AP Statistics Quiz: Sampling For Differences In Sample Means

What this quiz covers

This quiz focuses on Sampling For Differences In Sample Means, 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. It becomes 4 times as large because the sample sizes are 4 times as large.
2. It becomes about half as large because each standard error term involves $\sqrt{n}$. (correct answer)
3. It stays the same because the population standard deviations do not change.
4. It becomes 4 times smaller because the sample sizes are 4 times as large.
5. It becomes 2 times larger because there are two groups instead of one.

Explanation: This question examines how sample size affects the standard deviation of the sampling distribution. When both sample sizes are quadrupled from 16 to 64, each term in the standard deviation formula $\sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}$ is divided by 4, making the entire expression half as large (since $\sqrt{1/4} = 1/2$). Option A incorrectly suggests it increases. Option C is wrong - sample size does affect spread. Option D has the wrong factor. Option E makes no sense in this context.

Question 2

1. It is centered at $\mu_{ER}-\mu_{UC}$, and its variability depends on both $n_1$ and $n_2$. (correct answer)
2. It is centered at $0$ because each sample mean is an unbiased estimator.
3. Its standard deviation depends only on the smaller sample size, $n_2=20$.
4. It has no variability because $\mu_{ER}$ and $\mu_{UC}$ are fixed values.
5. It is the distribution of individual wait-time differences between one ER patient and one Urgent Care patient.

Explanation: This question tests understanding of how the sampling distribution behaves with different sample sizes. The sampling distribution of $\bar{x}_{ER}-\bar{x}_{UC}$ is centered at $\mu_{ER}-\mu_{UC}$, and its standard deviation is $\sqrt{\frac{\sigma_{ER}^2}{100} + \frac{\sigma_{UC}^2}{20}}$, which depends on both sample sizes. Option B is incorrect - the center is the difference in population means, not zero. Option C is wrong because both sample sizes affect variability. Option D incorrectly claims no variability exists. Option E confuses the sampling distribution with individual patient differences.

Question 3

1. It is approximately normal only if the population distributions are exactly normal.
2. Its mean is $\mu_1-\mu_2$, regardless of the sample sizes. (correct answer)
3. Its standard deviation is $\sigma_1/\sqrt{n_1}-\sigma_2/\sqrt{n_2}$ because the statistic subtracts the means.
4. It has less variability when $n_1$ increases but is unaffected by $n_2$.
5. It has no variability if the same number of deliveries is sampled from each warehouse.

Explanation: This question asks about properties of the sampling distribution for the difference in sample means. The mean of the sampling distribution of $\bar{x}_1-\bar{x}_2$ is always $\mu_1-\mu_2$, regardless of sample sizes, because sample means are unbiased estimators. This is a fundamental property that holds for any sample sizes. Option A is incorrect because the Central Limit Theorem allows for approximate normality with large samples even if populations aren't normal. Option C gives an incorrect formula for the standard deviation. Option D is wrong because both sample sizes affect variability. Option E is incorrect because sampling variability always exists when taking samples.

Question 4

1. It is centered at $\mu_A-\mu_B$, and its standard deviation decreases when either $n_1$ or $n_2$ increases. (correct answer)
2. It is centered at $0$ whenever the two sample sizes are equal.
3. It has no variability because the same two programs are being compared each time.
4. Its standard deviation is $\sigma_A-\sigma_B$ because the statistic is a difference.
5. It describes the distribution of individual score differences between one student from A and one student from B.

Explanation: This question tests understanding of the sampling distribution of differences in sample means. The sampling distribution of $\bar{x}_A-\bar{x}_B$ is centered at the difference in population means, $\mu_A-\mu_B$, because each sample mean is an unbiased estimator of its population mean. The standard deviation of this distribution is $\sqrt{\frac{\sigma_A^2}{n_1} + \frac{\sigma_B^2}{n_2}}$, which decreases as either sample size increases. Option B is incorrect because the center depends on population means, not sample sizes. Option C is wrong because sampling variability exists even when comparing the same programs repeatedly. Option D incorrectly states the standard deviation formula. Option E confuses the sampling distribution with individual differences.

Question 5

1. The variability depends only on $n_Q$ because $\bar{x}_Q$ is listed first in the difference.
2. If both sample sizes were doubled, the sampling distribution of $\bar{x}_Q-\bar{x}_M$ would typically have less spread. (correct answer)
3. The sampling distribution has the same spread as the distribution of puzzle times for individuals in condition Q.
4. There is no spread because each sample mean equals the corresponding population mean.
5. The spread increases as sample sizes increase because more data create more possible values of $\bar{x}_Q-\bar{x}_M$.

Explanation: This question tests understanding of how sample size affects the variability of x̄_Q - x̄_M. The standard deviation of this sampling distribution is √(σ_Q²/n_Q + σ_M²/n_M), which decreases as sample sizes increase. Doubling both sample sizes would reduce each variance term by half, thereby reducing the overall standard deviation and spread of the sampling distribution. Choice A incorrectly ignores the contribution of the second sample. Choice C confuses the sampling distribution with the population distribution. Choice D wrongly claims no spread exists. Choice E reverses the relationship - larger samples actually reduce spread in the sampling distribution.

