Difference of Two Population Proportions (Test) - AP Statistics

A school district compared the proportion of students who pass a math proficiency exam in two independent random samples: 44 of 80 students from School X passed and 70 of 140 students from School Y passed. A two-proportion $z$ test was conducted at $\alpha=0.05$ to test $H_0: p_X=p_Y$ versus $H_a: p_X>p_Y$ , where $p_X$ and $p_Y$ are the true passing proportions at Schools X and Y. The p-value was 0.12. What conclusion is appropriate?
Answer: Because the p-value is 0.12, there is not convincing evidence that School X's true passing proportion is higher than School Y's. This question tests interpretation of a one-sided two-proportion z-test comparing math proficiency rates. With a p-value of 0.12 and α = 0.05, we fail to reject the null hypothesis since 0.12 > 0.05. The alternative hypothesis was $H_a$ : $p_X$ > $p_Y$ (testing if School X has a higher proportion), but the large p-value means there's not convincing evidence for this claim. Choice B correctly states this conclusion. Choice A incorrectly claims evidence when the p-value exceeds α. For one-sided tests, failing to reject H₀ means we lack evidence for the specific directional claim in $H_a$ . The p-value represents the probability of observing our data or more extreme if the null hypothesis were true.
A company tested whether a new website layout changes the proportion of visitors who make a purchase. A random sample of 400 visitors shown the old layout had 52 purchases; a separate random sample of 420 visitors shown the new layout had 70 purchases. A two-proportion $z$ test was performed for $H_0: p_{old}=p_{new}$ versus $H_a: p_{old}\ne p_{new}$ , and the p-value was $0.041$ . At the $\alpha=0.05$ level, what conclusion is appropriate?
Answer: Reject $H_0$ ; there is sufficient evidence that the population purchase proportions differ between the old and new layouts. This question tests interpretation of a two-tailed hypothesis test for difference in proportions. With p-value = 0.041 < α = 0.05, we reject the null hypothesis. The correct conclusion is that there's sufficient evidence that the population purchase proportions differ between the old and new layouts (choice B). Choice C incorrectly claims causation and specifies a direction when the test was two-tailed. Choice D incorrectly refers to sample proportions rather than population proportions. Choice E would be correct if the p-value were greater than 0.05. When conducting hypothesis tests, we make inferences about population parameters based on sample data, and two-tailed tests only allow us to conclude that a difference exists, not which direction or causation.
A teacher compares pass rates on a certification exam for two review programs. From a random sample of 120 students who used Program A, 78 passed. From a separate random sample of 140 students who used Program B, 100 passed. A two-proportion $z$ test was conducted for $H_0: p_A=p_B$ versus $H_a: p_A\ne p_B$ , resulting in a p-value of 0.27. Using $\alpha=0.05$ , what conclusion is appropriate?
Answer: Fail to reject $H_0$ ; there is not sufficient evidence that the population pass rates differ between Program A and Program B. 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.
A city council wants to know whether support for a new recycling policy is higher among homeowners than renters. In a random sample of 300 homeowners, 201 support the policy. In a random sample of 260 renters, 156 support the policy. A two-proportion $z$ test for $H_0: p_H=p_R$ versus $H_a: p_H>p_R$ gave a p-value of 0.006. At the 5% significance level, what conclusion is appropriate?
Answer: 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. 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.
A fitness app company compared the proportion of users who remain active after 30 days for two independent random samples: 92 of 160 users who received daily reminders stayed active, and 75 of 155 users who did not receive reminders stayed active. A two-proportion $z$ test at $\alpha=0.10$ tested $H_0: p_R=p_N$ versus $H_a: p_R>p_N$ , where $p_R$ is the true active-after-30-days proportion with reminders and $p_N$ is the true proportion with no reminders. The p-value was 0.07. What conclusion is appropriate?
Answer: Because the p-value is 0.07, there is convincing evidence that reminders increase the true proportion of users who remain active after 30 days. This question involves a one-sided test examining whether reminders increase user retention. With a p-value of 0.07 and α = 0.10, we reject the null hypothesis since 0.07 < 0.10. The alternative hypothesis $H_a$ : $p_R$ > $p_N$ tests whether reminders lead to a higher active proportion. Since we reject H₀, there is convincing evidence that reminders increase the true proportion of active users. Choice A correctly states this conclusion. Choice B incorrectly claims no evidence when the p-value is below the significance level. Note that α = 0.10 is less stringent than the typical 0.05, but conclusions are based on the stated significance level. Statistical significance at α = 0.10 still provides evidence for the alternative hypothesis.
