AP Statistics Quiz: Sampling For Differences In Sample Proportions

What this quiz covers

This quiz focuses on Sampling For Differences In Sample Proportions, 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. The sampling distribution of $\hat p_1-\hat p_2$ describes how $p_1-p_2$ varies from sample to sample.
2. The sampling distribution of $\hat p_1-\hat p_2$ describes how $\hat p_1-\hat p_2$ varies from sample to sample. (correct answer)
3. The sampling distribution of $\hat p_1-\hat p_2$ is the distribution of individual responses (satisfied vs not) pooled across both groups.
4. The sampling distribution of $\hat p_1-\hat p_2$ is the same as the distribution of $\hat p_1$ alone because both are proportions.
5. The sampling distribution of $\hat p_1-\hat p_2$ cannot be centered at $p_1-p_2$ unless $n_1=n_2$.

Explanation: This question distinguishes between what varies in a sampling distribution. The sampling distribution of $\hat{p}_1 - \hat{p}_2$ describes how the sample statistic $\hat{p}_1 - \hat{p}_2$ varies from sample to sample when we repeat the sampling process. The population parameters $p_1$ and $p_2$ are fixed and don't vary (eliminating A). It's not about individual responses but about the difference in proportions (eliminating C). The distributions of $\hat{p}_1$ alone and $\hat{p}_1 - \hat{p}_2$ are different (eliminating D). The center is $p_1 - p_2$ regardless of whether sample sizes are equal (eliminating E).

Question 2

1. If $p_1-p_2$ is positive, then $\hat p_1-\hat p_2$ must be positive in every sample.
2. The sampling distribution of $\hat p_1-\hat p_2$ is centered at $p_1-p_2$, and its spread depends on $p_1$, $p_2$, $n_1$, and $n_2$. (correct answer)
3. The sampling distribution is centered at $\hat p_1-\hat p_2$ and its spread depends only on $n_1+n_2$.
4. The sampling distribution describes the difference between the two populations, not the difference between the two sample proportions.
5. The sampling distribution cannot be used unless the two populations have the same size.

Explanation: This question addresses both center and spread of the sampling distribution. The sampling distribution of $\hat{p}_1 - \hat{p}_2$ is centered at the population difference $p_1 - p_2$, and its spread (standard deviation) depends on all four values: $p_1$, $p_2$, $n_1$, and $n_2$ through the formula $\sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}$. Even if $p_1 - p_2 > 0$, sampling variability means some samples could yield negative differences (eliminating A). The center is not the sample statistic (eliminating C). The distribution describes sample statistics, not populations (eliminating D). Population sizes aren't relevant to the sampling distribution (eliminating E).

Question 3

1. With large sample sizes, $\hat p_1-\hat p_2$ will equal $p_1-p_2$ in every repetition.
2. The sampling distribution of $\hat p_1-\hat p_2$ is approximately normal if the success–failure condition is met in both groups. (correct answer)
3. The sampling distribution is approximately normal only when $p_1$ and $p_2$ are both close to 0.5.
4. The sampling distribution is approximately normal because the population distributions must be normal.
5. The sampling distribution has no spread because the sample sizes are equal and large.

Explanation: This question tests understanding of normality conditions for sampling distributions. The sampling distribution of $\hat{p}_1 - \hat{p}_2$ is approximately normal when the success-failure condition is met in both groups (typically $np \geq 10$ and $n(1-p) \geq 10$ for each group). Large samples don't eliminate all variability (eliminating A). Normality doesn't require proportions near 0.5 (eliminating C). The population distributions don't need to be normal for the sampling distribution to be approximately normal (eliminating D). There is still spread due to sampling variability regardless of sample size (eliminating E).

Question 4

1. The mean of the sampling distribution of $\hat p_1-\hat p_2$ is $\hat p_1-\hat p_2$ from this one set of samples.
2. The sampling distribution of $\hat p_1-\hat p_2$ has no variability because $n_1$ and $n_2$ are fixed.
3. The mean of the sampling distribution of $\hat p_1-\hat p_2$ is $p_1-p_2$. (correct answer)
4. The standard deviation of $\hat p_1-\hat p_2$ is $\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2}}$ exactly, for all samples.
5. The sampling distribution of $\hat p_1-\hat p_2$ is centered at 0 whenever $n_1\neq n_2$.

