Inference for a Difference in Two Proportions — AP Statistics Study Guide

For: AP Statistics candidates sitting AP Statistics.

Covers: This chapter covers sampling distribution of the difference in two sample proportions, conditions for inference, confidence intervals for a difference in two population proportions, and hypothesis tests for a difference in two proportions including pooled vs unpooled procedures.

You should already know: Inference for a single population proportion, basics of confidence intervals and hypothesis testing, properties of sampling distributions for categorical data.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Statistics style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Inference for a Difference in Two Proportions?

Inference for a difference in two proportions is the set of statistical methods we use to compare the proportion of successes (a binary outcome) between two independent populations or two treatment groups in an experiment. This topic makes up roughly 30-40% of Unit 6 (Inference for Categorical Data: Proportions), which is 12-15% of the total AP Statistics exam, so it appears regularly on both multiple choice (MCQ) and free response (FRQ) sections, often as a full 4-5 point FRQ question. The core goal is to use data from two independent samples to make a claim about the true difference between the two population proportions. Standard notation uses $p_1$ for the true proportion of successes in population 1, $p_2$ for the true proportion in population 2, ${\hat{p}}_1=X_1/n_1$ for the sample proportion from group 1, and ${\hat{p}}_2=X_2/n_2$ for the sample proportion from group 2. We are almost always interested in the difference $p_1-p_2$, estimated by ${\hat{p}}_1-{\hat{p}}_2$. Synonyms for this topic include two-sample inference for proportions, comparison of two proportions, and inference for the difference between population proportions.

2. Sampling Distribution and Conditions for Inference

For valid inference, the sampling distribution of ${\hat{p}}_1-{\hat{p}}_2$ must meet three key properties: it is centered at the true difference $p_1-p_2$, it has a standard deviation that depends on the sample sizes and population proportions, and it is approximately normal when samples are large enough. The mean and standard deviation of the sampling distribution are given by: $${\mu }_{{\hat{p}}_1-{\hat{p}}_2}=p_1-p_2$$ $${\sigma }_{{\hat{p}}_1-{\hat{p}}_2}=\sqrt{\frac{p_1(1-p_1)}{n_1}+\frac{p_2(1-p_2)}{n_2}}$$ When the conditions below are met, the sampling distribution is approximately Normal, which allows us to use z-based inference. All inference requires four key conditions, split by inference type for the large counts check:

1. Random: Both groups come from either two independent random samples from their populations, or a randomized controlled experiment with two treatments. This ensures unbiased estimates.
2. 10% Condition: When sampling without replacement from a finite population, each sample size must be no more than 10% of its population to ensure independence of observations. This can be skipped for randomized experiments.
3. Independent Groups: The two samples/groups are independent of each other (no pairing or matching).
4. Large Counts (Normality): For confidence intervals, all four observed counts $n_1{\hat{p}}_1,n_1(1-{\hat{p}}_1),n_2{\hat{p}}_2,n_2(1-{\hat{p}}_2)$ are at least 10. For hypothesis tests with $H_0:p_1=p_2$, the pooled counts $(n_1+n_2){\hat{p}}_{pooled}$ and $(n_1+n_2)(1-{\hat{p}}_{pooled})$ are at least 10.

Worked Example

A community organizer wants to compare the proportion of registered voters who voted in the last election between two neighborhoods: Neighborhood 1 (population 1800) and Neighborhood 2 (population 2500). They take an SRS of 120 voters from Neighborhood 1 and 180 voters from Neighborhood 2 to construct a confidence interval for the difference in turnout. In their sample, 78 voters from Neighborhood 1 and 99 voters from Neighborhood 2 voted. Check if all conditions for inference are met.

Solution:

1. Random: The problem states that both are independent simple random samples, so the random condition is satisfied.
2. 10% Condition: $n_1=120<0.1(1800)=180$, and $n_2=180<0.1(2500)=250$, so the 10% condition for independence within groups is satisfied.
3. Independent Groups: Samples from the two neighborhoods are independent, so this condition is met.
4. Large Counts (Confidence Interval): Calculate counts: $n_1{\hat{p}}_1=78$, $n_1(1-{\hat{p}}_1)=42$, $n_2{\hat{p}}_2=99$, $n_2(1-{\hat{p}}_2)=81$. All are ≥ 10, so normality is satisfied. All conditions for inference are met.

