AP Statistics Topic 3.13: Carrying Out a Test for the Difference Between Two Population Proportions

01 · From an observed difference to evidence

The concept in 60 seconds

Independent random samples find that $124$ of $200$ students at School A and $69$ of $150$ at School B prefer digital study notes. The sample rates are $62\%$ and $46\%$ : A is ahead by $16$ percentage points.

The research question, chosen before examining the data, is: Is the population preference proportion higher at A? A two-proportion $z$ -test asks how unusual this sample difference would be if the $2$ population proportions were actually equal.

Visual guide 1: follow the question through the test

QuestionIs A’s population rate higher?

$H_{0}: p_{1}=p_{2}$ ; $H_{a}: p_{1} > p_{2}$ . Define the same preference outcome and keep A minus B.

Evidence under equality $z\approx 2.978$ ; $p\approx 0.00145$

The positive sample difference is far into the right tail of the approximate null model.

Decision in contextReject at $\alpha =0.05$

Convincing evidence of a higher population preference proportion at A; the study does not establish a causal school effect.

The sample difference is an estimate, the $p$ -value is evidence under the equality null, and the conclusion concerns the population claim. Each step has a different job.

The school result: $z\approx 2.978$ and the right-sided $p$ -value is approximately $0.00145$ . At $\alpha =0.05$ , reject the equality null. There is convincing evidence that A’s population preference proportion is higher. The small $p$ -value measures evidence against equality; it does not measure the probability that equality is true.

This lesson shows where those numbers come from and how to explain them. All numerical studies on this page are fictional teaching examples.

Quick check: is a difference of

16

percentage points automatically convincing evidence?

No. Its strength as evidence also depends on sample sizes, variability, the alternative and whether the study meets the test conditions. A justified test measures the difference relative to its null standard error.

02 · Keep the setup visible

State and check the test plan

Let $p_{1}$ be the proportion of all School A students who prefer digital notes and $p_{2}$ the corresponding proportion at School B. Keep A minus B as the subtraction order throughout.

$H_0:p_1-p_2=0$ .

$H_a:p_1-p_2\gt0$ .

Use a two-sample $z$ -test for the difference between $2$ population proportions. For the school example, the problem specifies separate independent SRSs, sampled without replacement. The populations contain $10{,}000\text{ and }8{,}000$ students.

Random sampling and independence: the independent SRS design is given; the same students are not measured $2$ times.
$10\%$ checks: $200\le 0.10(10{,}000)=1{,}000$ and $150\le 0.10(8{,}000)=800$ .
Normality: with $\hat{p}_{c}=\frac{193}{350}$ , the $4$ pooled expected counts are approximately $110.29, 89.71, 82.71,\text{ and }67.29$ . All are $\ge10$ .

School inputs: compare the same preference outcome in $2$ independent samples.
Quantity	School A: group 1	School B: group 2
Success definition	Prefers digital notes	Prefers digital notes
Success count $x$	$124$	$69$
Sample size $n$	$200$	$150$
Other responses $n-x$	$76$	$81$
Population size $N$	$10{,}000$	$8{,}000$
Observed proportion $\hat{p}$	$0.62$	$0.46$
Pooled expected successes	$110.29$	$82.71$
Pooled expected failures	$89.71$	$67.29$

Before calculating: justify the design, both $10\%$ checks when sampling without replacement, and all $4$ pooled expected counts. The sampling $10\%$ condition is not required merely for random assignment. A separate finite-population sampling stage can still require its own check. A calculator output does not repair paired responses, biased recruitment or failed expected counts.

Quick check: can I skip the conditions if software gives a very small

p

-value?

No. The usual normal-approximation $p$ -value is useful only when the procedure is appropriate. Show why the design and count conditions support the test before treating its output as evidence.

03 · Measure the distance from equality

Calculate the pooled $z$ -statistic

The statistic compares the observed sample difference with the null difference of $0$ , measured in estimated null standard errors.

$\hat p_1=\frac{x_1}{n_1}$ , $\hat p_2=\frac{x_2}{n_2}$ .

\hat p_c=\frac{x_1+x_2}{n_1+n_2}

\operatorname{SE}_0=\sqrt{\hat p_c(1-\hat p_c)\left(\frac1{n_1}+\frac1{n_2}\right)}

z=\frac{(\hat p_1-\hat p_2)-0}{\operatorname{SE}_0}

$\operatorname{SE}_{0}$ is the estimated standard deviation of the sample difference under the equality null. Pool because $H_{0}$ assumes a common success rate. Combining counts estimates that rate; it does not prove equality.