Explanation: This question tests understanding of the sampling distribution of the difference in sample proportions. The sampling distribution of $\hat{p}_1 - \hat{p}_2$ describes how this difference varies across many repeated samples. Its mean (center) is the true population difference $p_1 - p_2$, not the observed sample difference from one particular sample (eliminating A). The distribution does have variability even with fixed sample sizes because different samples yield different proportions (eliminating B). The standard deviation formula uses population proportions $p_1$ and $p_2$, not sample proportions (eliminating D). The center is $p_1 - p_2$ regardless of whether sample sizes are equal (eliminating E).

Question 5

1. A normal approximation is appropriate only if $p_A=p_B$.
2. A normal approximation is appropriate if each sample has at least 10 expected successes and 10 expected failures: $n_Ap_A,\,n_A(1-p_A),\,n_Bp_B,\,n_B(1-p_B)\ge 10$. (correct answer)
3. A normal approximation is appropriate whenever $n_A+n_B\ge 30$.
4. A normal approximation is never appropriate for a difference of two sample proportions.
5. A normal approximation is appropriate if $\hat{p}_A$ and $\hat{p}_B$ from one sample are both between 0.4 and 0.6.

Explanation: This question assesses conditions for normality in the sampling distribution of proportion differences, vital in AP Statistics for valid inferences. The normal approximation holds if each group has at least 10 expected successes and failures, ensuring the individual proportion distributions are mound-shaped. Choice C is a distractor, oversimplifying to total sample size >=30 without checking success-failure counts per group. Mini-lesson: for (\hat{p}_A - \hat{p}_B) from independent samples, normality requires (n p geq 10) and (n(1-p) geq 10) for each, allowing the difference to be approximately normal centered at (p_A - p_B) with calculable spread. This isn't guaranteed by equal proportions or observed values alone. Proper checks prevent skewed distributions in analysis.

Question 6

1. The sampling distribution would have the same center but typically larger standard deviation (more spread). (correct answer)
2. The sampling distribution would shift its center from $p_W-p_M$ to $\hat{p}_W-\hat{p}_M$.
3. The sampling distribution would have smaller spread because smaller samples are less variable.
4. The sampling distribution would become centered at 0 regardless of $p_W-p_M$.
5. The sampling distribution would have no change because the populations did not change.

Explanation: This question investigates how changing sample sizes affects the sampling distribution of proportion differences, a practical AP Statistics concept for study design adjustments. Reducing sizes keeps the center at (p_W - p_M) but increases spread, as the SD grows with smaller n in the formula. Choice C is a distractor, incorrectly stating smaller samples reduce variability, when the opposite is true due to less information. Mini-lesson: for independent samples, the difference distribution's center is invariant to sample size, but spread inversely relates to n, so halving sizes widens it without shifting to observed values or zero. Populations staying the same doesn't negate size effects. This informs trade-offs in precision versus feasibility.

Question 7

1. Its mean is 0, but it will still vary from sample to sample. (correct answer)
2. Its mean is 0, and it will always equal 0 in repeated samples.
3. Its mean equals $\hat{p}_X-\hat{p}_Y$ from the first pair of samples.
4. Its mean must be positive because sample proportions are always between 0 and 1.
5. Its mean is 1 because the two sample proportions add to 1.

Explanation: This question probes the properties of the sampling distribution for proportion differences when population proportions are equal, relevant in AP Statistics for null hypothesis scenarios. With (p_X = p_Y = 0.5), the center is 0, but spread exists due to sampling variability, so (\hat{p}_X - \hat{p}_Y) fluctuates around 0 in repeated samples. Choice B is a distractor, falsely implying no variability when the mean is 0, ignoring that even equal proportions yield differing sample outcomes by chance. Mini-lesson: for independent samples, the distribution of (\hat{p}_1 - \hat{p}_2) is centered at ($p_1$ - $p_2$) (here 0) with standard deviation reflecting sample sizes and proportions, always showing spread unless samples are infinite. This variability is key for understanding p-values in tests of equal proportions. Proportions being between 0 and 1 doesn't force the difference to be positive.

