Inference for Categorical Data: Proportions — AP Statistics Study Guide

For: AP Statistics candidates sitting AP Statistics.

Covers: AP Statistics Unit 6 — confidence intervals for one and two proportions, hypothesis tests (1-sample z and 2-sample z), assumption checks, Type I/II errors, power.

You should already know: Sampling distributions (Unit 5), probability (Unit 4).

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 papers.

1. Why Inference for Proportions Matters

Unit 6 is the first inference unit — you finally use sampling distributions (Unit 5) to draw conclusions about populations from samples. About 12-15% of AP Stats covers Unit 6 directly. The two-sample test of proportions is one of the most-tested FRQ scenarios.

The unifying logic: take a sample, compute a statistic, ask "how likely is this statistic if the null hypothesis is true?" — and use the answer to support or reject claims about the population proportion.

2. Confidence interval for one proportion

When estimating an unknown population proportion $p$: $$\hat{p}\pm z^{\ast }\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$$

Where $\hat{p}$ is the sample proportion, $n$ is sample size, $z^{\ast }$ is the critical value (1.96 for 95%, 1.645 for 90%, 2.576 for 99%).

Conditions (always state and check on FRQ):

1. Random sample.
2. 10% condition: $n\le 0.10N$.
3. Large Counts: $n\hat{p}\ge 10$ AND $n(1-\hat{p})\ge 10$.

Interpretation of a 95% CI: "We are 95% confident that the true population proportion is between [lower] and [upper]." Wrong: "There is a 95% probability that $p$ is in this interval" — once computed, $p$ is fixed, the interval is what's random.

3. Hypothesis test for one proportion

Setup:

• $H_0:p=p_0$ (null hypothesis, status quo).
• $H_a:p>p_0$, $p<p_0$, or $p\ne p_0$ (alternative).
• Significance level $\alpha$ (usually 0.05).

Test statistic (1-sample z): $$z=\frac{\hat{p}-p_0}{\sqrt{p_0(1-p_0)/n}}$$

Note: standard error uses $p_0$ (null value), not $\hat{p}$.

Conditions same as for CI but with Large Counts using $p_0$: $n{p}_0\ge 10$ AND $n(1-p_0)\ge 10$.

P-value: probability of observing a test statistic as extreme as $z$ if $H_0$ is true.

• Two-sided: $P(|Z|\ge |z|)$.
• Right-tail: $P(Z\ge z)$.
• Left-tail: $P(Z\le z)$.

Decision: reject $H_0$ if $p\text{-value}<\alpha$. Otherwise fail to reject.

Conclusion in context: "We have/do not have evidence at the 5% level that the true proportion of [context] differs from [claim]."

4. Two-sample inference for proportions

Estimating difference $p_1-p_2$ (CI): $$({\hat{p}}_1-{\hat{p}}_2)\pm z^{\ast }\sqrt{\frac{{\hat{p}}_1(1-{\hat{p}}_1)}{n_1}+\frac{{\hat{p}}_2(1-{\hat{p}}_2)}{n_2}}$$

Conditions: each sample random, each independent, each $n_i\le 0.10N_i$, both samples meet Large Counts.

Hypothesis test: $H_0:p_1=p_2$ (equivalent to $p_1-p_2=0$). $$z=\frac{({\hat{p}}_1-{\hat{p}}_2)-0}{\sqrt{{\hat{p}}_c(1-{\hat{p}}_c)\left(\frac{1}{n_1}+\frac{1}{n_2}\right)}}$$

where the pooled proportion ${\hat{p}}_c=\frac{x_1+x_2}{n_1+n_2}$. Under $H_0$, both populations have the same proportion, so we pool the data for a better estimate.

5. Type I and Type II errors

Power = probability of correctly rejecting a false $H_0$. Increased by:

• Larger sample size $n$.
• Larger effect (true $p$ farther from $p_0$).
• Larger $\alpha$ (but this raises Type I risk too).

6. Worked Example

A medical study claims that 60% of patients respond to a new drug. A trial of 200 patients shows 105 respond. Test at the 5% level whether the true response rate is below 60%.

Solution.

Step 1: state hypotheses. $H_0:p=0.60$, $H_a:p<0.60$, $\alpha =0.05$.

Step 2: check conditions.

• Random ✓ (assumed from study design).
• 10% condition: 200 ≤ 0.10 × population (vast — all patients with this condition).
• Large Counts: $n{p}_0=200\cdot 0.60=120\ge 10$ ✓; $n(1-p_0)=80\ge 10$ ✓.

Step 3: compute. $\hat{p}=105/200=0.525$. SE = $\sqrt{0.60\cdot 0.40/200}=\sqrt{0.0012}=0.0346$. $z=(0.525-0.60)/0.0346=-2.165$.

Step 4: p-value. Left-tail: $P(Z\le -2.165)\approx 0.0152$.

Step 5: decision. $p$-value 0.0152 < $\alpha$ 0.05 → reject $H_0$.

Step 6: conclusion. We have evidence at the 5% level that the true response rate is below 60%.

7. Common Pitfalls

• Wrong SE for CI vs test: CI uses $\hat{p}$ in SE; one-sample test uses $p_0$. Two-sample test uses pooled ${\hat{p}}_c$.
• Sign of test statistic vs alternative direction: $z=-2.165$ in our example; $H_a:p<0.60$ → left tail. Don't double-tail or right-tail mistakenly.
• "Accept $H_0$": never say this. Either "reject $H_0$" or "fail to reject $H_0$". A failure to reject is not proof of $H_0$.
• Random vs random assignment: random sampling tells us about population (inference); random assignment tells us about cause-effect. AP problems specify which.
• Conditions on FRQ: must state AND check, not just state. Lose points for either omission.

8. Practice Questions (CED Style)

1. A poll of 1500 voters shows 825 favour candidate A. Construct a 95% CI for the true proportion supporting A. Interpret in context.
2. A factory claims at most 2% of widgets are defective. A sample of 500 has 15 defective. Test the claim at $\alpha =0.05$.
3. In a clinical trial, 40 of 200 patients in the treatment group recovered, vs 30 of 200 in the placebo group. Test whether the treatment is more effective at $\alpha =0.05$.

9. Quick Reference Cheatsheet

• 1-sample CI for $p$: $\hat{p}\pm z^{\ast }\sqrt{\hat{p}(1-\hat{p})/n}$.
• 1-sample test for $p$: $z=(\hat{p}-p_0)/\sqrt{p_0(1-p_0)/n}$.
• 2-sample CI for $p_1-p_2$: $({\hat{p}}_1-{\hat{p}}_2)\pm z^{\ast }\sqrt{{\hat{p}}_1(1-{\hat{p}}_1)/n_1+{\hat{p}}_2(1-{\hat{p}}_2)/n_2}$.
• 2-sample test for $p_1=p_2$: $z=({\hat{p}}_1-{\hat{p}}_2)/\sqrt{{\hat{p}}_c(1-{\hat{p}}_c)(1/n_1+1/n_2)}$.
• Conditions: Random / 10% / Large Counts (≥10 each direction).
• $z^{\ast }$ for CI: 90%→1.645; 95%→1.96; 99%→2.576.
• Type I = false reject. Type II = false fail-to-reject.

10. What's Next

Inference for proportions is the prototype for Unit 7 (Inference for Means) — the same logic applies but with t-distribution instead of z (because sample $\sigma$ replaces population). Unit 8 (Chi-Squared) generalises to multi-category. Use Ollie for any specific test set-up: "Should I use a 1-sample or 2-sample test here?" or "Why is my pooled SE different from my unpooled SE?"