A city evaluated whether the proportion of commuters who use public transit differs between two years. In a random sample of 350 commuters surveyed in 2023, 98 reported using public transit; in a separate random sample of 360 commuters surveyed in 2025, 125 reported using public transit. A two-proportion $z$ test was conducted for $H_0: p_{2023}=p_{2025}$ versus $H_a: p_{2023}\ne p_{2025}$ , and the p-value was $0.009$ . Using $\alpha=0.05$ , what conclusion is appropriate?
Answer: Reject $H_0$ ; there is sufficient evidence that the population public-transit use proportions differ between 2023 and 2025. This problem involves a two-tailed test comparing public transit use between years. With p-value = 0.009 < α = 0.05, we reject the null hypothesis. The correct conclusion is that there's sufficient evidence that the population public-transit use proportions differ between 2023 and 2025 (choice B). Choice A would be correct if p-value > 0.05. Choice C incorrectly implies causation from temporal data. Choice D incorrectly refers to sample proportions. Choice E specifies a direction when the test was two-tailed. The small p-value provides strong evidence of a change in population transit use between the years. Two-tailed tests allow us to conclude a difference exists but not which year had higher usage without examining the sample data.
A political scientist compared the proportion of registered voters who approve of a policy in two independent random samples. Among 500 voters in Region North, 255 approved; among 450 voters in Region South, 207 approved. A two-proportion $z$ test at $\alpha=0.05$ tested $H_0: p_N=p_S$ versus $H_a: p_N\ne p_S$ , where $p_N$ and $p_S$ are the true approval proportions in the two regions. The p-value was 0.18. What conclusion is appropriate?
Answer: 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. 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.
A doctor compared the proportion of patients who experience a side effect for two medications using independent random samples. For Medication M, 26 of 130 patients had the side effect; for Medication P, 18 of 120 patients had the side effect. A two-proportion $z$ test at $\alpha=0.05$ tested $H_0: p_M=p_P$ versus $H_a: p_M\ne p_P$ , where $p_M$ and $p_P$ are the true side-effect proportions for the two medications. The p-value was 0.28. What conclusion is appropriate?
Answer: Because the p-value is 0.28, there is not convincing evidence of a difference in the true side-effect proportions between Medication M and Medication P. This question tests understanding of two-proportion z-test conclusions when comparing medication side effects. With a p-value of 0.28 and α = 0.05, we fail to reject the null hypothesis since 0.28 > 0.05. This means there is not convincing evidence of a difference in the true side-effect proportions between the medications. Choice B correctly states this conclusion. Choice A incorrectly claims evidence of difference when the large p-value indicates otherwise. Even though Medication M had a higher observed proportion (26/130 vs 18/120), the p-value tells us this difference could reasonably occur by chance. Statistical tests account for sample sizes and variability, not just raw counts.
A hospital tests whether a new discharge checklist reduces the proportion of patients readmitted within 30 days. Under the old process, a random sample of 260 patients had 52 readmissions. Under the new checklist, a separate random sample of 240 patients had 34 readmissions. A two-proportion $z$ test was performed for $H_0: p_{old}=p_{new}$ versus $H_a: p_{old}>p_{new}$ , and the p-value was 0.047. Using $\alpha=0.05$ , what conclusion is appropriate?
Answer: Reject $H_0$ ; there is sufficient evidence that the old process has a higher population readmission proportion than the new checklist. 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.
A company tests two website designs to see which yields a higher proportion of visitors who make a purchase. From independent random samples, 142 of 1000 visitors purchased on Design 1 and 130 of 1000 purchased on Design 2. A two-proportion $z$ test for $H_0: p_1-p_2=0$ versus $H_a: p_1-p_2\ne 0$ produced a p-value of $0.42$ . At $\alpha=0.05$ , what conclusion is appropriate?
Answer: Fail to reject $H_0$ ; there is not convincing evidence of a difference in the population purchase proportions. This problem involves a two-tailed test for the difference between two population proportions. The p-value (0.42) is much larger than α = 0.05, so we fail to reject the null hypothesis. This means there is not convincing evidence of a difference in the population purchase proportions between the two website designs. Choice E is incorrect because failing to reject H₀ does not prove the proportions are equal—it only means we lack evidence they differ. Choice D incorrectly interprets the large p-value as showing causation. When we fail to reject H₀, we conclude there is insufficient evidence to support the alternative hypothesis, not that the null hypothesis is proven true.