Question 8

1. The sampling distribution of $\hat{p}_1-\hat{p}_2$ is centered at 0 whenever $n_1\neq n_2$.
2. The sampling distribution of $\hat{p}_1-\hat{p}_2$ is centered at $p_1-p_2$. (correct answer)
3. Because $n_2$ is larger than $n_1$, the sampling distribution depends only on $n_2$.
4. The sampling distribution has no spread if the samples are independent.
5. The center of the sampling distribution is $\hat{p}_1-\hat{p}_2$ for any particular pair of samples.

Explanation: This question evaluates knowledge of the sampling distribution for differences in sample proportions, essential in AP Statistics for comparing success rates between groups like subscription renewals. The distribution is centered at ($p_1$ - $p_2$) with spread determined by (\sqrt{\frac{p_1(1-$p_1$)}{n_1} + \frac{p_2(1-$p_2$)}{n_2}}), which accounts for unequal sample sizes without the larger one dominating entirely. Choice E is a distractor, wrongly claiming the center is the observed difference, which varies per sample while the true center remains the population parameter. Mini-lesson: the sampling distribution of (\hat{p}_1 - \hat{p}_2) from independent samples models how this statistic behaves over many repetitions, always centering on the fixed ($p_1$ - $p_2$) but with variability that combines the uncertainties from each sample. Larger samples reduce spread, improving reliability for hypothesis tests or confidence intervals. Even with ($n_2$ > $n_1$), both contribute to the overall variance.

Question 9

1. The sampling distribution will be more variable because $n_R$ is large.
2. The sampling distribution's variability is driven more by the smaller sample size $n_S$ than by the larger $n_R$. (correct answer)
3. The sampling distribution depends only on $p_R-p_S$, not on $n_R$ or $n_S$.
4. Because $n_R\neq n_S$, the sampling distribution cannot be approximately normal.
5. The sampling distribution is centered at $\hat{p}_R-\hat{p}_S$ regardless of the true $p_R-p_S$.

Explanation: This question examines how unequal sample sizes influence the sampling distribution of proportion differences, a critical AP Statistics concept for real-world studies with varying group sizes. The spread is more affected by the smaller sample ((n_S = 50)) because its term in the standard deviation formula contributes more variance relative to the larger (n_R = 500). Choice A distracts by claiming more variability due to the large sample, which is backward since larger samples reduce individual variance contributions. Mini-lesson: in difference distributions from independent samples, the total spread combines variances additively, with smaller samples driving more uncertainty, while the center remains at (p_R - p_S). Unequal sizes don't prevent normality if conditions are met. This highlights the importance of bolstering smaller groups for balanced precision.

Question 10

1. The sampling distribution of $\hat{p}_A-\hat{p}_B$ is centered at the observed sample difference $\hat{p}_A-\hat{p}_B$ from this one set of samples.
2. If $n_1$ and $n_2$ are increased, the sampling distribution of $\hat{p}_A-\hat{p}_B$ becomes wider because larger samples vary more.
3. The sampling distribution of $\hat{p}_A-\hat{p}_B$ is centered at the true difference in population proportions $p_A-p_B$. (correct answer)
4. Because two samples are taken, $\hat{p}_A-\hat{p}_B$ has no sampling variability and will be the same in repeated samples.
5. The standard deviation of $\hat{p}_A-\hat{p}_B$ depends only on $n_1$ and $n_2$, not on $p_A$ or $p_B$.

Explanation: This question assesses understanding of the sampling distribution for the difference in two sample proportions, a key concept in AP Statistics for comparing categorical data from two groups. The sampling distribution of $\hat{p}_A - \hat{p}_B$ is centered at the true population difference $(p_A - p_B)$, not at the observed sample difference, and its spread is given by the standard deviation formula that incorporates both population proportions and sample sizes. A common distractor is choice A, which incorrectly suggests the distribution is centered at the observed difference from one sample, confusing the sample statistic with the parameter. In a mini-lesson on difference distributions: when taking independent random samples from two populations, the difference in sample proportions is an unbiased estimator of the true difference, meaning repeated samples will produce values varying around $(p_A - p_B)$ with a normal shape under large sample conditions. The spread decreases as sample sizes increase, reflecting more precise estimates. This centering at the parameter ensures that inferences about the population difference are valid.

