Sampling for Differences in Sample Proportions
Help Questions
AP Statistics › Sampling for Differences in Sample Proportions
A university compares the proportion of students who report being satisfied with dining services between on-campus and off-campus students. An SRS of $n_1=90$ on-campus students and an independent SRS of $n_2=110$ off-campus students are surveyed; $\hat p_1$ and $\hat p_2$ are the sample proportions satisfied. The sampling distribution of $\hat p_1-\hat p_2$ is modeled for repeated sampling. Which statement is correct?
The sampling distribution of $\hat p_1-\hat p_2$ describes how $\hat p_1-\hat p_2$ varies from sample to sample.
The sampling distribution of $\hat p_1-\hat p_2$ is the distribution of individual responses (satisfied vs not) pooled across both groups.
The sampling distribution of $\hat p_1-\hat p_2$ cannot be centered at $p_1-p_2$ unless $n_1=n_2$.
The sampling distribution of $\hat p_1-\hat p_2$ describes how $p_1-p_2$ varies from sample to sample.
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.
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).
A tech company compares the proportion of users who click a new button design on two versions of an app. Version A is shown to an independent random sample of $n_1=250$ users and Version B to $n_2=250$ users; $\hat p_1$ and $\hat p_2$ are the sample click proportions. The company considers the sampling distribution of $\hat p_1-\hat p_2$ over many repetitions. Which statement is correct?
The sampling distribution cannot be used unless the two populations have the same size.
The sampling distribution describes the difference between the two populations, not the difference between the two sample proportions.
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 $n_2$.
The sampling distribution is centered at $\hat p_1-\hat p_2$ and its spread depends only on $n_1+n_2$.
If $p_1-p_2$ is positive, then $\hat p_1-\hat p_2$ must be positive in every sample.
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).
A political analyst compares the proportion of voters who favor Candidate X in two counties. An independent random sample of $n_1=500$ registered voters from County 1 and $n_2=500$ from County 2 is taken; $\hat p_1$ and $\hat p_2$ are the sample proportions favoring Candidate X. Over repeated sampling, consider the sampling distribution of $\hat p_1-\hat p_2$. Which statement is correct?
The sampling distribution is approximately normal only when $p_1$ and $p_2$ are both close to 0.5.
With large sample sizes, $\hat p_1-\hat p_2$ will equal $p_1-p_2$ in every repetition.
The sampling distribution has no spread because the sample sizes are equal and large.
The sampling distribution is approximately normal because the population distributions must be normal.
The sampling distribution of $\hat p_1-\hat p_2$ is approximately normal if the success–failure condition is met in both groups.
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).
A school district wants to compare support for a new start-time policy between two groups of parents. A random sample of $n_1=80$ elementary-school parents and an independent random sample of $n_2=120$ high-school parents are surveyed; in each group, the sample proportion who support the policy is recorded as $\hat p_1$ and $\hat p_2$. If these sampling methods were repeated many times, the distribution of $\hat p_1-\hat p_2$ would be approximately normal with mean $p_1-p_2$ and standard deviation $\sqrt{\frac{p_1(1-p_1)}{n_1}+\frac{p_2(1-p_2)}{n_2}}$ (assuming conditions are met). Which statement is correct?
The sampling distribution of $\hat p_1-\hat p_2$ has no variability because $n_1$ and $n_2$ are fixed.
The mean of the sampling distribution of $\hat p_1-\hat p_2$ is $p_1-p_2$.
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.
The sampling distribution of $\hat p_1-\hat p_2$ is centered at 0 whenever $n_1\neq n_2$.
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.
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).
A researcher compares the proportion of plants that survive under two fertilizers. Fertilizer A: random sample of $n_A=40$ plants; Fertilizer B: independent random sample of $n_B=40$ plants. The statistic is $\hat{p}_A-\hat{p}_B$. Consider the condition for using a normal approximation for the sampling distribution of $\hat{p}_A-\hat{p}_B$. Which statement is correct?
A normal approximation is appropriate if $\hat{p}_A$ and $\hat{p}_B$ from one sample are both between 0.4 and 0.6.
A normal approximation is never appropriate for a difference of two sample proportions.
A normal approximation is appropriate only if $p_A=p_B$.
A normal approximation is appropriate if each sample has at least 10 expected successes and 10 expected failures: $n_Ap_A,,n_A(1-p_A),,n_Bp_B,,n_B(1-p_B)\ge 10$.