Question 6

1. It will be approximately normal whenever the two population distributions are approximately normal. (correct answer)
2. It will be exactly normal for any population distributions because the sample sizes are equal.
3. It cannot be approximately normal unless $n_B$ and $n_N$ are both at least 100.
4. It will be approximately normal only if the population means are equal.
5. It will be approximately normal because a difference of two sample means always looks normal, regardless of sample size and population shape.

Explanation: This question asks about conditions for approximate normality of x̄_B - x̄_N with small samples (n_B = n_N = 10). Since the samples are too small for the Central Limit Theorem to apply, we need the populations themselves to be approximately normal for the sampling distribution of the difference to be approximately normal. This makes choice A correct. Choice B wrongly claims equal sample sizes guarantee normality. Choice C sets an unnecessarily high bar - samples of 30+ often suffice for the CLT. Choice D incorrectly links normality to equal population means. Choice E makes the false claim that differences of sample means are always approximately normal.

Question 7

1. It is centered at $\mu_L-\mu_S$. (correct answer)
2. It is centered at $\bar{x}_L-\bar{x}_S$ from the most recent sample.
3. It is centered at $0$ because sampling makes the two groups equal on average.
4. It is centered at $\mu_L+\mu_S$ because both groups contribute to the total.
5. It is centered at the median of the combined population because means are sensitive to skew.

Explanation: This question asks about the center of the sampling distribution for x̄_L - x̄_S. A fundamental property of sampling distributions is that the expected value (mean) of x̄_L - x̄_S equals μ_L - μ_S, the difference in population means. This holds true regardless of sample sizes, population shapes, or variability. Choice B incorrectly uses a single observed difference instead of the theoretical center. Choice C wrongly assumes sampling equalizes the groups. Choice D incorrectly adds the means instead of subtracting. Choice E confuses the mean with the median and incorrectly considers the combined population.

Question 8

1. It becomes 4 times as large because the sample sizes are 4 times as large.
2. It becomes about half as large because each standard error term involves $\sqrt{n}$. (correct answer)
3. It stays the same because the population standard deviations do not change.
4. It becomes 4 times smaller because the sample sizes are 4 times as large.
5. It becomes 2 times larger because there are two groups instead of one.

Explanation: This question examines how sample size affects the standard deviation of the sampling distribution. When both sample sizes are quadrupled from 16 to 64, each term in the standard deviation formula $\sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}$ is divided by 4, making the entire expression half as large (since $\sqrt{1/4} = 1/2$). Option A incorrectly suggests it increases. Option C is wrong - sample size does affect spread. Option D has the wrong factor. Option E makes no sense in this context.

Question 9

1. Increasing $n_2$ (from 48 to a larger value) will reduce the spread of $\bar{x}_A-\bar{x}_B$, even if $n_1$ stays 12. (correct answer)
2. Increasing $n_2$ will not change the spread because the smaller sample size $n_1=12$ controls all variability.
3. Increasing $n_2$ will shift the center of the sampling distribution to 0.
4. Increasing $n_2$ will eliminate sampling variability entirely because City B's mean becomes known.
5. Changing either sample size affects only the center, not the spread, of the sampling distribution.

Explanation: This question examines how unequal sample sizes affect the sampling distribution. The standard deviation is $\sqrt{\frac{\sigma_A^2}{12} + \frac{\sigma_B^2}{48}}$. Increasing $n_2$ from 48 will reduce the second term under the square root, thereby reducing the overall standard deviation of $\bar{x}_A-\bar{x}_B$. Option B is incorrect - both sample sizes matter, not just the smaller one. Option C is wrong - sample size doesn't affect the center. Option D incorrectly claims variability is eliminated. Option E is false - sample sizes affect spread, not center.

Question 10

1. It must be exactly normal because it is a difference of two means.
2. It is likely to be closer to normal if both sample sizes were increased. (correct answer)
3. Its center is $\bar{x}_A-\bar{x}_B$ because the statistic determines the mean of its own sampling distribution.
4. It describes the distribution of differences in height for pairs of individual plants, one from each fertilizer group.
5. It has no variability because the populations are fixed and known to be skewed.

Explanation: This question examines the effect of small sample sizes and skewed populations on the sampling distribution. With small samples (n=10 each) from strongly skewed populations, the sampling distribution of $\bar{x}_A-\bar{x}_B$ may not be approximately normal. However, increasing both sample sizes would help the distribution become more normal due to the Central Limit Theorem. Option A is too restrictive - exact normality isn't required. Option C incorrectly identifies the center. Option D misinterprets what the sampling distribution represents. Option E is wrong because sampling variability exists regardless of population shape.

Question 11

1. It must be skewed because $n_1\ne n_2$.
2. It will be approximately normal because both population distributions are roughly symmetric and the sample sizes are moderate to large. (correct answer)
3. It will be exactly normal only if $\sigma_1=\sigma_2$.
4. It has no variability because $n_2$ is large.
5. It will be approximately uniform because percentages are bounded between 0 and 100.

