AP Statistics · Unit 3 · 15-25% of the exam · ~30 class periods

Inference for Categorical Data: Proportions

AP Statistics Unit 3 covers inference for categorical data: confidence intervals and significance tests for one proportion and two proportions, plus chi-square tests. It is 15-25% of the multiple-choice section and about 30 class periods, the longest unit.

AP Statistics: Unit 3 (topics 3.1 Estimators, 3.2 Sampling Distributions for Sample Proportions, 3.3 Constructing a Confidence Interval for a Population Proportion, 3.4 Justifying a Claim Based on a Confidence Interval for a Population Proportion, 3.5 Setting Up a Test for a Population Proportion, 3.6 p-Values, 3.7 Carrying Out a Test for a Population Proportion, 3.8 Potential Errors When Performing Tests, 3.9 Sampling Distributions for the Difference Between Sample Proportions, 3.10 Constructing a Confidence Interval for the Difference Between Two Population Proportions, 3.11 Justifying a Claim Based on a Confidence Interval for the Difference Between Two Population Proportions, 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, 3.14 Setting Up a Chi-Square Test for Homogeneity or Independence, 3.15 Carrying Out a Chi-Square Test for Homogeneity or Independence). This is Unit 3 of the Fall 2026 AP Statistics course (15-25% of the multiple-choice section, about 30 class periods); it keeps the chi-square tests for homogeneity and independence while the chi-square goodness-of-fit test was removed from the exam.

What Unit 3 covers

Unit 3 is where AP Statistics turns from describing data and computing probabilities to statistical inference: using a sample to draw conclusions about a population. The categorical data here are counts of successes and failures, summarized as proportions. You estimate a population proportion pp (the true fraction of a population with some trait) using a sample proportion p^\hat{p}, read "p-hat."

Unit 3 is 15-25% of the multiple-choice section and about 30 class periods, which makes it the longest of the five units. It has 15 topics that build from inference for one proportion up to the difference between two proportions and then chi-square tests for two-way tables.

The official College Board framework frames the unit around questions like how to measure changing opinions over time and how to decide whether a difference between two groups is statistically significant.

Estimators and the sampling distribution of p-hat (topics 3.1 to 3.2)

3.1 Estimators. A sample statistic is a point estimator of the matching population parameter. An estimator is unbiased if, on average, it neither underestimates nor overestimates the parameter, and p^\hat{p} is an unbiased point estimator of pp.

3.2 Sampling distributions for sample proportions. When the sampled values are independent, the sampling distribution of p^\hat{p} has mean μp^=p\mu_{\hat{p}} = p (the Greek letter mu, μ\mu, denotes a mean) and standard deviation p(1p)n\sqrt{\frac{p(1-p)}{n}}, where nn is the sample size. That distribution is approximately normal when np10np \ge 10 and n(1p)10n(1-p) \ge 10. Sampling without replacement also needs a random sample and the 10% condition, meaning the population is at least 10 times the sample size.

Confidence intervals for one proportion (topics 3.3 to 3.4)

3.3 Constructing a confidence interval. The procedure is a one-sample zz-interval for a population proportion, built as point estimate ±\pm (critical value)(standard error), or p^±zp^(1p^)n\hat{p} \pm z^{*}\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}. The standard error (SE) estimates the standard deviation of the sampling distribution, and zz^{*} is the critical value that encloses the middle C%C\% of the standard normal curve. The three conditions are randomization, the 10% condition, and a normality condition that checks the observed successes and observed failures are each at least 10.

3.4 Justifying a claim. In repeated random sampling with the same nn, about C%C\% of intervals built this way capture the true pp. You interpret a C%C\% interval as "we are C%C\% confident the interval from aa to bb captures the population proportion" in context. Raising the confidence level widens the interval, while raising the sample size narrows it, with the width roughly proportional to 1n\frac{1}{\sqrt{n}}. The how to calculate a confidence interval guide walks through the arithmetic.

Significance tests for one proportion (topics 3.5 to 3.8)

3.5 Setting up the test. The procedure is a one-sample zz-test for a population proportion. The null hypothesis H0:p=p0H_0: p = p_0 states the status-quo value p0p_0, and the alternative HaH_a is one-sided (<< or >>) or two-sided (\ne). Here the normality condition uses the hypothesized value: np010np_0 \ge 10 and n(1p0)10n(1-p_0) \ge 10.

3.6 p-values. The p-value is the probability, computed assuming H0H_0 is true, of getting a test statistic as extreme or more extreme than the one observed, in the direction of HaH_a. Small p-values mean the observed result would be unusual under H0H_0 and give evidence for HaH_a; a large p-value does not prove H0H_0 is true. See what does a p-value mean.

3.7 Carrying out the test. The test statistic is a standardized statistic of the form (statistic minus parameter) over standard error:

z=p^p0p0(1p0)nz = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1 - p_0)}{n}}}

Compare the p-value to the significance level α\alpha: if the p-value is below α\alpha, reject H0H_0; otherwise fail to reject. Rejecting gives convincing evidence for HaH_a, but failing to reject never proves H0H_0.

