Justifying Claims Based on Confidence Interval - AP Statistics

Two brands of batteries are tested for the proportion that last at least 10 hours. A 95% confidence interval for the difference in proportions, defined as $p_{X} - p_{Y}$ (Brand X minus Brand Y), is $(-0.20, -0.05)$ . A consumer claims, "Brand X has a lower proportion lasting at least 10 hours than Brand Y." Is the claim supported by the confidence interval?
Answer: Yes, because the entire interval is below 0, suggesting $p_X < p_Y$ . The skill here is justifying claims based on confidence intervals for proportion differences, determining if the interval supports a claim that one brand has a lower durability proportion than another. The 95% CI for $p_X$ - $p_Y$ is (-0.20, -0.05), entirely negative and excluding zero, suggesting $p_X$ < $p_Y$ , which supports the consumer's claim. This means Brand X likely has a lower proportion lasting at least 10 hours. Distractor choice D wrongly claims a negative interval means $p_X$ > $p_Y$ , confusing the sign of the difference. Mini-lesson: to claim p₁ < p₂ using CI for p₁ - p₂, the interval must be entirely below zero; if it straddles zero, no strong evidence for inequality. Pay attention to which proportion is subtracted to avoid reversing conclusions. This approach aligns with hypothesis testing at the corresponding significance level.
A survey compares the proportion of adults who approve of a policy in 2024 versus 2025. A 95% confidence interval for the difference in proportions, defined as $p_{2024} - p_{2025}$ , is $( -0.04,\ 0.10)$ . A reporter claims, "Approval was higher in 2024 than in 2025." Is the claim supported by the confidence interval?
Answer: No, because the interval includes 0, so higher approval in 2024 is not clearly supported. This question tests understanding of confidence intervals that span both negative and positive values. The 95% confidence interval (-0.04, 0.10) for $p_{2024}$ - $p_{2025}$ includes negative values, zero, and positive values. The presence of 0 in the interval means "no difference in approval" is a plausible scenario. Additionally, the negative values suggest that approval could have been lower in 2024 than in 2025. Since the interval includes cases where $p_{2024}$ ≤ $p_{2025}$ , the claim that "Approval was higher in 2024 than in 2025" is not clearly supported. For a directional claim to be supported, the entire confidence interval must exclude 0 and be on the appropriate side (positive for "higher," negative for "lower").
A school compares the proportion of students who prefer online homework in Grade 9 versus Grade 10. A 95% confidence interval for the difference in proportions, defined as $p_9 - p_{10}$ , is $(0.03,\ 0.17)$ . A student claims, "Grade 9 has a higher proportion of students who prefer online homework than Grade 10." Is the claim supported by the confidence interval?
Answer: Yes, because the entire interval is above 0, so $p_9 > p_{10}$ is plausible. This question tests whether you can justify claims about comparing two proportions using a confidence interval. The confidence interval (0.03, 0.17) represents plausible values for p₉ - p₁₀, where p₉ is the proportion for Grade 9 and p₁₀ is the proportion for Grade 10. Since the entire interval is positive (above 0), this means p₉ - p₁₀ > 0, which is equivalent to p₉ > p₁₀. The claim that "Grade 9 has a higher proportion" is indeed supported because all plausible values for the difference show Grade 9 higher than Grade 10. When a confidence interval for a difference does not contain 0, we have evidence of a significant difference in the direction indicated by the interval's sign.
A sports analyst compares the proportion of free throws made by players using two different training routines. Routine A: 150 made out of 200 attempts; Routine B: 138 made out of 200 attempts. A 95% confidence interval for $p_A-p_B$ is $(0.01,\ 0.13)$ . The analyst claims Routine A leads to a higher free-throw success rate than Routine B. Is the claim supported by the confidence interval?
Answer: Yes, because the entire interval is above 0, supporting $p_A>p_B$ . This question tests understanding of how a confidence interval entirely above zero supports a comparative claim. The 95% confidence interval for $p_A$ - $p_B$ is (0.01, 0.13), where $p_A$ represents the free-throw success rate for Routine A and $p_B$ for Routine B. Since every value in this interval is positive (the lower bound is 0.01 and upper bound is 0.13), we have convincing evidence that $p_A$ - $p_B$ > 0, which means $p_A$ > $p_B$ . This directly supports the analyst's claim that Routine A leads to a higher free-throw success rate than Routine B. The key principle is that when a confidence interval for a difference does not include 0 and is entirely positive, it provides statistical evidence for the first proportion being larger than the second.