A normal approximation is appropriate whenever $n_A+n_B\ge 30$.
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.
A tech company compares the proportion of users who enable two-factor authentication on two platforms. Platform W: SRS of $n_W=1000$ users; Platform M: independent SRS of $n_M=1000$ users. The statistic is $\hat{p}_W-\hat{p}_M$. Suppose $p_W$ and $p_M$ stay the same over time and the sampling method stays the same. Which statement is correct about what would happen to the sampling distribution if both sample sizes were reduced to $n_W=n_M=100$?
The sampling distribution would shift its center from $p_W-p_M$ to $\hat{p}_W-\hat{p}_M$.
The sampling distribution would have smaller spread because smaller samples are less variable.
The sampling distribution would have the same center but typically larger standard deviation (more spread).
The sampling distribution would have no change because the populations did not change.
The sampling distribution would become centered at 0 regardless of $p_W-p_M$.
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.
Two independent random samples are taken to compare the proportion of voters who support a ballot measure in two counties. County X: $n_X=60$; County Y: $n_Y=60$. The statistic is $\hat{p}_X-\hat{p}_Y$. Suppose the true population proportions are $p_X=0.50$ and $p_Y=0.50$. Over many repetitions, which statement is correct about the sampling distribution of $\hat{p}_X-\hat{p}_Y$?
Its mean is 0, and it will always equal 0 in repeated samples.
Its mean equals $\hat{p}_X-\hat{p}_Y$ from the first pair of samples.
Its mean is 0, but it will still vary from sample to sample.
Its mean is 1 because the two sample proportions add to 1.
Its mean must be positive because sample proportions are always between 0 and 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.
A company compares the proportion of customers who renew a subscription under two email campaigns. From Campaign 1, an SRS of $n_1=80$ customers is selected; from Campaign 2, an independent SRS of $n_2=320$ customers is selected. The statistic of interest is $\hat{p}_1-\hat{p}_2$. Over many repetitions, the sampling distribution of $\hat{p}_1-\hat{p}_2$ is centered at $p_1-p_2$ and has standard deviation $\sqrt{\frac{p_1(1-p_1)}{n_1}+\frac{p_2(1-p_2)}{n_2}}$. Which statement is correct?
The center of the sampling distribution is $\hat{p}_1-\hat{p}_2$ for any particular pair of samples.
Because $n_2$ is larger than $n_1$, the sampling distribution depends only on $n_2$.
The sampling distribution has no spread if the samples are independent.
The sampling distribution of $\hat{p}_1-\hat{p}_2$ is centered at $p_1-p_2$.
The sampling distribution of $\hat{p}_1-\hat{p}_2$ is centered at 0 whenever $n_1\neq n_2$.
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.
A streaming service compares the proportion of users who finish a new series in two regions. Region R: an SRS of $n_R=500$ users; Region S: an independent SRS of $n_S=50$ users. The statistic is $\hat{p}_R-\hat{p}_S$. In repeated sampling, the sampling distribution is centered at $p_R-p_S$ and its standard deviation is influenced by both sample sizes. Which statement is correct?
Because $n_R\neq n_S$, the sampling distribution cannot be approximately normal.
The sampling distribution’s variability is driven more by the smaller sample size $n_S$ than by the larger $n_R$.
The sampling distribution depends only on $p_R-p_S$, not on $n_R$ or $n_S$.
The sampling distribution will be more variable because $n_R$ is large.
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.
A public health researcher wants to compare the proportion of adults who got a flu shot this year in two cities. A simple random sample of $n_1=200$ adults is taken from City A and $n_2=200$ adults is taken from City B, and the statistic $\hat{p}_A-\hat{p}_B$ is computed. If these samples were repeatedly taken the same way, the sampling distribution of $\hat{p}_A-\hat{p}_B$ would be approximately normal and centered at $p_A-p_B$, with a standard deviation that depends on $p_A$, $p_B$, $n_1$, and $n_2$. Which statement is correct?
The sampling distribution of $\hat{p}_A-\hat{p}_B$ is centered at the true difference in population proportions $p_A-p_B$.
The standard deviation of $\hat{p}_A-\hat{p}_B$ depends only on $n_1$ and $n_2$, not on $p_A$ or $p_B$.
Because two samples are taken, $\hat{p}_A-\hat{p}_B$ has no sampling variability and will be the same in repeated samples.
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.
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.
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.