3.8 Potential errors. A Type I error rejects a true H0H_0, and its probability equals α\alpha. A Type II error fails to reject a false H0H_0, with probability 11 - power, where power is the chance of correctly rejecting a false H0H_0. Power increases when the sample size grows, the standard error shrinks, the true value sits farther from H0H_0, or α\alpha increases. See Type I vs Type II errors.

Inference for two proportions (topics 3.9 to 3.13)

3.9 Sampling distribution of the difference. For two independent samples, p^1p^2\hat{p}_1 - \hat{p}_2 has mean p1p2p_1 - p_2 and standard deviation p1(1p1)n1+p2(1p2)n2\sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}. It is approximately normal when the expected successes and expected failures in both samples are each at least 10.

3.10 and 3.11 Two-sample z-interval. The interval estimates p1p2p_1 - p_2 as (p^1p^2)±zp^1(1p^1)n1+p^2(1p^2)n2(\hat{p}_1 - \hat{p}_2) \pm z^{*}\sqrt{\frac{\hat{p}_1(1-\hat{p}_1)}{n_1} + \frac{\hat{p}_2(1-\hat{p}_2)}{n_2}}. If the interval contains 0, there is insufficient evidence of a difference; if it excludes 0, there is evidence of a difference between the two population proportions.

3.12 and 3.13 Two-sample z-test. The null hypothesis is H0:p1=p2H_0: p_1 = p_2. Because H0H_0 assumes the proportions are equal, the test pools the two samples into a combined proportion p^c=n1p^1+n2p^2n1+n2\hat{p}_c = \frac{n_1\hat{p}_1 + n_2\hat{p}_2}{n_1 + n_2} and uses it in the standard error:

z=p^1p^2p^c(1p^c)(1n1+1n2)z = \frac{\hat{p}_1 - \hat{p}_2}{\sqrt{\hat{p}_c(1 - \hat{p}_c)\left(\frac{1}{n_1} + \frac{1}{n_2}\right)}}

To choose between the one-proportion and two-proportion procedures, read one-proportion vs two-proportion z-test and practice with the proportion z-test calculator.

Chi-square tests for two-way tables (topics 3.14 to 3.15)

3.14 Setting up the test. The chi-square statistic (the Greek letter chi, written χ2\chi^2) measures the distance between observed and expected counts relative to expected counts. Use a chi-square test for homogeneity to compare the distribution of one categorical variable across two or more populations or treatments, and a chi-square test for independence to check whether two categorical variables are associated within a single sampled population. The conditions are randomization, the 10% condition, and an expected-counts condition that every expected count is at least 5.

3.15 Carrying out the test. Each expected count is row total×column totaltable total\frac{\text{row total} \times \text{column total}}{\text{table total}}, and the statistic sums over every cell of the table:

χ2=(ObservedExpected)2Expected\chi^2 = \sum \frac{(\text{Observed} - \text{Expected})^2}{\text{Expected}}

The degrees of freedom equal (rows1)(columns1)(\text{rows} - 1)(\text{columns} - 1), and a larger χ2\chi^2 gives a smaller p-value. Both tests share this arithmetic; the sampling design and the wording of the hypotheses are what differ. See chi-square tests explained and the chi-square calculator.

Conditions and how to study for the exam

Every procedure in Unit 3 begins by naming the method and verifying its conditions in context. The AP framework warns that students lose points for checking conditions superficially, such as writing "SRS" instead of stating that a simple random sample was taken. State each condition using the actual numbers from the problem.

Language matters as much as the arithmetic. Interpret a confidence interval as capturing a parameter, not a statistic, and never say a test "accepts" or "proves" H0H_0, because a test can only reject or fail to reject it. When a chi-square p-value is large, say the data do not give convincing evidence of an association rather than claiming there is none.

One change from the older 9-unit course: the chi-square goodness-of-fit test was removed from the AP exam, while the tests for homogeneity and independence stay in Unit 3. Goodness-of-fit is still standard in most college introductory statistics courses.

One-proportion z-interval (95% confidence)

A random sample of 200 registered voters finds that 120 support a ballot measure. Construct a 95% confidence interval for the population proportion pp who support the measure.

  1. Find the sample proportion: p^=120200=0.60\hat{p} = \frac{120}{200} = 0.60.

  2. Check the normality condition: successes =120= 120 and failures =80= 80 are each at least 10. Assume a random sample and that there are more than 10×200=200010 \times 200 = 2000 voters.

  3. Compute the standard error: SE=0.60(0.40)200=0.24200=0.0012=0.03464SE = \sqrt{\frac{0.60(0.40)}{200}} = \sqrt{\frac{0.24}{200}} = \sqrt{0.0012} = 0.03464.

  4. Use the critical value for 95% confidence: z=1.96z^{*} = 1.96.

  5. Compute the margin of error: 1.96×0.03464=0.06791.96 \times 0.03464 = 0.0679.

  6. Build the interval: 0.60±0.0679=(0.5321,0.6679)0.60 \pm 0.0679 = (0.5321, 0.6679).

We are 95% confident that the interval from 0.5321 to 0.6679 captures the true proportion of registered voters who support the ballot measure.