Question 11

1. If $p_1-p_2$ is negative, then $\hat{p}_1-\hat{p}_2$ must be negative in every sample.
2. The sampling distribution of $\hat{p}_1-\hat{p}_2$ can include both positive and negative values even if $p_1-p_2\ne 0$. (correct answer)
3. The sampling distribution cannot be centered at a negative value because proportions are nonnegative.
4. The sampling distribution is centered at $\hat{p}_1-\hat{p}_2$ because that is the best estimate of $p_1-p_2$.
5. If $n_2>n_1$, then $\hat{p}_1-\hat{p}_2$ is biased toward negative values.

Explanation: This question tests understanding of variability in the sampling distribution of proportion differences, important in AP Statistics for interpreting possible outcomes. Even if ($p_1$ - $p_2$ eq 0), sampling error allows (\hat{p}_1 - \hat{p}_2) to be positive or negative, reflecting the distribution's spread around the center. Choice A distracts by claiming a negative parameter forces negative statistics every time, underestimating variability. Mini-lesson: the difference distribution from independent samples centers at ($p_1$ - $p_2$) (which can be negative) with normal shape, but values span both sides due to chance, not biased by sample size differences. Proportions' nonnegativity doesn't restrict the difference's sign. This variability underpins confidence intervals crossing zero.

Question 12

1. It is $\frac{p_1-p_2}{n_1+n_2}$.
2. It is $\sqrt{\frac{p_1(1-p_1)}{n_1}+\frac{p_2(1-p_2)}{n_2}}$. (correct answer)
3. It is $\sqrt{\frac{\hat{p}_1(1-\hat{p}_1)}{n_1}+\frac{\hat{p}_2(1-\hat{p}_2)}{n_2}}$ for the population distribution.
4. It equals 0 when $n_1=n_2$.
5. It depends only on $p_1-p_2$ and not on $p_1$ and $p_2$ separately.

Explanation: This question focuses on the standard deviation formula for the sampling distribution of proportion differences, a core AP Statistics tool for quantifying variability in comparisons. The correct spread is (\sqrt{\frac{p_1(1-$p_1$)}{n_1} + \frac{p_2(1-$p_2$)}{n_2}}), depending on both population proportions and sample sizes. Choice C distracts by using sample proportions for the population distribution, which is for estimation, not the true SD. Mini-lesson: the difference distribution from independent samples centers at ($p_1$ - $p_2$) with this SD formula, which doesn't zero out for equal n or depend solely on the difference. It separates effects of p and n, guiding sample size planning. Understanding this prevents underestimating variability in equal-sized groups.

Question 13

1. Independence guarantees that $\hat{p}_A-\hat{p}_B$ equals $p_A-p_B$ in every sample.
2. Independence is not needed; the standard deviation formula for $\hat{p}_A-\hat{p}_B$ is the same even if the samples overlap.
3. Independence allows the variances from the two samples to add when finding the standard deviation of $\hat{p}_A-\hat{p}_B$. (correct answer)
4. Independence makes the sampling distribution uniform rather than approximately normal.
5. Independence implies $p_A=p_B$.

Explanation: This question explores the role of independence in the sampling distribution of proportion differences, essential in AP Statistics for correct variance calculations. Independence allows adding the variances of the individual proportions to get the SD of the difference, ensuring accurate spread quantification. Choice A is a distractor, wrongly stating independence makes the statistic exactly equal the parameter every time, ignoring sampling variability. Mini-lesson: for independent samples, the distribution of (\hat{p}_A - \hat{p}_B) is centered at (p_A - p_B) with SD from added variances, approximately normal under conditions, but without independence, covariance terms complicate it. This doesn't imply equal proportions or change the shape to uniform. Independence is key for valid two-sample inferences.

