AP Statistics Quiz: Difference Of Two Population Proportions Test

What this quiz covers

This quiz focuses on Difference Of Two Population Proportions Test, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Statistics.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. Because the p-value is 0.018, there is sufficient evidence that the population proportions of adults who received a flu shot are different in City 1 and City 2. (correct answer)
2. Because the p-value is 0.018, there is sufficient evidence that City 2 has a higher population proportion of adults who received a flu shot than City 1.
3. Because the p-value is 0.018, we can conclude that the samples have different proportions, so the cities must differ.
4. Because the p-value is 0.018, we should fail to reject $H_0$ because 0.018 is greater than 0.05.
5. Because the p-value is 0.018, living in City 1 causes people to be more likely to get a flu shot.

Explanation: This question evaluates understanding of a two-sided two-proportion z-test for differences in flu shot proportions between two cities. The p-value of 0.018 is less than alpha=0.05, so we reject the null hypothesis, concluding there is sufficient evidence of a difference in the population proportions. Choice B is a distractor because it specifies a direction (City 2 higher), but the two-sided alternative only supports a difference without indicating which is larger. Conclusions in two-proportion tests should emphasize inference to populations, not just observed sample differences, and avoid assuming causation from observational data. For two-sided tests, rejection means evidence of inequality, but further analysis like confidence intervals would be needed to determine direction. This highlights the importance of matching the conclusion to the test type and significance level.

Question 2

1. Reject $H_0$; there is sufficient evidence that the population renewal proportion is higher for Plan X than for Plan Y. (correct answer)
2. Reject $H_0$; there is sufficient evidence that the population renewal proportion is higher for Plan Y than for Plan X.
3. Fail to reject $H_0$; there is not sufficient evidence that the population renewal proportion for Plan X is higher than for Plan Y.
4. Reject $H_0$; Plan X causes members to renew at a higher rate than Plan Y.
5. Reject $H_0$; there is sufficient evidence that the two samples have different renewal proportions.

Explanation: This problem involves a one-tailed test with H_a: p_X > p_Y, testing if Plan X has a higher renewal proportion. With p-value = 0.012 < α = 0.05, we reject H₀. The correct conclusion is that there's sufficient evidence that the population renewal proportion is higher for Plan X than for Plan Y (choice A). Choice B reverses the direction. Choice C would be correct if we failed to reject H₀. Choice D incorrectly implies causation from an observational comparison. Choice E refers to sample proportions and doesn't specify direction. In one-tailed tests, the conclusion must match the alternative hypothesis direction, and we conclude about population parameters. The small p-value provides strong evidence supporting the claim that Plan X has a higher population renewal rate.

Question 3

1. Because the p-value is 0.18, there is convincing evidence that approval is different in the two regions.
2. Because the p-value is 0.18, there is not convincing evidence of a difference in the true approval proportions between Region North and Region South. (correct answer)
3. Because the p-value is 0.18, we conclude that 255 approvals in North and 207 approvals in South means North approves more in the population.
4. Because the p-value is 0.18, there is convincing evidence that Region South has a higher true approval proportion than Region North.
5. Because the p-value is 0.18, the policy causes voters in Region North to approve more than voters in Region South.

Explanation: This question tests understanding of two-proportion z-test conclusions for voter approval rates. With a p-value of 0.18 and α = 0.05, we fail to reject the null hypothesis since 0.18 > 0.05. This means there is not convincing evidence of a difference in the true approval proportions between the regions. Choice B correctly states this conclusion. Choice A incorrectly claims evidence of difference when the large p-value indicates otherwise. When we fail to reject H₀, we cannot conclude the proportions are different, regardless of the observed sample proportions. The two-sided test (≠) examines whether any difference exists, and the high p-value suggests the observed difference could reasonably occur by chance alone.