An environmental group compares the proportion of households that recycle in two neighborhoods. Neighborhood A had 58 recyclers out of 100 households; Neighborhood B had 64 recyclers out of 100 households. A 95% confidence interval for $p_A-p_B$ is $(-0.18,\ 0.06)$ . A volunteer claims that Neighborhood A has a lower recycling rate than Neighborhood B. Is the claim supported by the confidence interval?
Answer: No, because the interval includes 0, so it does not support a definite difference in either direction. This question examines whether a confidence interval supports a claim about one proportion being lower than another. The 95% confidence interval for $p_A$ - $p_B$ is (-0.18, 0.06), which includes both negative and positive values, including 0. The volunteer claims that Neighborhood A has a lower recycling rate than Neighborhood B, which would require $p_A$ < $p_B$ , or equivalently, $p_A$ - $p_B$ < 0. However, since the interval includes positive values (up to 0.06), it's plausible that $p_A$ could actually be greater than $p_B$ . The inclusion of 0 means we cannot rule out equal recycling rates either. Therefore, the interval does not provide convincing evidence to support the claim that $p_A$ < $p_B$ . To support such a claim, the entire interval would need to be below 0.
A company tests two website designs for the proportion of visitors who make a purchase: Design 1 and Design 2. A 90% confidence interval for the difference $p_1 - p_2$ is $(-0.02, 0.07)$ . A designer claims, "Design 1 leads to a higher purchase rate than Design 2." Is the claim supported by the confidence interval?
Answer: No, because the interval includes 0, so there is not convincing evidence that $p_1 > p_2$ . This question tests justifying claims via confidence intervals for purchase proportion differences, checking if one design yields higher rates. The 90% CI for p₁ - p₂ is (-0.02, 0.07), including zero, so not convincing evidence that p₁ > p₂, unsupported claim. The interval allows for Design 1 being slightly better, worse, or equal. Distractor choice D incorrectly notes the interval as entirely positive, but it's not. Mini-lesson: to justify p₁ > p₂, ensure the CI for p₁ - p₂ excludes zero and is positive throughout; zero inclusion suggests no clear winner. Confirm the difference calculation fits the claim. Lower confidence levels narrow intervals, potentially supporting claims more easily.
A university compares the proportion of applicants who accept an admission offer for two majors. Let $p_S$ be the true acceptance proportion for STEM offers and $p_H$ for Humanities offers. A 95% confidence interval for $p_H-p_S$ is $(-0.22,\ -0.05)$ . An advisor claims, "A higher proportion of students accept STEM offers than Humanities offers." Is the claim supported by the interval?
Answer: Yes, because $p_H-p_S$ is negative throughout the interval, so $p_S>p_H$ is supported. This question requires careful interpretation of a negative interval and the order of subtraction. The 95% confidence interval for $p_H$ - $p_S$ is (-0.22, -0.05), which is entirely negative. Since the interval is entirely below 0, this means $p_H$ - $p_S$ < 0, which translates to $p_H$ < $p_S$ , or equivalently, $p_S$ > $p_H$ . The advisor claims "A higher proportion of students accept STEM offers than Humanities offers," which means $p_S$ > $p_H$ . This claim is indeed supported by the interval. When a confidence interval for a difference is entirely negative, it indicates the second proportion (the one being subtracted) is larger than the first, which in this case supports the claim about STEM having higher acceptance rates.
A school compares the proportion of students who passed a math exam in two teaching programs. Program A had 84 passes out of 120 students, and Program B had 70 passes out of 120 students. A 95% confidence interval for the difference in population proportions, $p_A-p_B$ , is $(0.02,\ 0.22)$ . A teacher claims that the pass rate is higher in Program A than in Program B. Is the claim supported by the confidence interval?
Answer: Yes, because the entire interval is above 0, so $p_A>p_B$ is plausible. This question tests your ability to interpret a confidence interval for the difference in proportions to justify a claim. The 95% confidence interval for $p_A$ - $p_B$ is (0.02, 0.22), which means we are 95% confident that the true difference between Program A's pass rate and Program B's pass rate lies between 0.02 and 0.22. Since the entire interval is above 0 (all positive values), this provides convincing evidence that $p_A$ > $p_B$ , meaning Program A has a higher pass rate than Program B. The key insight is that when a confidence interval for $p_1$ - $p_2$ contains only positive values, it supports the claim that $p_1$ > $p_2$ . Choice C incorrectly interprets positive values as supporting $p_B$ > $p_A$ , which would be true if the interval were for $p_B$ - $p_A$ instead.