Exam tip: Always write out each condition explicitly when asked to check, AP graders award a point for each condition checked correctly, don’t skip any.

3. Confidence Intervals for a Difference in Two Proportions

A confidence interval for a difference in two proportions gives a range of plausible values for the true difference $p_1-p_2$ between two population proportions. Unlike hypothesis testing, we do not assume that $p_1=p_2$ when estimating the difference, so we always use an unpooled standard error that relies on the individual sample proportions. The formula for a confidence interval is: $$({\hat{p}}_1-{\hat{p}}_2)\pm z^{\ast }\sqrt{\frac{{\hat{p}}_1(1-{\hat{p}}_1)}{n_1}+\frac{{\hat{p}}_2(1-{\hat{p}}_2)}{n_2}}$$ Where $z^{\ast }$ is the critical value for the desired confidence level from the standard Normal distribution (e.g., $z^{\ast }=1.96$ for 95% confidence, $z^{\ast }=1.645$ for 90% confidence). If 0 is inside the interval, it means 0 is a plausible value for the true difference, so we do not have statistically significant evidence of a difference at the corresponding significance level (α = 1 - confidence level). Always interpret the interval in the context of the problem.

Worked Example

Using the voting example from the previous section: $n_1=120$, ${\hat{p}}_1=78/120=0.65$ (Neighborhood 1 turnout), $n_2=180$, ${\hat{p}}_2=99/180=0.55$ (Neighborhood 2 turnout). Construct and interpret a 95% confidence interval for the difference $p_1-p_2$. Conditions are already checked.

Solution:

1. Calculate the difference in sample proportions: ${\hat{p}}_1-{\hat{p}}_2=0.65-0.55=0.10$.
2. Find the critical value: For 95% confidence, $z^{\ast }=1.96$.
3. Calculate unpooled standard error: $$SE=\sqrt{\frac{0.65(0.35)}{120}+\frac{0.55(0.45)}{180}}=\sqrt{0.001896+0.001375}\approx 0.0572$$
4. Calculate margin of error: $ME=1.96\ast 0.0572\approx 0.112$.
5. Construct the interval: $0.10\pm 0.112=(-0.012,0.212)$.
6. Interpretation: We are 95% confident that the true difference in voter turnout between Neighborhood 1 and Neighborhood 2 is between -0.012 and 0.212.

Exam tip: If you swap $p_1$ and $p_2$, the interval bounds will flip sign but the conclusion (does 0 fall in the interval?) stays the same—just make sure your interpretation matches the order of your difference.

4. Hypothesis Tests for a Difference in Two Proportions

We use hypothesis tests for a difference in two proportions to test a claim about whether the two population proportions differ. Most commonly, the null hypothesis is that there is no difference between the two proportions: $H_0:p_1-p_2=0$, with a one- or two-sided alternative hypothesis. Because the null hypothesis assumes that $p_1=p_2=p$, we can pool the two samples to get a single estimate of the common population proportion ${\hat{p}}_{pooled}$, which gives a more accurate standard error for the test. The pooled proportion is calculated as: $${\hat{p}}_{pooled}=\frac{X_1+X_2}{n_1+n_2}$$ Where $X_1$ and $X_2$ are the number of successes in each sample. The pooled standard error and test statistic are: $$S{E}_{pooled}=\sqrt{{\hat{p}}_{pooled}(1-{\hat{p}}_{pooled})\left(\frac{1}{n_1}+\frac{1}{n_2}\right)}$$ $$z=\frac{({\hat{p}}_1-{\hat{p}}_2)-0}{S{E}_{pooled}}$$ We then calculate the p-value from the standard Normal distribution and compare it to our significance level α to draw a conclusion in context.

Worked Example

A college wants to test whether a new early registration process increases the proportion of students who register for classes before the deadline. 250 students were randomly assigned to the new process (group 1) and 250 to the old process (group 2). 205 students in the new group registered early, compared to 180 in the old group. Test at α = 0.05 whether the new process increases the proportion of early registration.

