Statistics: set up hypotheses, use the pooled proportion in the standard error, compute z, find the p-value, and conclude in context.

What you'll do in this lesson

A voice-first session with the Crimsora tutor on U6.10 Two-Proportion z-Test, then targeted practice and FRQs — with the tutor adapting to where you get stuck.

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What this lesson covers

You already know how to build a two-proportion confidence interval. Now we flip to hypothesis testing: is there real evidence that two population proportions differ? The two-proportion z-test answers exactly that. The one twist that trips students up is the standard error — because the null hypothesis assumes the two proportions are equal, we combine the samples into a single pooled proportion before computing the SE. This lesson walks through hypotheses, the pooled SE, the test statistic, the p-value, and the conclusion, all in the four-step format the exam rewards.

Setting Up the Hypotheses

A two-proportion z-test compares the true proportions of two independent populations,

p_1

and

p_2

. Because we are testing for a difference, we write hypotheses about

p_1 - p_2

.

The null hypothesis always states there is no difference:

H_0: p_1 - p_2 = 0

, equivalently

p_1 = p_2

. The alternative reflects the research question and comes in three flavors:

Claim	Alternative
Proportions differ	$H_a: p_1 - p_2 \neq 0$
Group 1 higher	$H_a: p_1 - p_2 > 0$
Group 1 lower	$H_a: p_1 - p_2 < 0$

Define your groups clearly in words (for example, "

p_1

= true proportion of treated patients who recover") so the direction of

H_a

is unambiguous. A common mistake is writing hypotheses about sample proportions

\hat{p}_1

and

\hat{p}_2

; hypotheses are always about population parameters. State your significance level, commonly

\alpha = 0.05

, before looking at the data. On the exam, points are awarded for correct parameter definitions and correctly-directed hypotheses, so never skip this step.

Why We Pool the Proportion

Here is the defining feature of the test. Under

H_0

we assume

p_1 = p_2

, meaning both samples are estimating the same common proportion. So it makes sense to combine all the successes from both samples into a single best estimate, the pooled (combined) proportion:

\hat{p}_c = \frac{x_1 + x_2}{n_1 + n_2}

Here

x_1

and

x_2

are the counts of successes and

n_1

n_2

are the sample sizes. We then use

\hat{p}_c

in the standard error for the test statistic:

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

This is the single biggest difference between the two-proportion z-test and the two-proportion z-interval. The confidence interval does NOT pool — it uses each sample's own

\hat{p}

because it is not assuming the proportions are equal. The test pools because the null hypothesis explicitly assumes equality. Mixing these up is one of the most frequent errors on FRQs. If a count

x_i

is not given directly, compute it as

x_i = n_i \hat{p}_i

and round to a whole number of successes.

Computing z and the p-value

With the pooled SE in hand, the test statistic measures how many standard errors the observed difference sits from zero:

z = \frac{(\hat{p}_1 - \hat{p}_2) - 0}{\sqrt{\hat{p}_c(1-\hat{p}_c)\left(\frac{1}{n_1} + \frac{1}{n_2}\right)}}

The numerator uses the actual sample proportions

\hat{p}_1

and

\hat{p}_2

; only the SE in the denominator uses the pooled value. The

-0

comes from the null value of the difference.

Once you have

z

, find the p-value from the standard normal distribution. For a two-sided

H_a

, the p-value is

2P(Z \ge |z|)

. For a one-sided test, use just the appropriate tail. On a calculator this is the 2-PropZTest function, which reports

z

and the p-value directly — but you must still show the formula and inputs to earn full credit.

Before trusting the result, verify conditions: independent random samples (or random assignment), the 10% condition if sampling without replacement, and the large-counts condition. For the test, check the expected counts using the pooled proportion:

n_1\hat{p}_c

n_1(1-\hat{p}_c)

n_2\hat{p}_c

, and

n_2(1-\hat{p}_c)

should all be at least 10.

Concluding in Context

A conclusion is a two-part sentence linking the p-value to the significance level and then to the real-world question.

If the p-value is less than or equal to

\alpha

, reject

H_0

: there is convincing statistical evidence of a difference. If the p-value is greater than

\alpha

, fail to reject

H_0

: there is not convincing evidence of a difference. Notice you never "accept"

H_0

and never "prove" anything.

A full-credit conclusion template: "Because the p-value of ___ is (less than / greater than)

\alpha

, we (reject / fail to reject)

H_0

. We (do / do not) have convincing evidence that [the proportion of ... differs from the proportion of ...]."

Always name the two groups and the context in the final clause — a bare "reject

H_0

" earns little. Watch the direction: if

H_a

was one-sided, your conclusion should reference "greater than" or "less than," not just "different." Finally, remember that failing to reject does not mean the proportions are equal; it means the data did not provide enough evidence to distinguish them.

Key terms

Two-Proportion z-Test.: A significance test that assesses whether two independent population proportions differ, using a z statistic based on the difference of sample proportions.
Pooled Proportion ( $\hat{p}_c$ ).: The combined estimate $\hat{p}_c = \frac{x_1 + x_2}{n_1 + n_2}$ used in the standard error because the null hypothesis assumes the two proportions are equal.
Null Hypothesis ( $H_0$ ).: The claim of no difference, $H_0: p_1 - p_2 = 0$ , meaning the two population proportions are equal.
Standard Error (pooled).: $SE = \sqrt{\hat{p}_c(1-\hat{p}_c)\left(\frac{1}{n_1}+\frac{1}{n_2}\right)}$ , the estimated standard deviation of the difference under $H_0$ .
p-value.: The probability, assuming $H_0$ is true, of getting a difference in sample proportions at least as extreme as the observed one.
Large-Counts Condition.: For the test, all of $n_1\hat{p}_c$ , $n_1(1-\hat{p}_c)$ , $n_2\hat{p}_c$ , $n_2(1-\hat{p}_c)$ must be at least 10 to justify the normal approximation.