A city surveys residents about support for building a new park. In a random sample, 198 of 300 residents in Neighborhood 1 support the park and 176 of 320 residents in Neighborhood 2 support the park. A 90% confidence interval for $p_1 - p_2$ is $(0.04,\ 0.18)$ . A council member claims that support is higher in Neighborhood 2 than in Neighborhood 1. Is the claim supported by the confidence interval?
Answer: No, because the interval is entirely positive, so $p_1 > p_2$ , not $p_2 > p_1$ . This question evaluates the ability to justify claims using a confidence interval for the difference in proportions, here p1 - p2 for park support in two neighborhoods. The 90% interval (0.04, 0.18) is entirely positive, meaning p1 > p2, which contradicts the claim that support is higher in Neighborhood 2 (p2 > p1). Thus, the claim is not supported, as the interval suggests the opposite. A frequent distractor is choice A, which misinterprets the positive interval as supporting p2 > p1, ignoring the order of subtraction. Mini-lesson: For a CI of p1 - p2, an entirely positive interval supports p1 > p2, an entirely negative one supports p1 < p2, and one including 0 does not support either being definitively higher. The correct answer highlights the importance of matching the claim to the interval's sign.
A researcher compares the proportion of defective items from two manufacturing lines. Line A: 18 defectives out of 300; Line B: 30 defectives out of 300. A 95% confidence interval for $p_A-p_B$ is $(-0.08,\ -0.01)$ . The researcher claims the defect rate is lower on Line A than on Line B. Is the claim supported by the confidence interval?
Answer: Yes, because the entire interval is below 0, supporting $p_A<p_B$ . This question tests interpretation of a confidence interval that is entirely below zero. The 95% confidence interval for $p_A$ - $p_B$ is (-0.08, -0.01), where $p_A$ is the defect rate for Line A and $p_B$ for Line B. Since every value in this interval is negative (below 0), this means $p_A$ - $p_B$ < 0, which is equivalent to $p_A$ < $p_B$ . This directly supports the researcher's claim that the defect rate is lower on Line A than on Line B. When a confidence interval for a difference contains only negative values, it provides convincing evidence that the first proportion is smaller than the second. The fact that the interval doesn't include 0 strengthens this conclusion, as we can rule out the possibility of equal defect rates.
A sports scientist compares the proportion of athletes who report improved performance after using two different training plans. In random samples, 54 of 90 athletes using Plan A report improvement and 62 of 95 athletes using Plan B report improvement. A 95% confidence interval for $p_A - p_B$ is $(-0.18,\ 0.05)$ . The scientist claims that Plan A leads to a higher improvement rate than Plan B. Is the claim supported by the confidence interval?
Answer: No, because the interval includes $0$ , so the data do not clearly support $p_A > p_B$ . The skill is evaluating claims with a confidence interval for $p_A$ - $p_B$ in athlete improvement rates. The 95% interval (-0.18, 0.05) includes 0, spanning negative and positive values, so it does not support the claim that Plan A has a higher rate ( $p_A$ > $p_B$ ) definitively. Possible differences include no difference or even $p_A$ < $p_B$ . Choice A distracts by suggesting positive values alone confirm $p_A$ higher, ignoring negatives and 0. Mini-lesson on comparative claims: A CI for p1 - p2 entirely above 0 supports p1 > p2, entirely below supports p1 < p2, but including 0 means the claim lacks strong evidence. Samples (54/90 = 0.60 for A, 62/95 ≈ 0.65 for B) reflect this ambiguity.
A researcher compares the proportion of voters who support Candidate Z in two age groups. In random samples, 168 of 280 voters ages 18–34 support Z and 210 of 350 voters ages 35+ support Z. A 95% confidence interval for $p_{18\text{–}34} - p_{35+}$ is $( -0.03,\ 0.11)$ . The researcher claims that support is higher among voters ages 18–34 than among voters ages 35+. Is the claim supported by the confidence interval?