Visual guide 2: build the statistic from

3

quantities

Observed minus null $0.16-0=0.16$

Use the sample difference A minus B; equality supplies the $0$ null difference.

Null variability $\operatorname{SE}_{0}\approx 0.0537197$

Pool $193$ successes from $350$ students. Use both sample sizes in the pooled standard error.

Standardized distance $\frac{0.16}{\operatorname{SE}_{0}}\approx 2.97842$

A positive $z$ means the observed difference is above $0$ ; the alternative determines its evidence direction.

Both the numerator and the standard error use proportion units. Their ratio is a unit-free standardized statistic. The null standard error is not the unpooled confidence-interval standard error.

School calculation

$\hat{p}_{1}=\frac{124}{200}=0.62$ ; $\hat{p}_{2}=\frac{69}{150}=0.46$ ; $\hat{p}_{c}=\frac{124 + 69}{200 + 150}=\frac{193}{350}$ .

\operatorname{SE}_0=\sqrt{\frac{193}{350}\cdot\frac{157}{350}\left(\frac1{200}+\frac1{150}\right)}\approx0.0537197

z=\frac{0.62-0.46}{\sqrt{\frac{193}{350}\cdot\frac{157}{350}\left(\frac1{200}+\frac1{150}\right)}}\approx2.97842

The observed difference is about $2.98$ null standard errors above $0$ . $\operatorname{SE}_{0}$ is about $0.05372$ in proportion units, or $5.372$ percentage points. Use proportions consistently: $\frac{0.16}{0.05372}$ and $\frac{16}{5.372}$ give the same $z$ ; $\frac{16}{0.05372}$ does not.

Keep precision until the end. Use the full pooled fraction and stored $\operatorname{SE}$ when finding $z$ and the $p$ -value. Display rounded numbers for readability. The confidence-interval standard error uses separate $\hat{p}_{1}$ and $\hat{p}_{2}$ ; do not substitute that unpooled expression into this equality-null test.

Calculator check: TI-84 family

Open STAT → TESTS → $\texttt{2-PropZTest}$ . Enter $x_{1}=124$ , $n_{1}=200$ , $x_{2}=69$ and $n_{2}=150$ ; select $p_{1} > p_{2}$ , then Calculate. Expect $z\approx 2.9784$ and $p\approx 0.0014487$ . Enter success counts, rather than sample proportions. Keep the group order and chosen alternative consistent. Menu wording can vary by model; TI’s official command reference identifies this test.

Quick check: what does a negative

z

tell us?

The observed group 1 minus group 2 difference is below the null difference of $0$ . Whether that is evidence for $H_{a}$ depends on its direction. A negative $z$ can give a small left-sided $p$ -value and a large right-sided $p$ -value.

04 · Let

H_a

choose the tail

Find the correct $p$ -value

Under a justified equality-null model, the standardized statistic is approximately standard normal. Let $Z$ represent a value from that model and $z_{\mathrm{obs}}$ the statistic from the data.

Tail rules for the standard equality-null two-proportion $z$ -test.
Alternative for group 1 minus group 2	$p$ -value rule	Results counted
$H_{a}: p_{1}-p_{2} > 0$	$P(Z\ge z_{\mathrm{obs}})$	At or above the observed $z$
$H_{a}: p_{1}-p_{2} < 0$	$P(Z\le z_{\mathrm{obs}})$	At or below the observed $z$
$H_{a}: p_{1}-p_{2} \ne 0$	$2P(Z\ge \lvert z_{\mathrm{obs}}\rvert)$	At least as far from $0$ in either direction

Visual guide 3: choose the tail from the alternative

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Carrying Out a Test for the Difference Between Two Population Proportions

By the end of this lesson, you should be able to:

The concept in 60 seconds

State and check the test plan

Calculate the pooled $z$ -statistic

School calculation

Calculator check: TI-84 family

Find the correct $p$ -value

Main Menu

Quick Links

The concept in 60 seconds

State and check the test plan

Calculate the pooled zz-statistic

School calculation

Calculator check: TI-84 family

Find the correct pp-value

Main Menu

Quick Links​

Calculate the pooled $z$ -statistic

Find the correct $p$ -value

Quick Links