Question 4

1. Fail to reject $H_0$; there is not sufficient evidence that the population pass rates differ between Program A and Program B.
2. Reject $H_0$; there is sufficient evidence that the population pass rate for Program A differs from the population pass rate for Program B. (correct answer)
3. Reject $H_0$; there is sufficient evidence that Program A causes a higher pass rate than Program B.
4. Reject $H_0$; there is sufficient evidence that the sample pass rates differ between the two samples.
5. Reject $H_0$; there is sufficient evidence that the population pass rate for Program B differs from the population pass rate for Program A, because $p_B>p_A$.

Explanation: This question tests understanding of hypothesis testing for the difference of two population proportions. With a p-value of 0.006, which is less than α = 0.05, we reject the null hypothesis. The correct interpretation is that there is sufficient evidence that the population pass rates differ between Program A and Program B (choice B). Choice C incorrectly claims causation and specifies direction when we used a two-tailed test. Choice D incorrectly refers to sample proportions rather than population proportions. Choice E incorrectly uses sample notation (p_B > p_A) when discussing population parameters. When rejecting H₀ in a two-proportion test, we conclude there's evidence of a difference in population proportions, not sample proportions or causal relationships.

Question 5

1. Reject $H_0$; there is sufficient evidence that the population weekly recycling proportions differ between Town A and Town B.
2. Fail to reject $H_0$; there is not sufficient evidence that the population weekly recycling proportions differ between Town A and Town B. (correct answer)
3. Fail to reject $H_0$; therefore, Town A and Town B have exactly the same population weekly recycling proportion.
4. Fail to reject $H_0$; there is not sufficient evidence that the sample weekly recycling proportions differ between the two samples.
5. Reject $H_0$; there is sufficient evidence that Town B's population weekly recycling proportion is greater than Town A's.

Explanation: This problem involves a two-tailed test for comparing recycling proportions between towns. With p-value = 0.29 > α = 0.05, we fail to reject the null hypothesis. The correct conclusion is that there's not sufficient evidence that the population weekly recycling proportions differ between Town A and Town B (choice B). Choice C incorrectly claims the proportions are exactly equal - failing to reject H₀ doesn't prove equality. Choice D incorrectly refers to sample proportions. Choice E would require rejecting H₀ and having a one-tailed test. When we fail to reject H₀, we cannot conclude the null hypothesis is true; we can only say there's insufficient evidence to support the alternative hypothesis. This distinction is crucial in hypothesis testing.

Question 6

1. Because the p-value is 0.002, there is convincing evidence that Method B has a higher population proportion of daily homework completion than Method A.
2. Because the p-value is 0.002, there is convincing evidence that the population proportion of students who complete homework daily is higher with Method A than with Method B. (correct answer)
3. Because the p-value is 0.002, we can conclude that Method A caused more of the sampled students to complete homework daily.
4. Because the p-value is 0.002, we can conclude that 126/180 is greater than 120/200 in these samples.
5. Because the p-value is 0.002, there is not sufficient evidence of a difference in the population proportions between the two methods.

Explanation: This question tests the skill of interpreting a two-proportion z-test for comparing population proportions of homework completion between two methods. With a p-value of 0.002 less than the typical alpha of 0.05, we reject the null hypothesis, providing convincing evidence that the population proportion is higher for Method A than Method B, as stated in the alternative hypothesis. A common distractor is choice A, which reverses the direction by claiming Method B is higher, but the sample data and alternative hypothesis support Method A having the higher proportion. In drawing conclusions from a two-proportion z-test, we focus on population parameters rather than sample statistics alone and avoid causal claims unless the study design supports it, such as through randomization. Here, the test allows us to infer a difference in populations but not that one method causes better completion rates. Remember, the p-value indicates the strength of evidence against the null, and rejecting it aligns with the direction specified in the alternative hypothesis.