Answer: No, because the interval includes $0$ , so it does not clearly support $p_{18\text{–}34} > p_{35+}$ . This question evaluates using a CI for p_{18–34} - $p_{35+}$ to justify claims about voter support. The 95% interval (-0.03, 0.11) includes 0, covering negative and positive differences, so it does not support the claim that support is higher in ages 18–34 ( $p_{18–34}$ > $p_{35+}$ ). Possible true differences include no difference or the opposite. Distractor choice A assumes positive values confirm the claim, ignoring negatives and 0. Mini-lesson: A CI for p1 - p2 entirely positive supports p1 > p2, entirely negative supports p1 < p2, but including 0 means the data do not conclusively favor one over the other. Sample proportions (168/280 = 0.60 for 18–34, 210/350 = 0.60 for 35+) indicate similarity, consistent with the interval.
A researcher compares the proportion of left-handed students in School 1 and School 2. A 99% confidence interval for the difference in proportions, defined as $p_1 - p_2$ , is $(0.00, 0.09)$ . The researcher claims, "School 1 has a higher proportion of left-handed students than School 2." Is the claim supported by the confidence interval?
Answer: No, because the interval includes 0, so there is not convincing evidence that $p_1 > p_2$ . This problem evaluates the skill of using confidence intervals to justify claims about proportion differences, specifically if the interval supports one school having more left-handed students. The 99% CI for p₁ - p₂ is (0.00, 0.09), which includes zero, indicating no convincing evidence that p₁ > p₂ since zero (no difference) is plausible. Therefore, the researcher's claim is not supported. Choice A is a distractor, incorrectly assuming including zero proves p₁ > p₂, which it doesn't. Mini-lesson on comparative claims: for p₁ > p₂, the CI for p₁ - p₂ must be fully above zero; including zero means the data don't rule out equality. Note the confidence level affects interval width but not the zero-check principle. Always confirm the subtraction order matches the claim.
A company tests two website designs and records the proportion of visitors who make a purchase. Design 1 had 96 purchases out of 400 visitors; Design 2 had 88 purchases out of 400 visitors. A 90% confidence interval for $p_1-p_2$ is $(-0.005,\ 0.065)$ . A manager claims Design 1 has a higher purchase rate than Design 2. Is the claim supported by the confidence interval?
Answer: No, because the interval includes 0, so it does not provide convincing evidence that $p_1>p_2$ . This question requires understanding how the inclusion of 0 in a confidence interval affects hypothesis testing. The 90% confidence interval for $p_1$ - $p_2$ is (-0.005, 0.065), which includes 0. When a confidence interval for a difference includes 0, it means the true difference could be zero (no difference between the populations), positive ( $p_1$ > $p_2$ ), or negative ( $p_1$ < $p_2$ ). Therefore, the interval does not provide convincing evidence to support the manager's claim that Design 1 has a higher purchase rate than Design 2. The presence of 0 in the interval indicates that we cannot rule out the possibility of no difference or even Design 2 being better. To support a directional claim like $p_1$ > $p_2$ , the entire interval must be above 0.
A school compares the proportion of students who prefer online homework in two grades. Let $p_9$ be the true proportion for 9th graders and $p_{12}$ for 12th graders. A 95% confidence interval for the difference in proportions, computed as $p_9-p_{12}$ , is $(0.04,\ 0.18)$ . A student claims, "A higher proportion of 9th graders than 12th graders prefer online homework." Is the claim supported by the interval?
Answer: Yes, because the entire interval is above $0$ , so $p_9>p_{12}$ is plausible. This question tests understanding of confidence intervals for comparing two proportions. The interval (0.04, 0.18) represents plausible values for p₉ - p₁₂, where p₉ is the proportion of 9th graders and p₁₂ is the proportion of 12th graders who prefer online homework. Since the entire interval is positive (above 0), this means p₉ - p₁₂ > 0, which translates to p₉ > p₁₂. The student's claim that "a higher proportion of 9th graders than 12th graders prefer online homework" is indeed supported by the interval. When a confidence interval for a difference does not contain 0 and is entirely positive, it provides evidence that the first proportion is larger than the second.
A company compares the proportion of defective items from Machine A and Machine B. A 90% confidence interval for the difference in proportions, defined as $p_A - p_B$ , is $(0.01,\ 0.06)$ . An engineer claims, "Machine A produces a higher proportion of defective items than Machine B." Is the claim supported by the confidence interval?
Answer: Yes, because the entire interval is above 0, so $p_A > p_B$ is supported. This question asks about a 90% confidence interval (0.01, 0.06) for the difference $p_A$ - $p_B$ in defective proportions. The entire interval is positive (above 0), which means all plausible values for $p_A$ - $p_B$ are greater than 0. This translates to $p_A$ > $p_B$ , indicating Machine A produces a higher proportion of defective items than Machine B. The claim is supported because every value in the confidence interval supports the same conclusion. Note that while the interval is "small" in magnitude, what matters for comparative claims is whether 0 is included, not the size of the difference. As long as the entire interval excludes 0 and is positive, we have evidence that Machine A has a higher defect rate.
