Unit 6 of AP Statistics: Inference for Proportions.
This unit covers confidence intervals for proportions, hypothesis tests for proportions and two-proportion z-test — essential concepts for AP Statistics. Use our interactive study games to test your understanding, or review questions in traditional format below.
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This unit covers confidence intervals for proportions, hypothesis tests for proportions and two-proportion z-test — essential concepts for AP Statistics. Use our interactive study games to test your understanding, or review questions in traditional format below.
Key Concepts Breakdown
1 Confidence Intervals For Proportions
A one-proportion z-interval estimates the true population proportion p using sample data. Students must be able to check all three conditions (Random, Normal/Large Counts, Independence), construct the interval, and interpret it correctly in context. The margin of error equals z* times the standard error of p-hat.
Key Points
- Formula: p̂ ± z*(√(p̂(1-p̂)/n)); use p̂ (not p) in the standard error since p is unknown
- Large Counts condition: np̂ ≥ 10 AND n(1-p̂) ≥ 10 — verify both, show work
- Independence condition: sample ≤ 10% of population (10% condition) for sampling without replacement
- Correct interpretation: 'We are 95% confident the true proportion of [context] is between [L] and [U]' — never say 'probability'
A random sample of 150 high school seniors found that 96 plan to attend a 4-year college. Construct and interpret a 95% confidence interval for the true proportion of all seniors who plan to attend a 4-year college.
First compute p̂ = 96/150 = 0.64 and verify conditions: Random (stated), Large Counts (150×0.64 = 96 ≥ 10 and 150×0.36 = 54 ≥ 10 ✓), Independence (150 < 10% of all seniors ✓). The interval is 0.64 ± 1.96√(0.64×0.36/150) = 0.64 ± 0.077, giving (0.563, 0.717). Interpretation: We are 95% confident the true proportion of all high school seniors who plan to attend a 4-year college is between 0.563 and 0.717.
2 Hypothesis Tests For Proportions
A one-proportion z-test evaluates evidence against a null hypothesis H₀: p = p₀ using a standardized test statistic and P-value. Students must correctly state hypotheses in terms of the population parameter p, use p₀ (not p̂) in the standard error calculation, and make a conclusion that links the P-value to the significance level with context.
Key Points
- Test statistic: z = (p̂ − p₀) / √(p₀(1−p₀)/n); use p₀ in denominator (H₀ assumed true)
- Large Counts condition for tests: np₀ ≥ 10 AND n(1−p₀) ≥ 10 (use p₀, not p̂)
- P-value interpretation: probability of getting a result as extreme or more extreme than observed, assuming H₀ is true
- Conclusion template: 'Because p-value [< or ≥] α, we [reject / fail to reject] H₀. We [do / do not] have convincing evidence that [Ha in context].'
A cereal company claims 20% of boxes contain a prize. A consumer group checks 200 randomly selected boxes and finds prizes in 32. At α = 0.05, is there evidence the true proportion is less than 20%?
H₀: p = 0.20 vs Hₐ: p < 0.20 (left-tailed). Conditions: np₀ = 200(0.20) = 40 ≥ 10 and n(1−p₀) = 160 ≥ 10 ✓. Compute p̂ = 32/200 = 0.16 and z = (0.16 − 0.20)/√(0.20×0.80/200) = −0.04/0.02828 ≈ −1.41; P-value = P(Z < −1.41) ≈ 0.079. Since 0.079 > 0.05, we fail to reject H₀ — there is not convincing evidence that the true proportion of prize boxes is less than 20%.
3 Two-Proportion Z-Test
A two-proportion z-test compares population proportions from two independent groups to determine whether they differ. Students must use the pooled proportion p̂_c (combined) in the standard error because H₀ assumes the two proportions are equal, and must check conditions for both groups separately. Confidence intervals for the difference use unpooled standard error instead.
Key Points
- H₀: p₁ = p₂ (equivalently p₁ − p₂ = 0); pooled proportion p̂_c = (x₁ + x₂)/(n₁ + n₂)
- Test statistic: z = (p̂₁ − p̂₂) / √(p̂_c(1−p̂_c)(1/n₁ + 1/n₂))
- Large Counts: check n₁p̂_c, n₁(1−p̂_c), n₂p̂_c, n₂(1−p̂_c) all ≥ 10 — four separate checks
- For a two-proportion confidence interval (not a test), use unpooled SE: √(p̂₁(1−p̂₁)/n₁ + p̂₂(1−p̂₂)/n₂)
In a study, 45 of 180 men and 60 of 150 women reported experiencing headaches after a treatment. At α = 0.01, is there a significant difference in the proportion of men and women who experience headaches?
H₀: p_m = p_w vs Hₐ: p_m ≠ p_w (two-tailed). Compute p̂_m = 45/180 = 0.25, p̂_w = 60/150 = 0.40, and p̂_c = (45+60)/(180+150) = 105/330 ≈ 0.318. All four Large Counts checks exceed 10 ✓. The test statistic is z = (0.25 − 0.40)/√(0.318×0.682×(1/180 + 1/150)) ≈ −0.15/0.0514 ≈ −2.92; P-value = 2×P(Z < −2.92) ≈ 0.0035. Since 0.0035 < 0.01, reject H₀ — there is convincing evidence of a difference in headache rates between men and women.
Questions, answered.
What is Inference for Proportions?
Inference for Proportions is Unit 6 of AP Statistics, covering confidence intervals for proportions, hypothesis tests for proportions and two-proportion z-test.
How to study for AP Statistics Unit 6?
Start with the Quick Summary above, review the Key Concepts, then test yourself with our interactive study games. Aim for 80%+ accuracy before moving on.
How many questions are in this unit?
This unit has 28+ review questions across 5 different game modes.