Question 7

1. Reject $H_0$; there is sufficient evidence that the population proportion of teens who drink soda daily is higher in Region A than in Region B. (correct answer)
2. Fail to reject $H_0$; there is not sufficient evidence that the population proportion is higher in Region A than in Region B.
3. Reject $H_0$; there is sufficient evidence that the population proportion is higher in Region B than in Region A.
4. Because the p-value is 0.084, Region A causes teens to drink soda daily more often than Region B.
5. Because the p-value is 0.084, we can conclude that the two sample proportions are different, so the population proportions must be different.

Explanation: This question assesses a one-sided two-proportion z-test for daily soda drinking proportions between regions. The p-value of 0.084 is less than alpha=0.10, so we reject the null, finding sufficient evidence that Region A has a higher population proportion than Region B. Choice B distracts by suggesting failure to reject, but 0.084 falls below the given 0.10 threshold. In conclusions, focus on populations and note that different alphas can change decisions—here, it would fail at 0.05 but rejects at 0.10. Avoid causation, as regions likely involve observational data. This shows the role of significance level in decision-making.

Question 8

1. Reject $H_0$; there is convincing evidence that the population proportion reporting reduced pain is higher for Program P than for Program Q.
2. Fail to reject $H_0$; there is not convincing evidence that Program P has a higher population reduced-pain proportion than Program Q. (correct answer)
3. Reject $H_0$; there is convincing evidence that the population proportion reporting reduced pain is higher for Program Q than for Program P.
4. Because the p-value is 0.11, Program P and Program Q work equally well for all athletes.
5. Fail to reject $H_0$; therefore 58/100 and 49/100 are not meaningfully different in these samples.

Explanation: This question tests whether Program P has a higher population proportion of athletes reporting reduced pain than Program Q. The p-value (0.11) exceeds α = 0.05, so we fail to reject H₀. This means there is not convincing evidence that Program P has a higher population reduced-pain proportion than Program Q, despite the sample showing 58/100 versus 49/100. Choice D incorrectly interprets the p-value as proving the programs work equally well. Choice E misunderstands the purpose of hypothesis testing, which is to make inferences about populations. When we fail to reject H₀, we're saying the sample difference could reasonably occur by chance if the population proportions were equal.

Question 9

1. Reject $H_0$; there is convincing evidence that Method A results in a higher population proportion of A grades than Method B.
2. Fail to reject $H_0$; there is not convincing evidence that Method A has a higher population A-rate than Method B. (correct answer)
3. Reject $H_0$; there is convincing evidence that Method B results in a higher population proportion of A grades than Method A.
4. Because students were randomly assigned, the p-value of 0.09 proves Method A causes a higher A-rate.
5. Fail to reject $H_0$; therefore the two population proportions of A grades are exactly equal.

Explanation: This problem involves testing whether Method A leads to a higher proportion of A grades than Method B. The p-value (0.09) is greater than α = 0.05, so we fail to reject the null hypothesis. This means there is not convincing evidence that Method A has a higher population A-rate than Method B, despite the sample showing 33/80 versus 25/80. Choice D incorrectly claims the p-value proves causation. Choice E incorrectly states that failing to reject H₀ proves the proportions are exactly equal. When the p-value exceeds the significance level, we lack sufficient evidence to support the alternative hypothesis, but this doesn't prove the null hypothesis is true.

Question 10

1. Because the p-value is 0.004, there is convincing evidence that Town A's true support proportion is greater than Town B's. (correct answer)
2. Because the p-value is 0.004, there is not convincing evidence that Town A's true support proportion is greater than Town B's.
3. Because the p-value is 0.004, there is convincing evidence that Town B's true support proportion is greater than Town A's.
4. Because the p-value is 0.004, we conclude that 150 households in Town A will always support the ordinance.
5. Because the p-value is 0.004, living in Town A causes households to support the ordinance more than living in Town B.