One-proportion z-test (one-sided)

A delivery company claims 90% of its packages arrive on time. A consumer group takes a random sample of 150 packages and finds 126 on time. Test at α=0.05\alpha = 0.05 whether the true on-time proportion is less than 0.90.

  1. State the hypotheses: H0:p=0.90H_0: p = 0.90 versus Ha:p<0.90H_a: p < 0.90, where pp is the true proportion of on-time packages.

  2. Check conditions: random sample; np0=150(0.90)=13510np_0 = 150(0.90) = 135 \ge 10 and n(1p0)=150(0.10)=1510n(1-p_0) = 150(0.10) = 15 \ge 10; more than 10×150=150010 \times 150 = 1500 packages.

  3. Find the sample proportion: p^=126150=0.84\hat{p} = \frac{126}{150} = 0.84.

  4. Compute the standard error under H0H_0: 0.90(0.10)150=0.09150=0.0006=0.02449\sqrt{\frac{0.90(0.10)}{150}} = \sqrt{\frac{0.09}{150}} = \sqrt{0.0006} = 0.02449.

  5. Compute the test statistic: z=0.840.900.02449=0.060.02449=2.45z = \frac{0.84 - 0.90}{0.02449} = \frac{-0.06}{0.02449} = -2.45.

  6. Find the p-value: P(Z<2.45)0.0071P(Z < -2.45) \approx 0.0071 from the standard normal table.

  7. Decide: 0.0071<0.050.0071 < 0.05, so reject H0H_0.

There is convincing statistical evidence that the true proportion of on-time packages is less than 0.90.

Chi-square test for independence (2 by 2 table)

A researcher takes a single random sample of 100 people and records whether each person exercises regularly and whether each sleeps at least 7 hours. Test for independence at α=0.05\alpha = 0.05.

GroupSleeps 7+ hrsSleeps under 7 hrsTotal
Exercises302050
Does not exercise203050
Total5050100
  1. State the hypotheses: H0H_0 is that exercise and sleep are independent in the population, and HaH_a is that they are associated.

  2. Compute expected counts: each cell =row total×column totaltable total=50×50100=25= \frac{\text{row total} \times \text{column total}}{\text{table total}} = \frac{50 \times 50}{100} = 25, so all four expected counts are 25, each at least 5.

  3. Compute the statistic: χ2=(3025)225+(2025)225+(2025)225+(3025)225=2525×4=4\chi^2 = \frac{(30-25)^2}{25} + \frac{(20-25)^2}{25} + \frac{(20-25)^2}{25} + \frac{(30-25)^2}{25} = \frac{25}{25} \times 4 = 4.

  4. Find the degrees of freedom: (21)(21)=1(2-1)(2-1) = 1.

  5. Find the p-value: with χ2=4\chi^2 = 4 and 1 degree of freedom, P(χ2>4)0.046P(\chi^2 > 4) \approx 0.046 from a chi-square table or technology.

  6. Decide: 0.046<0.050.046 < 0.05, so reject H0H_0.

There is convincing evidence of an association between exercising regularly and getting at least 7 hours of sleep in this population.

Frequently asked questions

How much of the AP Statistics exam is Unit 3?

Unit 3 counts for 15 to 25% of the multiple-choice section, and at about 30 class periods it is the longest of the five units. Inference also appears on the free-response section, where Question 3 is always an inference problem, either a hypothesis test or a confidence interval.

What is the difference between a chi-square test for homogeneity and a test for independence?

Use homogeneity to compare the distribution of one categorical variable across two or more separate populations or treatments. Use independence to check whether two categorical variables are associated within a single population that was sampled once. The arithmetic is identical (expected counts, the χ2\chi^2 statistic, degrees of freedom); the sampling design and hypothesis wording differ.

Do I still need to know the chi-square goodness-of-fit test?

No. For the Fall 2026 course, the chi-square goodness-of-fit test was removed from the AP Statistics exam. Only the chi-square tests for homogeneity and independence remain, both in Unit 3. Goodness-of-fit is still standard in most college introductory statistics courses.

Every topic in Unit 3

  1. 3.1Estimators
  2. 3.2Sampling Distributions for Sample Proportions
  3. 3.3Constructing a Confidence Interval for a Population Proportion
  4. 3.4Justifying a Claim Based on a Confidence Interval for a Population Proportion
  5. 3.5Setting Up a Test for a Population Proportion
  6. 3.6p-Values
  7. 3.7Carrying Out a Test for a Population Proportion
  8. 3.8Potential Errors When Performing Tests
  9. 3.9Sampling Distributions for the Difference Between Sample Proportions
  10. 3.10Constructing a Confidence Interval for the Difference Between Two Population Proportions
  11. 3.11Justifying a Claim Based on a Confidence Interval for the Difference Between Two Population Proportions
  12. 3.12Setting Up a Test for the Difference Between Two Population Proportions
  13. 3.13Carrying Out a Test for the Difference Between Two Population Proportions
  14. 3.14Setting Up a Chi-Square Test for Homogeneity or Independence
  15. 3.15Carrying Out a Chi-Square Test for Homogeneity or Independence