Why does the one proportion test use p₀ instead of p-hat in the standard error?

The test asks what would happen if H₀ were true, and H₀ says p = p₀. So the standard error is built from p₀. A confidence interval makes no such assumption, so it uses p-hat instead.

One-proportion vs two-proportion z-test setup

Q: When do I use the pooled proportion?

Use the pooled proportion p-hat_c only for a two proportion z test, because the null hypothesis assumes the two proportions are equal. Do not pool for a confidence interval for p₁ - p₂; that interval uses each sample's own proportion in the standard error.

Q: Can I use these tests if the counts are small?

No. If any expected success or failure count is below 10, the normal approximation is unreliable and the z test does not apply. The large counts condition is what lets you use the normal curve to find the p-value.

By Jude Wallis · Published July 28, 2026

Use a one proportion z test when you compare one sample to a single claimed value. Use a two proportion z test when you compare the proportions of two groups. The one-sample test builds its standard error from the claimed value; the two-sample test uses a pooled proportion.

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 tests are Unit 3 of the Fall 2026 AP Statistics course, with the one proportion test in topics 3.5 to 3.7 and the two proportion test in topics 3.12 to 3.13.

When to use each test

The choice comes down to how many groups you are asking about.

Use a one proportion z test when you have a single sample and want to compare its proportion to one fixed number, such as a claimed rate or a historical value. AP Statistics covers this in Unit 3 topics 3.5 to 3.7.

Use a two proportion z test when you have two independent groups and want to know whether their population proportions differ. The two groups can come from two separate random samples or from one randomized experiment with two treatment arms. This is Unit 3 topics 3.12 to 3.13.

A quick way to decide: if the null hypothesis names a specific number for the proportion, you have a one-sample problem. If the null hypothesis only says two groups are equal without naming a value, you have a two-sample problem.

First, some notation. $p$ is the true population proportion, the number you cannot see directly. $\hat{p}$ (read 'p-hat') is the sample proportion, the count of successes divided by the sample size $n$ . $p_0$ is the value the null hypothesis claims for $p$ .

When there are two groups you will also meet $\hat{p}_c$ , the pooled or combined sample proportion.

The two tests side by side

The structure is the same, but the parameter and the standard error change.

Feature	One proportion z test	Two proportion z test
Question	Is $p$ different from $p_0$ ?	Is $p_1$ different from $p_2$ ?
Parameter	$p$	$p_1 - p_2$
Standard error	$\sqrt{p_0(1-p_0)/n}$	$\sqrt{\hat{p}_c(1-\hat{p}_c)(1/n_1+1/n_2)}$
AP topics	3.5 to 3.7	3.12 to 3.13

The one proportion z test setup

You test a claim about a single population proportion $p$ .

State the hypotheses in terms of $p$ and the claimed value $p_0$ :

$H_0: p = p_0$
$H_a: p \ne p_0$ (or $p > p_0$ , or $p < p_0$ )

The key feature of this test is the standard error. Because the null hypothesis assumes $p = p_0$ is true, you build the standard error (SE) from $p_0$ , not from $\hat{p}$ :

SE = \sqrt{\frac{p_0(1 - p_0)}{n}}

The standardized test statistic follows the general form, (statistic minus parameter) divided by standard error:

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

You then convert $z$ to a p-value using the standard normal curve. The p-value calculator does this conversion for you.

The two proportion z test setup

Here you compare two population proportions, $p_1$ and $p_2$ , using two independent samples.

State the hypotheses in terms of the difference $p_1 - p_2$ :

$H_0: p_1 = p_2$ (equivalently $p_1 - p_2 = 0$ )
$H_a: p_1 \ne p_2$ (or $p_1 > p_2$ , or $p_1 < p_2$ )

Because $H_0$ says the two proportions are equal, you combine both samples into one estimate of that shared proportion. This is the pooled proportion $\hat{p}_c$ , the total successes divided by the total sample size:

\hat{p}_c = \frac{n_1\hat{p}_1 + n_2\hat{p}_2}{n_1 + n_2}

You then use $\hat{p}_c$ in the standard error for the test:

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

The test statistic compares the observed difference to the hypothesized difference of 0:

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

The pooled proportion is only for the test. A confidence interval for $p_1 - p_2$ does not pool; it uses each sample's own proportion in the standard error.

Decide which group is group 1 before you compute, and keep that order for both the difference and the alternative hypothesis. If you swap the groups, the sign of $z$ flips, so a one-sided $H_a$ must point the same way as your subtraction.

Conditions to check first

Both tests need the same three conditions, checked slightly differently.

Random. The data come from a random sample or a randomized experiment. The two proportion test needs two independent random samples or random assignment to two groups.

10%. When you sample without replacement, each sample should be no more than 10% of its population. For the two proportion test, check this for each sample separately.

Large counts. The sampling distribution is close enough to normal only when the successes and failures are both large.

One proportion: use the expected counts under $H_0$ , so check $np_0 \ge 10$ and $n(1 - p_0) \ge 10$ .
Two proportions: use the pooled proportion $\hat{p}_c$ (the combined success rate under $H_0$ ), so check $n_1\hat{p}_c$ , $n_1(1 - \hat{p}_c)$ , $n_2\hat{p}_c$ , and $n_2(1 - \hat{p}_c)$ are all at least 10.

The four-step structure

Both tests follow the same four steps, which match the AP free-response rubric.