Explanation: This question involves a one-sided test comparing recycling ordinance support between towns. With a p-value of 0.004 and α = 0.05, we reject the null hypothesis since 0.004 < 0.05. The alternative hypothesis H_a: p_A > p_B specifically tests whether Town A has higher support than Town B. The very small p-value provides convincing evidence for this directional claim. Choice A correctly states this conclusion. Choice B incorrectly claims no evidence when the p-value is well below α. A p-value of 0.004 indicates the observed data would be very unlikely if the null hypothesis were true. In one-sided tests, small p-values provide strong evidence for the specific directional claim in the alternative hypothesis.

Question 11

1. Because the p-value is 0.008, there is convincing evidence that the new design causes a different purchase proportion than the old design. (correct answer)
2. Because the p-value is 0.008, there is not convincing evidence of a difference in the true purchase proportions.
3. Because the p-value is 0.008, we conclude that 95 of 620 is greater than 78 of 600, so the true proportions must be different.
4. Because the p-value is 0.008, there is convincing evidence that Design O has a higher true purchase proportion than Design N.
5. Because the p-value is 0.008, the samples prove that exactly $p_N-p_O$ equals the observed difference in sample proportions.

Explanation: This question involves a two-proportion z-test comparing purchase rates for different checkout designs. With a p-value of 0.008 and α = 0.01, we reject the null hypothesis since 0.008 < 0.01. This provides convincing evidence that the true purchase proportions differ between the designs. Choice A correctly states this conclusion. Choice B incorrectly claims no evidence of difference when the small p-value indicates strong evidence. The two-sided alternative hypothesis (≠) means we can conclude the proportions differ but cannot specify which is higher without additional analysis. Remember that rejecting H₀ in a two-sided test only establishes that a difference exists, not its direction.

Question 12

1. Because the p-value is 0.21, there is convincing evidence that City A has a higher true flu-shot proportion than City B.
2. Because the p-value is 0.21, there is not convincing evidence of a difference in the true flu-shot proportions between City A and City B. (correct answer)
3. Because the p-value is 0.21, we conclude the sample proportions are equal in the two samples.
4. Because the p-value is 0.21, there is convincing evidence that City B has a different true flu-shot proportion than City A.
5. Because the p-value is 0.21, getting a flu shot causes adults to be more likely to live in City A than City B.

Explanation: This question tests understanding of two-proportion z-test conclusions when comparing flu shot rates between cities. With a p-value of 0.21 and α = 0.05, we fail to reject the null hypothesis since 0.21 > 0.05. This means there is not convincing evidence of a difference in the true flu-shot proportions between the cities. Choice B correctly states this conclusion. Choice A incorrectly claims evidence for City A having a higher proportion, but the two-sided test doesn't support directional claims and the p-value is too large anyway. When p-values exceed the significance level, we conclude there's insufficient evidence to claim the population proportions differ, not that they're equal or that causation exists.

Question 13

1. Because the p-value is 0.041, fail to reject $H_0$; there is not sufficient evidence that Treatment 1 has a higher population germination proportion than Treatment 2.
2. Because the p-value is 0.041, reject $H_0$; there is sufficient evidence that Treatment 1 has a higher population germination proportion than Treatment 2. (correct answer)
3. Because the p-value is 0.041, reject $H_0$; there is sufficient evidence that Treatment 2 has a higher population germination proportion than Treatment 1.
4. Because the p-value is 0.041, we can conclude that Treatment 1 causes more germination in all seeds.
5. Because the p-value is 0.041, we can conclude that the sample proportion 111/150 is greater than 104/160.

Explanation: This question focuses on a one-sided two-proportion z-test for germination proportions between treatments. The p-value of 0.041 is less than alpha=0.05, so we reject the null, finding sufficient evidence that Treatment 1 has a higher population germination proportion than Treatment 2. Choice C distracts by reversing the direction, claiming Treatment 2 is higher, which contradicts the alternative hypothesis and sample data. When concluding from such tests, we infer about populations, not causation, unless the study is experimental with randomization. Remember, rejection supports the alternative's direction, but does not mean the effect applies to all cases—it's about overall proportions. This underscores matching the conclusion to the hypothesis direction and avoiding overgeneralization.