Solution:

1. State hypotheses: Let $p_1$ = true proportion of early registration for new process, $p_2$ = true proportion for old process. $H_0:p_1-p_2=0$, $H_a:p_1-p_2>0$.
2. Check conditions: Random assignment (random condition met), 10% condition skipped for experiment, groups independent, pooled large counts: ${\hat{p}}_{pooled}=(205+180)/(250+250)=385/500=0.77$. $(500)(0.77)=385\ge 10$, $500(0.23)=115\ge 10$, so normality is met.
3. Calculate test statistic: ${\hat{p}}_1=205/250=0.82$, ${\hat{p}}_2=180/250=0.72$, difference = 0.10. $S{E}_{pooled}=\sqrt{0.77(0.23)(1/250+1/250)}\approx 0.0377$. $z=0.10/0.0377\approx 2.65$.
4. Find p-value: Right-tailed test, $p$-value = $P(Z>2.65)\approx 0.004$.
5. Conclusion: Since $0.004<0.05$, we reject $H_0$. There is sufficient evidence at the 0.05 significance level that the new early registration process increases the proportion of students who register before the deadline.

Exam tip: Only pool when your null hypothesis is $p_1-p_2=0$. If you are testing a null hypothesis that the difference is equal to some non-zero value (extremely rare on the AP exam), you use unpooled standard error.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Pooling the proportion when calculating a confidence interval for a difference in two proportions. Why: Students confuse the pooling rule for hypothesis tests with the rule for confidence intervals, incorrectly assuming pooling is always required. Correct move: Always use unpooled standard error for confidence intervals; only pool when testing $H_0:p_1-p_2=0$.
• Wrong move: Using population proportions $p_1,p_2$ instead of sample proportions ${\hat{p}}_1,{\hat{p}}_2$ when checking large counts for a confidence interval. Why: Students memorize the normality condition for the sampling distribution and forget that we use observed counts when population proportions are unknown. Correct move: For confidence intervals, always check that the four observed sample counts are all at least 10.
• Wrong move: Using two-sample inference for difference in proportions for paired (matched) proportion data. Why: Students see two proportions and automatically use two-sample methods, even when observations are matched (e.g., same subjects before/after a program). Correct move: If data is paired, use one-sample inference on the proportion of positive differences, not two-sample inference.
• Wrong move: Interpreting a 95% confidence interval as "there is a 95% chance that the true difference is between the bounds". Why: Students confuse the definition of confidence level with probability for a fixed interval. Correct move: State that 95% of all possible samples of this size would produce an interval that contains the true difference, and that you are 95% confident the true difference is in your calculated interval.
• Wrong move: Checking the 10% condition for a randomized experiment. Why: Students memorize the 10% condition as required for all inference and don't remember it only applies to sampling without replacement from finite populations. Correct move: Skip the 10% condition for experiments, only check it for random samples from finite populations.
• Wrong move: Failing to explicitly define $p_1$ and $p_2$ before starting inference, leading to incorrect direction of difference or misinterpretation. Why: Students skip this step to save time, but lose points for unclear reasoning. Correct move: Always write out what $p_1$ and $p_2$ represent in context before calculating anything.

6. Practice Questions (AP Statistics Style)

Question 1 (Multiple Choice)

A researcher wants to test whether the proportion of left-handed people differs between men and women, with a null hypothesis of no difference. He collects independent random samples of 100 men and 100 women, and finds 12 men and 8 women are left-handed. Which of the following gives the correct test statistic for this hypothesis test? A. $z=\frac{0.12-0.08}{\sqrt{\frac{0.12(0.88)}{100}+\frac{0.08(0.92)}{100}}}$ B. $z=\frac{0.12-0.08}{\sqrt{0.1(0.9)\left(\frac{1}{100}+\frac{1}{100}\right)}}$ C. $z=\frac{0.12-0.08}{\sqrt{0.12(0.88)\left(\frac{1}{100}+\frac{1}{100}\right)}}$ D. $z=\frac{0.12-0.08}{\sqrt{\frac{0.12}{100}+\frac{0.08}{100}}}$

Worked Solution: For a hypothesis test with $H_0:p_1-p_2=0$, we use the pooled proportion to calculate the standard error. The pooled proportion here is ${\hat{p}}_{pooled}=\frac{12+8}{100+100}=0.1$. The formula for the pooled standard error is $S{E}_{pooled}=\sqrt{{\hat{p}}_{pooled}(1-{\hat{p}}_{pooled})(1/n_1+1/n_2)}$, which matches the formula in option B. Option A uses unpooled standard error (incorrect for this test), and options C and D use incorrect variance calculations. The correct answer is B.