A botanist compares the germination success of two seed treatments. In a random sample of 150 seeds given Treatment 1, 111 germinated. In a separate random sample of 160 seeds given Treatment 2, 104 germinated. A two-proportion $z$ test was performed for $H_0: p_1=p_2$ versus $H_a: p_1>p_2$ and resulted in a p-value of 0.041. Using $\alpha=0.05$ , what conclusion is appropriate?
Answer: 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. 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.
A manufacturer compared defect rates for two production lines using independent random samples. Line 1 had 18 defects in a sample of 300 items; Line 2 had 9 defects in a sample of 250 items. A two-proportion $z$ test at $\alpha=0.05$ tested $H_0: p_1=p_2$ versus $H_a: p_1>p_2$ , where $p_1$ and $p_2$ are the true defect proportions for Lines 1 and 2. The p-value was 0.03. What conclusion is appropriate?
Answer: Because the p-value is 0.03, there is convincing evidence that Line 1 has a higher true defect proportion than Line 2. This question involves a one-sided test comparing defect rates between production lines. With a p-value of 0.03 and α = 0.05, we reject the null hypothesis since 0.03 < 0.05. The alternative hypothesis $H_a$ : $p_1$ > $p_2$ specifically tests whether Line 1 has a higher defect proportion than Line 2. Since we reject H₀, there is convincing evidence supporting this directional claim. Choice A correctly states this conclusion. Choice B incorrectly claims no evidence when the p-value is below α. In one-sided tests, rejecting H₀ provides evidence for the specific direction stated in $H_a$ . Statistical significance doesn't imply causation or guarantee exact replication of sample results.
A fitness center compared the proportion of members who renew their membership after one year under two pricing plans. In a random sample of 180 members on Plan X, 117 renewed; in a random sample of 200 members on Plan Y, 110 renewed. A two-proportion $z$ test was conducted for $H_0: p_X=p_Y$ versus $H_a: p_X>p_Y$ , and the p-value was $0.012$ . Using $\alpha=0.05$ , what conclusion is appropriate?
Answer: Reject $H_0$ ; there is sufficient evidence that the population renewal proportion is higher for Plan X than for Plan Y. 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.
A public health researcher compares the proportion of adults who get at least 7 hours of sleep on weeknights in two cities. In independent random samples, 96 of 200 adults in City X reported at least 7 hours, and 118 of 200 adults in City Y reported at least 7 hours. A two-proportion $z$ test was conducted for $H_0: p_X-p_Y=0$ versus $H_a: p_X-p_Y<0$ , yielding a p-value of $0.012$ . Using $\alpha=0.05$ , what conclusion is appropriate?
Answer: Reject $H_0$ ; there is convincing evidence that the population proportion with at least 7 hours of sleep is lower in City X than in City Y. This question tests a one-sided hypothesis where Ha: $p_X$ - $p_Y$ < 0, meaning we're testing if City X has a lower proportion than City Y. The p-value (0.012) is less than α = 0.05, so we reject H₀. Since the alternative hypothesis states $p_X$ < $p_Y$ , we conclude there is convincing evidence that the population proportion with at least 7 hours of sleep is lower in City X than in City Y. Choice D incorrectly claims causation from an observational study. Choice E incorrectly limits the conclusion to samples only. In hypothesis testing, rejecting H₀ with a left-tailed test (Ha: p₁ - p₂ < 0) provides evidence that the first population proportion is less than the second.
A company compares the proportion of defective items produced by two machines. In a random sample of 500 items from Machine 1, 18 were defective. In a random sample of 450 items from Machine 2, 27 were defective. A two-proportion $z$ test was performed for $H_0: p_1=p_2$ versus $H_a: p_1<p_2$ , and the p-value was 0.032. At the 5% significance level, what conclusion is appropriate?
Answer: Reject $H_0$ ; there is sufficient evidence that Machine 1 has a lower population defect proportion than Machine 2. 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.
A public health researcher compared the proportion of adults who received a flu shot in two counties. In a random sample of 250 adults from County 1, 155 reported getting a flu shot; in a random sample of 220 adults from County 2, 110 reported getting a flu shot. A two-proportion $z$ test was conducted for $H_0: p_1=p_2$ versus $H_a: p_1>p_2$ , and the p-value was $0.018$ . Using $\alpha=0.05$ , what conclusion is appropriate?