Question 14

1. The mean of $\hat p_1-\hat p_2$ is always 0 because both samples have the same size.
2. If the repeated-sampling conditions are met, $\hat p_1-\hat p_2$ is approximately normal and centered at $p_1-p_2$. (correct answer)
3. The sampling distribution must be skewed because proportions are bounded between 0 and 1.
4. The standard deviation of $\hat p_1-\hat p_2$ is $\frac{p_1-p_2}{\sqrt{n_1+n_2}}$.
5. Once $n_1=n_2$, there is no need for random sampling to justify the sampling distribution.

Explanation: This question examines conditions for the sampling distribution of differences. When proper conditions are met (random sampling, independence, and success-failure conditions), the sampling distribution of $\hat{p}_1 - \hat{p}_2$ is approximately normal and centered at $p_1 - p_2$. Equal sample sizes don't make the mean 0 unless $p_1 = p_2$ (eliminating A). The distribution can be approximately normal despite proportions being bounded (eliminating C). The standard deviation formula is $\sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}$, not the given expression (eliminating D). Random sampling is always necessary for valid inference (eliminating E).

Question 15

1. The sampling distribution of $\hat p_1-\hat p_2$ is centered at the population difference $p_1-p_2$. (correct answer)
2. The sampling distribution of $\hat p_1-\hat p_2$ is centered at the sample difference $\hat p_1-\hat p_2$, since that is what was measured.
3. The sampling distribution has standard deviation 0 because the same question is asked in both neighborhoods.
4. The sampling distribution cannot be described without knowing $\hat p_1$ and $\hat p_2$ from the first sample.
5. The sampling distribution is always normal for any sample sizes when comparing two proportions.

Explanation: This question tests the fundamental property that sampling distributions are centered at population parameters. The sampling distribution of $\hat{p}_1 - \hat{p}_2$ describes the behavior of this statistic over many repetitions and is centered at the true population difference $p_1 - p_2$. It is not centered at any particular sample's observed difference (eliminating B). The standard deviation is positive even when asking the same question, due to sampling variability (eliminating C). The distribution can be described theoretically without knowing specific sample values (eliminating D). Normality requires meeting conditions, not just comparing two proportions (eliminating E).

Question 16

1. If both sample sizes were doubled, the sampling distribution of $\hat{p}_1-\hat{p}_2$ would typically be less spread out. (correct answer)
2. If both sample sizes were doubled, the sampling distribution of $\hat{p}_1-\hat{p}_2$ would typically be more spread out.
3. Doubling both sample sizes changes the center of the sampling distribution from $p_1-p_2$ to $\hat{p}_1-\hat{p}_2$.
4. With equal sample sizes, $\hat{p}_1-\hat{p}_2$ has no sampling variability.
5. The sampling distribution cannot be approximately normal because it involves two proportions.

Explanation: This question tests comprehension of how sample size affects the sampling distribution of the difference in proportions, a fundamental AP Statistics skill for designing comparative studies. Doubling both sample sizes reduces the spread of the distribution while keeping the center at ($p_1$ - $p_2$), as the standard deviation formula shows an inverse relationship with (n). Choice B distracts by suggesting increased spread with larger samples, which is incorrect since larger samples provide more precise estimates and less variability. Mini-lesson: the difference (\hat{p}_1 - \hat{p}_2) from independent samples has a sampling distribution that's approximately normal, centered at the true difference, with spread shrinking as sample sizes grow, enabling tighter confidence intervals. Equal sample sizes don't eliminate variability; there's always sampling error unless populations are fully enumerated. This principle guides researchers in balancing precision with cost.

Question 17

1. The sampling distribution of $\hat p_1-\hat p_2$ has mean $p_1-p_2$, regardless of the sample sizes, assuming random sampling and independence. (correct answer)
2. The sampling distribution of $\hat p_1-\hat p_2$ has mean $\hat p_1-\hat p_2$ because that is the statistic being computed.
3. The sampling distribution has mean 0 because $\hat p_1$ and $\hat p_2$ are both proportions.
4. The sampling distribution has no variability when the samples are independent.
5. The sampling distribution is guaranteed to be normal for any values of $p_1$ and $p_2$ as long as $n_1+n_2\ge 30$.