Question 2 (Free Response)

A school district wants to compare the proportion of high school students who have accessed mental health support in two different schools: School A implemented a new on-campus counseling program, while School B kept its old off-campus program. Independent random samples of 120 students from School A and 150 students from School B are taken. 42 students from School A report accessing support, compared to 36 from School B. (a) Check the conditions for constructing a confidence interval for the difference in population proportions. (b) Construct and interpret a 90% confidence interval for the difference ($p_A-p_B$) in the proportion of students who accessed support. (c) Based on your interval, does the data provide evidence that the new program changes the proportion of students accessing support? Explain.

Worked Solution: (a) Random: Stated independent random samples, so random condition is satisfied. 10% Condition: The population of students at each high school is larger than 10 times the sample size (12010 = 1200, 15010 = 1500), so 10% condition is met. Large Counts: $n_A{\hat{p}}_A=42$, $n_A(1-{\hat{p}}_A)=78$, $n_B{\hat{p}}_B=36$, $n_B(1-{\hat{p}}_B)=114$, all ≥ 10, so normality is satisfied. All conditions are met. (b) ${\hat{p}}_A=42/120=0.35$, ${\hat{p}}_B=36/150=0.24$, difference = $0.35-0.24=0.11$. Critical $z^{\ast }$ for 90% confidence is 1.645. Unpooled SE = $\sqrt{\frac{0.35(0.65)}{120}+\frac{0.24(0.76)}{150}}\approx 0.0558$. Margin of error = $1.645\ast 0.0558\approx 0.092$. Interval = $0.11\pm 0.092=(0.018,0.202)$. Interpretation: We are 90% confident that the true difference in the proportion of students accessing mental health support between School A (new program) and School B (old program) is between 0.018 and 0.202. (c) 0 is not contained in the 90% confidence interval, so we have statistically significant evidence at the α = 0.10 significance level that the new program changes the proportion of students accessing support. All values in the interval are positive, so the new program is associated with a higher proportion of students accessing support.

Question 3 (Application / Real-World Style)

A plant biologist tests whether a new fungicide reduces the proportion of potato plants that develop blight. She plants 300 potato plants, randomly assigns 150 to get the new fungicide and 150 to get no treatment. After the growing season, 27 of the fungicide-treated plants get blight, and 42 of the untreated plants get blight. At the α = 0.05 significance level, is there evidence that the new fungicide reduces the proportion of blight? Conduct a full hypothesis test and interpret your conclusion in context.

Worked Solution:

1. Define parameters: $p_1$ = true proportion of blight for fungicide-treated plants, $p_2$ = true proportion for untreated plants. Hypotheses: $H_0:p_1-p_2=0$, $H_a:p_1-p_2<0$.
2. Check conditions: Random assignment confirms random and independent groups, 10% condition is skipped for this experiment. Pooled large counts: ${\hat{p}}_{pooled}=(27+42)/300=0.23$, $300\ast 0.23=69\ge 10$, $300\ast 0.77=231\ge 10$, so normality is met.
3. Calculate test statistic: ${\hat{p}}_1=27/150=0.18$, ${\hat{p}}_2=42/150=0.28$, difference = -0.10. $S{E}_{pooled}=\sqrt{0.23(0.77)(1/150+1/150)}\approx 0.0486$, $z=-0.10/0.0486\approx -2.06$.
4. P-value = $P(Z<-2.06)\approx 0.020$, which is less than α = 0.05. We reject the null hypothesis. There is statistically significant evidence at the 0.05 level that the new fungicide reduces the proportion of potato plants that develop blight.

7. Quick Reference Cheatsheet

8. What's Next

This topic is the foundation for comparing binary outcomes across two independent groups, which extends directly to chi-square inference for categorical data, coming up next in Unit 8. Without mastering the logic of comparing proportions, checking conditions for inference, and the pooled vs unpooled distinction, you will struggle to understand chi-square tests for homogeneity, which compare proportions across three or more groups. This topic also aligns with the design of randomized controlled experiments, where comparing two treatment proportions is a very common AP FRQ task. The same inferential logic you learn here (estimating a difference, testing for no difference) transfers directly to inference for a difference in two means, so mastering this topic makes that upcoming unit much more intuitive.

• Chi-Square Tests for Homogeneity
• Inference for a Difference in Two Means
• Inference for a Single Proportion
• Matched Pairs Inference