State. Define the parameter in words, write $H_0$ and $H_a$ , and give the significance level $\alpha$ (often 0.05). See null vs alternative hypothesis if the direction is unclear.
Plan. Name the procedure (one or two proportion z test) and check the three conditions with numbers.
Do. Compute $\hat{p}$ (and $\hat{p}_c$ if needed), the standard error, the $z$ statistic, and the p-value.
Conclude. Compare the p-value to $\alpha$ . If the p-value is at most $\alpha$ , reject $H_0$ ; otherwise fail to reject. Then state what that means in context.

How to read the result

The p-value is the probability of a sample result at least as extreme as yours, assuming $H_0$ is true. A small p-value means your data would be surprising if the claim were correct, which is evidence against $H_0$ .

Two reminders that AP graders look for:

You never accept $H_0$ . You either reject it or fail to reject it.
Your conclusion must mention the context and the parameter, not just 'reject $H_0$ '.

To check your arithmetic, the proportion z test calculator runs both the one and two sample versions.

One proportion z test: a support claim

A campaign claims that 60% of a city's voters support a ballot measure. A news outlet takes a random sample of 200 voters and finds 108 who support it. Test at $\alpha = 0.05$ whether the true proportion differs from 0.60.

State. Let $p$ be the true proportion of all city voters who support the measure. $H_0: p = 0.60$ and $H_a: p \ne 0.60$ , with $\alpha = 0.05$ .
Plan. One proportion z test. Random: the 200 voters were sampled at random. 10%: 200 is less than 10% of all city voters. Large counts: $np_0 = 200(0.60) = 120$ and $n(1 - p_0) = 200(0.40) = 80$ , both at least 10.
Find the sample proportion. $\hat{p} = 108/200 = 0.54$ .
Find the standard error from $p_0$ . $SE = \sqrt{\frac{0.60(0.40)}{200}} = \sqrt{\frac{0.24}{200}} = \sqrt{0.0012} = 0.034641$ .
Find the test statistic. $z = \frac{0.54 - 0.60}{0.034641} = \frac{-0.06}{0.034641} = -1.732$ .
Find the p-value. This is a two-sided test, so p-value $= 2 \times P(Z < -1.732) \approx 2(0.0416) = 0.083$ .
Conclude. Since $0.083 > 0.05$ , fail to reject $H_0$ . There is not convincing evidence that the true proportion of supporters differs from 0.60.

$z \approx -1.73$ , p-value $\approx 0.083$ . Fail to reject $H_0$ at $\alpha = 0.05$ ; there is no convincing evidence that the support rate differs from 60%.

Two proportion z test: comparing two success rates

A researcher runs a randomized experiment on a new study method. In the treatment group, 90 of 150 students pass a test. In the control group, 48 of 100 students pass. Test at $\alpha = 0.05$ whether the treatment group has a higher pass rate.

State. Let $p_1$ be the true pass proportion with the new method and $p_2$ the true pass proportion for the control. $H_0: p_1 = p_2$ and $H_a: p_1 > p_2$ , with $\alpha = 0.05$ .
Plan. Two proportion z test. Random: students were randomly assigned to the two groups. Independence: random assignment to the two groups makes them independent, and the 10% condition is not needed for a randomized experiment. Large counts: use the pooled proportion $\hat{p}_c = (90 + 48)/(150 + 100) = 0.552$ . The expected counts $150(0.552) = 82.8$ , $150(0.448) = 67.2$ , $100(0.552) = 55.2$ , and $100(0.448) = 44.8$ are all at least 10.
Find the sample proportions. $\hat{p}_1 = 90/150 = 0.60$ and $\hat{p}_2 = 48/100 = 0.48$ .
Find the pooled proportion. $\hat{p}_c = \frac{90 + 48}{150 + 100} = \frac{138}{250} = 0.552$ .
Find the standard error. $SE = \sqrt{0.552(0.448)\left(\frac{1}{150} + \frac{1}{100}\right)} = \sqrt{0.247296 \times 0.016667} = \sqrt{0.004122} = 0.06420$ .
Find the test statistic. $z = \frac{0.60 - 0.48}{0.06420} = \frac{0.12}{0.06420} = 1.869$ .
Find the p-value. This is a one-sided test, so p-value $= P(Z > 1.869) \approx 0.031$ .
Conclude. Since $0.031 < 0.05$ , reject $H_0$ . There is convincing evidence that the new study method produces a higher pass rate than the control.

$\hat{p}_c = 0.552$ , $z \approx 1.87$ , p-value $\approx 0.031$ . Reject $H_0$ at $\alpha = 0.05$ ; there is convincing evidence the new method has a higher pass rate.

Frequently asked questions

When do I use the pooled proportion?

Use the pooled proportion $\hat{p}_c$ only for a two proportion z test, because the null hypothesis assumes the two proportions are equal. Do not pool for a confidence interval for $p_1 - p_2$ ; that interval uses each sample's own proportion in the standard error.

Why does the one proportion test use $p_0$ instead of $\hat{p}$ in the standard error?

The test asks what would happen if $H_0$ were true, and $H_0$ says $p = p_0$ . So the standard error is built from $p_0$ . A confidence interval makes no such assumption, so it uses $\hat{p}$ instead.

Can I use these tests if the counts are small?

No. If any expected success or failure count is below 10, the normal approximation is unreliable and the z test does not apply. The large counts condition is what lets you use the normal curve to find the p-value.