Explanation: This question tests the fundamental property about the mean of a sampling distribution. The sampling distribution of $\hat{p}_1 - \hat{p}_2$ has mean equal to $p_1 - p_2$ regardless of sample sizes, as long as we have random sampling and independence. This is a key unbiased property of the difference in sample proportions. The mean is not the sample statistic itself (eliminating B). The mean is only 0 when $p_1 = p_2$, not just because both are proportions (eliminating C). Independence doesn't eliminate variability (eliminating D). The normality condition requires checking success-failure conditions, not just having $n_1 + n_2 \geq 30$ (eliminating E).

Question 18

1. If $p_1=p_2$, then the sampling distribution of $\hat p_1-\hat p_2$ is centered at 0. (correct answer)
2. If $p_1=p_2$, then $\hat p_1-\hat p_2$ will equal 0 in every sample.
3. The sampling distribution is centered at $\hat p_1-\hat p_2$ because that is the observed difference.
4. The sampling distribution is uniform because each sample proportion is between 0 and 1.
5. Changing the sample sizes cannot change the spread of $\hat p_1-\hat p_2$.

Explanation: This question focuses on properties of the sampling distribution when comparing two proportions. When the true population proportions are equal ($p_1 = p_2$), the difference $p_1 - p_2 = 0$, so the sampling distribution of $\hat{p}_1 - \hat{p}_2$ is centered at 0. However, individual samples will still show variation due to sampling variability, so $\hat{p}_1 - \hat{p}_2$ won't equal 0 in every sample (eliminating B). The center is always $p_1 - p_2$, not the observed difference (eliminating C). The distribution is approximately normal under proper conditions, not uniform (eliminating D). Changing sample sizes does affect the spread through the standard error formula (eliminating E).

Question 19

1. Because $n_2$ is smaller, the sampling distribution of $\hat p_1-\hat p_2$ will typically have greater spread than if $n_2$ were larger (all else equal). (correct answer)
2. Because $n_1$ is large, the spread of $\hat p_1-\hat p_2$ depends only on $n_1$ and not on $n_2$.
3. The sampling distribution of $\hat p_1-\hat p_2$ has mean $\hat p_1-\hat p_2$ and standard deviation $\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2}}$ exactly.
4. If $p_1=p_2$, then the sampling distribution has standard deviation 0.
5. If the samples are independent, then $\hat p_1-\hat p_2$ must be positive in most repetitions.

Explanation: This question focuses on how sample size affects the spread of the sampling distribution. The standard error formula $\sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}$ shows that smaller sample sizes lead to larger standard errors. Since $n_2 = 25$ is relatively small, it contributes more to the overall spread than if it were larger. The spread depends on both sample sizes, not just the larger one (eliminating B). The given formula in C uses sample proportions instead of population proportions (eliminating C). Equal population proportions don't eliminate variability (eliminating D). Independence doesn't determine the sign of the difference (eliminating E).

Question 20

1. Because $n_1$ is large, $\hat p_1-\hat p_2$ will be close to $p_1-p_2$ with no sampling variability.
2. The sampling distribution of $\hat p_1-\hat p_2$ is centered at $p_1-p_2$. (correct answer)
3. The sampling distribution of $\hat p_1-\hat p_2$ is centered at $\hat p_1-\hat p_2$ from the first repetition.
4. The standard deviation of $\hat p_1-\hat p_2$ depends only on $n_1$ and $n_2$, not on $p_1$ and $p_2$.
5. The distribution of $\hat p_1-\hat p_2$ cannot be approximately normal unless $n_1=n_2$.

Explanation: This question tests understanding of center and spread in sampling distributions for differences. The sampling distribution of $\hat{p}_1 - \hat{p}_2$ is centered at the true population difference $p_1 - p_2$, regardless of sample sizes. Even with large samples, there is still sampling variability (eliminating A). The center is not determined by the first sample's result (eliminating C). The standard deviation depends on both the sample sizes AND the population proportions through the formula $\sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}$ (eliminating D). The distribution can be approximately normal with unequal sample sizes if conditions are met (eliminating E).