Topic 3.13 Notes – Carrying Out a Test for the Difference Between Two Population Proportions

What a Two-Sample z-Test for Proportions Does

This test is for a binary categorical variable, where each person or item is a success or failure, measured in two independent groups.

• $p_1$ and $p_2$ are the true population proportions for group 1 and group 2.
  • Example: $p_1$ could be the true proportion of teens at School A who have a driver’s license.
  • $p_2$ could be the true proportion at School B.

The sample proportions are

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

where $x_1, x_2$ are successes and $n_1, n_2$ are sample sizes.

For this AP Stats test, the null always says no difference:

$$H_0:p_1=p_2 \qquad \text{or} \qquad H_0:p_1-p_2=0$$

Possible alternatives are

• $H_a:p_1>p_2$
• $H_a:p_1<p_2$
• $H_a:p_1\ne p_2$

You compare the observed difference $\hat p_1-\hat p_2$ to the null difference of 0. The order matters the whole way through. If you write group 1 minus group 2 in the hypotheses, you must keep that same order in the calculation and conclusion.

The two groups should be independent. A randomized experiment, like the one shown here, is a common setup where that happens.

Random assignment to treatment and control groups

This is not for matched pairs or paired data. If the same person is measured twice, or subjects are paired up, this procedure does not fit.

Conditions and Why the Test Pools

Under $H_0$, the test assumes the two population proportions are equal. That means both samples are estimating one common proportion, so we pool them:

$$\hat p_c=\frac{x_1+x_2}{n_1+n_2} =\frac{n_1\hat p_1+n_2\hat p_2}{n_1+n_2}$$

Pooling matters because the test’s standard error is built from the null claim that the proportions are the same.

• It is used only for the hypothesis test standard error
• It is not usually $(\hat p_1+\hat p_2)/2$, unless the sample sizes are equal
• It is different from a confidence interval, which uses an unpooled standard error

Conditions for the z-test:

• Randomization
  • Two independent random samples, or two groups from a randomized experiment
• Independence
  • The groups must be independent
  • If sampling without replacement, each sample must satisfy the 10% condition
• Large expected counts
  • $n_1\hat p_c\ge10$, $n_1(1-\hat p_c)\ge10$
  • $n_2\hat p_c\ge10$, $n_2(1-\hat p_c)\ge10$

A very common mistake is checking counts with $\hat p_1$ and $\hat p_2$. For this test, use the pooled proportion.

Carrying Out the Test

Here’s the calculation path:

1. Find $\hat p_1$ and $\hat p_2$
2. Find $\hat p_c$
3. Compute the null standard error

$$SE_0=\sqrt{\hat p_c(1-\hat p_c)\left(\frac1{n_1}+\frac1{n_2}\right)}$$

4. Compute the test statistic

$$z=\frac{(\hat p_1-\hat p_2)-0}{SE_0}$$

5. Use the standard normal distribution for the p-value

• Right-tailed: $P(Z\ge z_{\text{obs}})$
• Left-tailed: $P(Z\le z_{\text{obs}})$
• Two-sided: $2P(Z\ge |z_{\text{obs}}|)$

If $z$ is positive, then $\hat p_1>\hat p_2$. If $z$ is negative, then $\hat p_1<\hat p_2$.

The picture below is a good reminder that the tail comes from the alternative hypothesis, not from whether your sample difference ended up positive or negative.

Right-tailed, left-tailed, and two-tailed p-values on a z-curve

Quick example. Suppose group 1 has 48 successes out of 80, and group 2 has 36 out of 90.

• $\hat p_1=48/80=0.60$
• $\hat p_2=36/90=0.40$
• $\hat p_c=(48+36)/(80+90)=84/170\approx0.494$

Then use $\hat p_c$ in $SE_0$, then in $z$. On technology, this is 2-PropZTest with $x_1,n_1,x_2,n_2$, plus the correct alternative.

Interpreting the p-Value and Writing the Conclusion

The p-value is the probability, assuming $H_0$ is true, of getting a sample difference or z-statistic at least as extreme as the one observed, in the direction of $H_a$.

Decision rule:

• If p-value $\le \alpha$, reject $H_0$
• If p-value $> \alpha$, fail to reject $H_0$

A strong AP-style conclusion sounds like this:

Because the p-value is less than 0.05, we reject $H_0$. There is convincing evidence that the true proportion of ___ in group 1 is greater than the true proportion of ___ in group 2.

Keep these out of your answer:

• “accept $H_0$”
• “the p-value is the probability that $H_0$ is true”
• “proved”
• causal claims unless it was a randomized experiment