← Back to the Inference For Proportions worksheet
Last updated: October 22, 2026 · Written by the Math Made Easy math teachers
Standards: AP Statistics Unit 3 (2026-27) · Level: Grade 12 AP Statistics
Problem 1 Conditions for an Interval
Random: the sample was selected randomly. 10%: 180 is less than 10% of 4,200 (which is 420). Large counts: np̂ = 54 ≥ 10 and n(1 − p̂) = 126 ≥ 10. All conditions are met.
All three conditions are satisfied
Problem 2 A 90% Confidence Interval
p̂ = 134/300 = 0.4467. SE = √(0.4467 × 0.5533/300) = 0.02870. ME = 1.645 × 0.02870 = 0.0472. Interval: 0.4467 ± 0.0472.
(0.399, 0.494)
Problem 3 Interpreting an Interval
We are 95% confident that the true proportion of all museum visitors who prefer the planetarium show is between 0.390 and 0.503. The confidence level means that if many random samples were taken and an interval built from each, about 95% of those intervals would capture the true proportion.
95% confident the true proportion is in (0.390, 0.503)
Problem 4 Margin of Error
SE = √(0.62 × 0.38/500) = 0.02171. ME = 2.576 × 0.02171 = 0.0559.
ME ≈ 0.0559
Problem 5 Using Intervals to Justify a Claim
A majority is supported only if every plausible value is above 0.50. School A: the interval (0.42, 0.56) contains 0.50, so not convincing. School B: (0.53, 0.67) is entirely above 0.50, so convincing evidence of a majority. School C: (0.38, 0.46) is entirely below 0.50, so the evidence points to a minority.
A: no; B: yes; C: no (evidence of fewer than half)
Problem 6 Sample Size for a Margin of Error
(a) n = (1.96/0.03)2(0.5)(0.5) = 1067.1, round up to 1068. (b) n = (1.96/0.03)2(0.3)(0.7) = 896.4, round up to 897.
(a) 1068; (b) 897
Problem 7 Writing Hypotheses
(a) H0: p = 0.80, Ha: p < 0.80. (b) H0: p = 0.25, Ha: p > 0.25. (c) H0: p = 0.50, Ha: p ≠ 0.50, where p is the true proportion of the relevant outcome.
(a) p < 0.80; (b) p > 0.25; (c) p ≠ 0.50
Problem 8 One-Sided z Test
p̂ = 148/200 = 0.74. SE = √(0.8 × 0.2/200) = 0.0283. z = (0.74 − 0.80)/0.0283 = −2.12. p-value = P(Z < −2.12) = 0.0169.
z ≈ −2.12, p-value ≈ 0.0169
Problem 9 Two-Sided z Test
p̂ = 0.56, SE = √(0.5 × 0.5/250) = 0.0316, z = (0.56 − 0.50)/0.0316 = 1.897. Two-sided p-value = 2(1 − 0.9711) = 0.0578. Since 0.0578 > 0.05, we fail to reject H0: there is not convincing evidence that the proportion differs from 0.50.
z ≈ 1.90, p-value ≈ 0.0578; fail to reject H0
Problem 10 Interpreting a p-Value
If exactly 50% of customers preferred the new design, the probability of getting a sample proportion at least as far from 0.50 as 0.56 (in either direction) in a random sample of 250 customers is about 0.058.
About 0.058 of samples would be this extreme if p = 0.50
Problem 11 Large Counts for a Test
np0 = 60(0.90) = 54 ≥ 10, but n(1 − p0) = 60(0.10) = 6 < 10. The condition fails, so the z test is not appropriate for this sample size.
Condition not met: n(1 − p0) = 6 < 10
Problem 12 Decisions and Significance Levels
0.032: reject at 0.05 (0.032 ≤ 0.05), fail to reject at 0.01. 0.067: fail to reject at both levels. 0.0004: reject at both levels.
0.032: reject at 0.05 only; 0.067: fail to reject at both; 0.0004: reject at both
Problem 13 Type I and Type II Errors
Type I: concluding that the infection rate is greater than 4% when it is actually 4% (or less); consequence: unnecessary changes to procedures. Type II: failing to find evidence that the rate is above 4% when it truly is higher; consequence: patients continue to face a higher risk.
Type I: reject a true H0; Type II: fail to reject a false H0
Problem 14 Factors That Affect Power
(a) A larger sample reduces the standard error, so the power increases. (b) A smaller α makes rejection harder, so power decreases. (c) A true value farther from p0 is easier to detect, so power increases.
(a) increase; (b) decrease; (c) increase
Problem 15 A Complete One-Proportion Test
State: H0: p = 0.33, Ha: p > 0.33, where p is the true proportion who recycle. Plan: one-proportion z test; random sample, 350 is less than 10% of the population, and 350(0.33) = 115.5 and 350(0.67) = 234.5 are at least 10. Do: p̂ = 0.38, SE = 0.02513, z = 1.989, p-value = 0.0233. Conclude: since the p-value is below 0.05, reject H0; there is convincing evidence that more than 33% of residents recycle.
z ≈ 1.99, p-value ≈ 0.0233; reject H0
Problem 16 Interval for a Difference
Conditions: both samples are random and independent; all counts 96, 54, 78, 82 are at least 10. p̂1 = 0.6400, p̂2 = 0.4875, difference = 0.1525. SE = √(0.6400(0.3600)/150 + 0.4875(0.5125)/160) = 0.0557. ME = 1.96 × 0.0557 = 0.1091.
(0.043, 0.262)
Problem 17 Using the Interval to Decide
The interval contains only positive values and does not include 0, so there is convincing evidence that the pass rate is higher with the program. We are 95% confident that the true pass rate with the program is between 4.3 and 26.2 percentage points higher than without it.
Yes; 0 is not in the interval
Problem 18 Two-Proportion z Test
H0: pA = pB; Ha: pA ≠ pB. Pooled proportion = 75/230 = 0.3261. Conditions: random independent samples and all expected counts are at least 10. SE = √(0.3261(0.6739)(1/120 + 1/110)) = 0.0619. z = (0.3750 − 0.2727)/0.0619 = 1.653; two-sided p-value = 0.0984. Since 0.0984 > 0.05, fail to reject: not convincing evidence that the proportions differ.
z ≈ 1.65, p-value ≈ 0.098; fail to reject H0
Problem 19 Computing Power
Reject H0 when p̂ exceeds 0.40 + 1.645√(0.4 × 0.6/300) = 0.40 + 1.645(0.02828) = 0.4465. If p = 0.47, the standard deviation of p̂ is √(0.47 × 0.53/300) = 0.02882. Power = P(p̂ > 0.4465) = P(Z > −0.815) = 0.7923.
Power ≈ 0.792
Problem 20 Poll Planning and Claim
(a) SE = √(0.54 × 0.46/900) = 0.01661; ME = 1.96 × 0.01661 = 0.0326; interval (0.507, 0.573). (b) The interval is entirely above 0.50, so yes. (c) ME is proportional to 1/√n; halving it requires 4 times the sample size, or 3,600 voters.
(a) (0.507, 0.573); (b) yes; (c) 4 times, 3,600 voters
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