Question 14

1. Fail to reject $H_0$; there is not sufficient evidence that Machine 1 has a lower population defect proportion than Machine 2.
2. Reject $H_0$; there is sufficient evidence that Machine 1 has a lower population defect proportion than Machine 2. (correct answer)
3. Reject $H_0$; there is sufficient evidence that Machine 2 has a lower population defect proportion than Machine 1.
4. Because the p-value is 0.032, Machine 1 causes fewer defects than Machine 2.
5. Because the p-value is 0.032, we can conclude only that 18/500 is less than 27/450 in these samples.

Explanation: This question involves a one-sided two-proportion z-test for defect proportions between machines. With a p-value of 0.032 less than alpha=0.05, we reject the null, providing sufficient evidence that Machine 1 has a lower population defect proportion than Machine 2. Choice C is a distractor, wrongly claiming Machine 2 has the lower proportion, ignoring the alternative's direction. Conclusions should target population proportions and refrain from causation claims without experimental evidence. Rejection supports the specified inequality, but it's probabilistic, not absolute. This example highlights verifying the hypothesis direction in one-sided tests.

Question 15

1. Reject $H_0$; there is strong evidence that the population purchase rate is lower for Layout X than for Layout Y.
2. Fail to reject $H_0$; there is not sufficient evidence that the population purchase rate for Layout X is lower than for Layout Y. (correct answer)
3. Fail to reject $H_0$; therefore the population purchase rates for Layout X and Layout Y are equal.
4. Reject $H_0$; there is strong evidence that the population purchase rate is lower for Layout Y than for Layout X.
5. Because the p-value is 0.91, Layout Y causes fewer purchases than Layout X.

Explanation: This question assesses interpreting a one-sided two-proportion z-test for website layout purchase rates. With a p-value of 0.91 much greater than alpha=0.01, we fail to reject the null hypothesis, indicating insufficient evidence that Layout X has a lower population purchase rate than Layout Y. Choice A is a distractor as it incorrectly suggests rejecting the null despite the high p-value, which does not provide evidence against equality. In two-proportion z-tests, failing to reject means we lack evidence for the alternative, but it does not prove the null is true—populations could still differ slightly. Avoid confusing sample differences with population inferences, and note that causation might be implied if layouts were randomly assigned, but the conclusion stays probabilistic. This example shows how a p-value far above alpha leads to non-rejection, emphasizing careful comparison to the significance level.

Question 16

1. Reject $H_0$; there is sufficient evidence that the population proportions who favor Candidate Q differ between the two age groups.
2. Fail to reject $H_0$; there is not sufficient evidence that the population proportions who favor Candidate Q differ between the two age groups. (correct answer)
3. Fail to reject $H_0$; therefore the two age groups have exactly the same population proportion who favor Candidate Q.
4. Because the p-value is 0.19, voters ages 35 and older have a higher population proportion favoring Candidate Q than voters ages 18–34.
5. Because the p-value is 0.19, age causes differences in support for Candidate Q.

Explanation: This question evaluates a two-sided two-proportion z-test for candidate support proportions between age groups. With a p-value of 0.19 greater than alpha=0.05, we fail to reject the null, indicating insufficient evidence of differing population proportions. Choice A is a distractor, incorrectly advocating rejection despite the p-value exceeding alpha. Conclusions must stress that non-rejection does not equate to proof of equality, only lack of evidence for difference. Refrain from causal claims like age causing preferences without further evidence. This demonstrates interpreting non-significant results in two-sided tests.

Question 17

1. Because the p-value is 0.006, there is convincing evidence that the population proportion of homeowners who support the policy is higher than the population proportion of renters who support the policy. (correct answer)
2. Because the p-value is 0.006, there is convincing evidence that the population proportion of renters who support the policy is higher than the population proportion of homeowners who support the policy.
3. Because the p-value is 0.006, we can conclude that homeowners in the sample support the policy more than renters in the sample, so the policy will pass.
4. Because the p-value is 0.006, the recycling policy causes homeowners to support it more than renters.
5. Because the p-value is 0.006, we should fail to reject $H_0$ since 0.006 is less than 0.05.