A city surveys residents in the North and South districts about supporting a new park. A 95% confidence interval for the difference in proportions, defined as $p_{North} - p_{South}$ , is $(-0.21,\ -0.05)$ . A council member claims, "The North district has a lower proportion supporting the park than the South district." Is the claim supported by the confidence interval?
Answer: Yes, because the entire interval is below 0, so $p_{North} < p_{South}$ is plausible. This question tests understanding of negative confidence intervals for differences in proportions. The 95% confidence interval (-0.21, -0.05) for p_North - p_South is entirely negative, meaning all plausible values for the difference are below 0. When p_North - p_South < 0, this is equivalent to p_North < p_South, which means the North district has a lower proportion than the South district. The claim that "The North district has a lower proportion supporting the park than the South district" is supported by the interval. Students often confuse the direction when intervals are negative, but remember: if the difference A - B is negative, then A is smaller than B.
A nonprofit compares the proportion of donors who give again the next year between two outreach strategies. In random samples, 150 of 600 donors contacted with Strategy A gave again, and 196 of 620 donors contacted with Strategy B gave again. A 95% confidence interval for $p_A - p_B$ is $(-0.12,\ -0.01)$ . A director claims that Strategy A results in a higher repeat-donation rate than Strategy B. Is the claim supported by the confidence interval?
Answer: No, because the interval is entirely negative, indicating $p_A < p_B$ . This question tests using a CI for $p_A$ - $p_B$ to justify claims about repeat-donation rates. The 95% interval (-0.12, -0.01) is entirely negative, indicating $p_A$ < $p_B$ , which opposes the claim that Strategy A has a higher rate ( $p_A$ > $p_B$ ). Thus, the claim is not supported; the data suggest the opposite. Distractor choice A misinterprets the negative interval as supporting $p_A$ > $p_B$ . Mini-lesson: An entirely negative CI for p1 - p2 supports p1 < p2, positive supports p1 > p2, and one with 0 does not support a difference claim. Sample rates (150/600 = 0.25 for A, 196/620 ≈ 0.316 for B) align with this conclusion.
Two brands of batteries are tested for the proportion that last at least 10 hours. A 90% confidence interval for the difference in proportions, defined as $p_A - p_B$ , is $(-0.08,\ 0.04)$ . A manager claims, "Brand A has a higher proportion lasting at least 10 hours than Brand B." Is the claim supported by the confidence interval?
Answer: No, because the interval includes 0, so $p_A > p_B$ is not clearly supported. This question asks whether a confidence interval supports a claim about Brand A having a higher proportion than Brand B. The 90% confidence interval (-0.08, 0.04) for $p_A$ - $p_B$ contains both negative and positive values, including 0. When 0 is in the confidence interval for a difference, it means that "no difference" ( $p_A$ = $p_B$ ) is a plausible value. Since the interval includes scenarios where $p_A$ < $p_B$ (negative values), $p_A$ = $p_B$ (zero), and $p_A$ > $p_B$ (positive values), we cannot conclusively support the claim that Brand A is higher. The presence of 0 in the interval indicates we lack sufficient evidence to claim one proportion is definitively higher than the other.
A city compares the proportion of commuters who use public transit in two districts. District East: 132 transit users out of 220 commuters; District West: 140 transit users out of 220 commuters. A 95% confidence interval for $p_{East}-p_{West}$ is $(-0.12,\ 0.04)$ . A planner claims that public transit use is higher in the West district than in the East district. Is the claim supported by the confidence interval?
Answer: No, because the interval includes 0, so it does not provide convincing evidence that $p_{West}>p_{East}$ . This question examines whether a confidence interval supports a claim when it includes zero. The 95% confidence interval for p_East - p_West is (-0.12, 0.04), which includes negative values, zero, and positive values. The planner claims that public transit use is higher in the West district than in the East district, which would require p_West > p_East, or equivalently, p_East - p_West < 0. While the interval does include negative values that would support this claim, it also includes positive values (up to 0.04) that would contradict it. The inclusion of 0 means we cannot rule out equal transit use in both districts. Therefore, the interval does not provide convincing evidence to support the claim that p_West > p_East, as we cannot exclude the possibilities that p_West ≤ p_East.

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