Answer: Reject $H_0$ ; there is sufficient evidence that the population flu-shot proportion in County 1 is greater than in County 2. This problem involves a one-tailed test for comparing two population proportions. The alternative hypothesis $H_a$ : p₁ > p₂ indicates we're testing if County 1's proportion is greater than County 2's. With p-value = 0.018 < α = 0.05, we reject H₀. The correct conclusion is that there's sufficient evidence that the population flu-shot proportion in County 1 is greater than in County 2 (choice A). Choice C reverses the direction of the inequality. Choice D incorrectly implies causation from an observational study. Choice E refers to sample proportions instead of population proportions. In one-tailed tests, the conclusion must match the direction specified in the alternative hypothesis, and we always make inferences about population parameters, not sample statistics.
A wildlife biologist compares the proportion of a certain bird species that successfully nests in two different habitats. In independent random samples, 41 of 90 nests succeeded in Habitat 1 and 52 of 90 nests succeeded in Habitat 2. A two-proportion $z$ test was performed for $H_0: p_1-p_2=0$ versus $H_a: p_1-p_2\ne 0$ , with p-value $0.047$ . Using $\alpha=0.05$ , what conclusion is appropriate?
Answer: Reject $H_0$ ; there is convincing evidence that the population nesting success proportions differ between Habitat 1 and Habitat 2. This is a two-tailed test (Ha: p₁ - p₂ ≠ 0) comparing nesting success proportions between two habitats. The p-value (0.047) is less than α = 0.05, so we reject H₀. For a two-tailed test, rejecting H₀ means there is convincing evidence that the population proportions differ, but it doesn't specify which is higher. Choice C incorrectly specifies a direction when the two-tailed test only establishes a difference exists. Choice E incorrectly claims causation about habitats. When we reject H₀ in a two-tailed test, we conclude the population proportions are different, but additional analysis would be needed to determine which is larger.
A school compares the proportion of students who pass an Algebra end-of-course exam after using two different review programs. In a random sample, 78 of 120 students using Program A passed, and 64 of 120 students using Program B passed. A two-proportion $z$ test was performed for $H_0: p_A-p_B=0$ versus $H_a: p_A-p_B>0$ , and the p-value was $0.018$ . Using $\alpha=0.05$ , what conclusion is appropriate?
Answer: Reject $H_0$ ; there is convincing evidence that the population pass rate is higher with Program A than with Program B. This question tests understanding of hypothesis test conclusions for comparing two population proportions. Since the p-value (0.018) is less than α = 0.05, we reject the null hypothesis. The alternative hypothesis states $p_A$ - $p_B$ > 0, which means $p_A$ > $p_B$ , so we have convincing evidence that Program A has a higher population pass rate than Program B. Choice D incorrectly claims causation and refers only to the sample rather than making an inference about populations. Choice E incorrectly states we can only conclude about samples, not populations, but the entire purpose of hypothesis testing is to make inferences about populations. When we reject H₀ in a one-sided test with Ha: $p_A$ - $p_B$ > 0, we conclude there is convincing evidence that the first population proportion exceeds the second.
An online retailer compares two website layouts. Layout X was shown to a random sample of 400 visitors, and 92 made a purchase. Layout Y was shown to a separate random sample of 380 visitors, and 78 made a purchase. A two-proportion $z$ test for $H_0: p_X=p_Y$ versus $H_a: p_X<p_Y$ produced a p-value of 0.91. At the 1% significance level, what conclusion is appropriate?
Answer: Fail to reject $H_0$ ; there is not sufficient evidence that the population purchase rate for Layout X is lower than for Layout Y. 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.
A researcher compares the proportion of defective parts produced by two machines. From independent random samples, Machine M had 18 defectives out of 300 parts, and Machine N had 9 defectives out of 300 parts. A two-proportion $z$ test for $H_0: p_M-p_N=0$ versus $H_a: p_M-p_N>0$ yielded a p-value of $0.03$ . At $\alpha=0.05$ , what conclusion is appropriate?
Answer: Reject $H_0$ ; there is convincing evidence that the population defective proportion is higher for Machine M than for Machine N. This question compares defective proportions between two machines, testing if Machine M has a higher defective rate. The p-value (0.03) is less than α = 0.05, so we reject H₀. Since Ha: $p_M$ - $p_N$ > 0, we conclude there is convincing evidence that Machine M has a higher population defective proportion than Machine N. Choice D incorrectly claims the result applies to every sample rather than being a population inference. Choice E incorrectly suggests we can only conclude about samples, not populations. In quality control contexts, rejecting H₀ with a one-sided test provides evidence about the long-run (population) defect rates of the machines, not just the observed samples.

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