Explanation: This question examines a one-sided two-proportion z-test for support proportions of a recycling policy between homeowners and renters. With a p-value of 0.006 below alpha=0.05, we reject the null, concluding convincing evidence that homeowners have a higher population support proportion than renters. Choice B is a distractor, incorrectly stating renters have higher support, which opposes the alternative hypothesis. In two-proportion conclusions, emphasize population-level differences without claiming causation, as this appears observational. The test allows inference beyond samples, but predictions like policy passage require more context. Always align the conclusion with the alternative's direction when rejecting.

Question 18

1. Reject $H_0$; there is sufficient evidence that the population pass rates differ between Program A and Program B.
2. Fail to reject $H_0$; there is not sufficient evidence that the population pass rates differ between Program A and Program B. (correct answer)
3. Fail to reject $H_0$; therefore the two programs have exactly the same population pass rate.
4. Reject $H_0$; there is sufficient evidence that Program A has a higher population pass rate than Program B.
5. Because the p-value is 0.27, Program B causes students to pass at a higher rate than Program A.

Explanation: This question tests interpretation of a two-sided two-proportion z-test for certification pass rates between programs. The p-value of 0.27 exceeds alpha=0.05, leading us to fail to reject the null, with insufficient evidence of differing population pass rates. Choice C distracts by claiming failure to reject proves exact equality, but it only means lack of evidence for a difference. Two-proportion test conclusions should note that non-rejection does not confirm the null—subtle differences might exist undetected. Avoid causal language unless supported by design, and focus on populations, not samples. This illustrates how high p-values result in non-rejection in two-sided tests.

Question 19

1. Fail to reject $H_0$; there is not sufficient evidence that the old process has a higher population readmission proportion than the new checklist.
2. Reject $H_0$; there is sufficient evidence that the old process has a higher population readmission proportion than the new checklist. (correct answer)
3. Reject $H_0$; there is sufficient evidence that the new checklist has a higher population readmission proportion than the old process.
4. Because the p-value is 0.047, the new checklist causes fewer readmissions for every patient.
5. Because the p-value is 0.047, we can conclude only that 52/260 is greater than 34/240 in these samples.

Explanation: This question focuses on a one-sided two-proportion z-test for readmission proportions between discharge processes. The p-value of 0.047 is less than alpha=0.05, so we reject the null, concluding sufficient evidence that the old process has a higher population readmission proportion than the new checklist. Choice C distracts by reversing the direction, claiming the new has higher, against the alternative. In conclusions, infer to populations and note potential causation if the checklist was experimentally implemented, but avoid absolutes like 'every patient.' Rejection supports the alternative's inequality, emphasizing improved outcomes. This highlights careful direction in one-sided tests.

Question 20

1. Reject $H_0$; there is convincing evidence that the population return rates differ between Program R and Program S.
2. Fail to reject $H_0$; there is not convincing evidence of a difference in the population return rates between the programs. (correct answer)
3. Reject $H_0$; there is convincing evidence that Program R has a lower population return rate than Program S.
4. Fail to reject $H_0$; therefore the two programs have exactly the same population return rate.
5. Because the p-value is large, Program S caused the sample return rate to be similar to Program R.

Explanation: This is a two-tailed test comparing return rates between two residence hall programs. The p-value (0.68) is much larger than α = 0.05, so we fail to reject H₀. This means there is not convincing evidence of a difference in the population return rates between Program R and Program S. Choice D incorrectly interprets failing to reject H₀ as proof the rates are exactly equal—we simply lack evidence they differ. Choice E incorrectly attributes causation to the programs based on the p-value. When the p-value is large in a two-tailed test, the observed difference between samples (155/200 vs 150/200) is consistent with what we'd expect if the population proportions were equal.