Worked example

A researcher tests whether a new flu vaccine reduces infection rates. In a randomized experiment, 30 of 200 vaccinated people got the flu, while 50 of 200 unvaccinated people got the flu. At

\alpha = 0.05

, is there evidence that the vaccinated group has a lower infection proportion?

Let

p_1

= true infection proportion for vaccinated and

p_2

= true infection proportion for unvaccinated. Since we expect the vaccine to lower infection, use a one-sided test.

H_0: p_1 - p_2 = 0

and

H_a: p_1 - p_2 < 0

.

Sample proportions:

\hat{p}_1 = \frac{30}{200} = 0.15

and

\hat{p}_2 = \frac{50}{200} = 0.25

.

Pooled proportion:

\hat{p}_c = \frac{30 + 50}{200 + 200} = \frac{80}{400} = 0.20

.

Check conditions: random assignment (experiment); expected counts

200(0.20)=40

and

200(0.80)=160

for each group, all at least 10.

Standard error:

SE = \sqrt{0.20(0.80)\left(\frac{1}{200}+\frac{1}{200}\right)} = \sqrt{0.16(0.01)} = \sqrt{0.0016} = 0.04

.

Test statistic:

z = \frac{0.15 - 0.25}{0.04} = \frac{-0.10}{0.04} = -2.5

.

p-value: for

H_a < 0

, p-value

= P(Z \le -2.5) \approx 0.0062

.

Conclusion: Because

0.0062 < 0.05

, we reject

H_0

. There is convincing evidence that the true infection proportion is lower for vaccinated people than for unvaccinated people.

Practice questions

In a two-proportion z-test, why is the pooled proportion

\hat{p}_c

used in the standard error instead of the separate sample proportions?

Because it always produces a smaller standard error
Because the null hypothesis assumes the two population proportions are equal
Because it increases the p-value
Because the sample sizes are unequal

Answer: Because the null hypothesis assumes the two population proportions are equal

The test is conducted assuming

H_0: p_1 = p_2

is true. Under that assumption both samples estimate one common proportion, so combining them gives the best single estimate,

\hat{p}_c

, for the standard error. The confidence interval, which makes no equality assumption, does not pool.

A poll finds that 120 of 400 city residents and 90 of 360 suburban residents support a new tax. Set up the hypotheses and compute the pooled proportion and the test statistic for a two-sided test.

Answer: $\hat{p}_c = 0.276$ , $z \approx 1.54$ ; fail to reject $H_0$ .

Let

p_1

= city support proportion,

p_2

= suburban.

H_0: p_1 - p_2 = 0

H_a: p_1 - p_2 \neq 0

. Sample proportions:

\hat{p}_1 = 120/400 = 0.30

\hat{p}_2 = 90/360 = 0.25

. Pooled:

\hat{p}_c = \frac{120+90}{400+360} = \frac{210}{760} \approx 0.276

SE = \sqrt{0.276(0.724)\left(\frac{1}{400}+\frac{1}{360}\right)} \approx \sqrt{0.1998(0.00528)} \approx 0.0325

. Then

z = \frac{0.30 - 0.25}{0.0325} = \frac{0.05}{0.0325} \approx 1.54

. The two-sided p-value

\approx 0.12 > 0.05

, so fail to reject

H_0

: not enough evidence of a difference in support.

A student computes a two-proportion z-test and reports 'the p-value is 0.20, so we accept the null hypothesis that the proportions are equal.' Identify and correct the error.

Answer: You never accept $H_0$ ; you fail to reject it, and this does not prove the proportions are equal.

With a large p-value we conclude there is not convincing evidence of a difference, so we fail to reject

H_0

. This is different from proving

H_0

true — the data simply could not distinguish the proportions. A correct statement: 'Because 0.20 is greater than

\alpha

, we fail to reject

H_0

; we do not have convincing evidence that the two proportions differ.'

FAQ

When do I pool and when do I not pool the proportions?: Pool for the two-proportion z-test (hypothesis test), because $H_0$ assumes the proportions are equal. Do NOT pool for the two-proportion z-interval, because a confidence interval does not assume equality — it uses each sample's own $\hat{p}$ .
What conditions do I need to check for a two-proportion z-test?: Independence (two independent random samples or two groups from random assignment), the 10% condition if sampling without replacement, and the large-counts condition using the pooled proportion: $n_1\hat{p}_c$ , $n_1(1-\hat{p}_c)$ , $n_2\hat{p}_c$ , and $n_2(1-\hat{p}_c)$ must all be at least 10.
How do I find the counts if the problem only gives proportions?: Multiply each sample proportion by its sample size: $x_i = n_i \hat{p}_i$ , then round to the nearest whole number since counts of successes must be integers. Use these counts in the pooled proportion formula.
How is my conclusion graded on the AP exam?: You must connect the p-value to $\alpha$ , state reject or fail to reject $H_0$ , and describe the result in the context of the two groups. A conclusion without context, or one that says you 'accept' or 'prove' the null, loses credit.

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