Explanation: This question addresses the shape of the sampling distribution with unequal sample sizes. With roughly symmetric population distributions and sample sizes of 20 and 80, the Central Limit Theorem applies reasonably well to both sample means. Since each sample mean is approximately normal, their difference is also approximately normal. Choice A incorrectly claims unequal sample sizes cause skewness. Choice C wrongly requires equal population standard deviations for approximate normality. Choice D incorrectly states that a large n₂ eliminates all variability. Choice E misunderstands the nature of percentage data - while bounded, the sampling distribution of means can still be approximately normal. The key is that both samples are large enough for the CLT to apply given the symmetric populations.

Question 12

1. It must be strongly right-skewed because both populations are right-skewed.
2. It will be approximately normal only if both population distributions are normal.
3. It will have mean $\mu_1-\mu_2$ regardless of skewness, but with $n_1=n_2=15$ it may not be approximately normal. (correct answer)
4. It has no variability because the same sample size is used in both cities.
5. Its mean is $0$ because differences average out over repetitions.

Explanation: This question addresses sampling distributions when populations are skewed and sample sizes are small. The mean of the sampling distribution of $\bar{x}_1 - \bar{x}_2$ is always $\mu_1 - \mu_2$, regardless of the shape of the population distributions. However, with strongly right-skewed populations and small sample sizes (n=15), the Central Limit Theorem may not fully apply, so the sampling distribution may not be approximately normal. Choice A incorrectly assumes the sampling distribution inherits the population's skewness directly. Choice D wrongly claims equal sample sizes eliminate variability. Choice E incorrectly states the mean is 0. The key insight is that while the mean is predictable, the shape may not be normal with small samples from skewed populations.

Question 13

1. The variability depends only on $n_1$ because Neighborhood 1 has the larger sample.
2. The variability depends on both $n_1$ and $n_2$, and is driven more by the smaller sample size $n_2=50$. (correct answer)
3. The variability is $0$ because the samples are large.
4. The variability is $\sigma$ because the standard deviation of a mean equals the population standard deviation.
5. The variability is larger when $n_1$ increases because larger samples create more possible means.

Explanation: This question examines how different sample sizes affect variability in the sampling distribution. The standard deviation of $\bar{x}_1 - \bar{x}_2$ is $\sqrt{\sigma^2/n_1 + \sigma^2/n_2} = \sigma\sqrt{1/200 + 1/50} = \sigma\sqrt{0.005 + 0.02} = \sigma\sqrt{0.025}$. The term $\sigma^2/50$ contributes four times as much to the variance as $\sigma^2/200$, so the smaller sample size (n=50) drives more of the variability. Choice A incorrectly ignores the second sample. Choice C wrongly claims large samples eliminate variability. Choice D confuses the standard deviation of the sampling distribution with the population standard deviation. Choice E incorrectly suggests larger samples increase variability, when they actually decrease it.

Question 14

1. The sampling distribution of $\bar{x}_1-\bar{x}_2$ has mean $\mu_1-\mu_2$ and is approximately normal. (correct answer)
2. The sampling distribution of $\bar{x}_1-\bar{x}_2$ has mean $0$ because sample means fluctuate around each other.
3. The sampling distribution is not approximately normal because $\sigma_1\ne\sigma_2$.
4. The standard deviation is $\sigma_1/n_1+\sigma_2/n_2$.
5. There is no sampling variability because $n_1=n_2$.

Explanation: This question examines the sampling distribution when population standard deviations differ. The sampling distribution of $\bar{x}_1 - \bar{x}_2$ has mean $\mu_1 - \mu_2$ and standard deviation $\sqrt{\sigma_1^2/n_1 + \sigma_2^2/n_2} = \sqrt{256/64 + 100/64} = \sqrt{4 + 1.5625} = \sqrt{5.5625} \approx 2.36$. With sample sizes of 64 each and approximately normal populations, the Central Limit Theorem ensures the sampling distribution is approximately normal. Choice B incorrectly claims the mean is 0. Choice C wrongly suggests unequal population standard deviations prevent normality. Choice D shows the wrong formula (should have squares under the radical). Choice E incorrectly claims equal sample sizes eliminate variability.

Question 15

1. The standard deviation of $\bar{x}_H-\bar{x}_E$ is $\sqrt{\sigma_H^2/n_H+\sigma_E^2/n_E}$, so it depends on both population standard deviations and both sample sizes. (correct answer)
2. The standard deviation of $\bar{x}_H-\bar{x}_E$ is $\sigma_H-\sigma_E$ because standard deviations subtract when means subtract.
3. The mean of $\bar{x}_H-\bar{x}_E$ is $\bar{x}_H-\bar{x}_E$ because the sampling distribution is built from the observed samples.
4. The sampling distribution has no variability if the two samples are the same size.
5. The standard deviation of $\bar{x}_H-\bar{x}_E$ is $\sqrt{\sigma_H^2+\sigma_E^2}$ because the two groups are independent.

Explanation: This question tests understanding of the standard deviation formula for differences in sample means. For independent samples, the standard deviation of $\bar{x}_H - \bar{x}_E$ is $\sqrt{\sigma_H^2/n_H + \sigma_E^2/n_E}$, which depends on both population standard deviations and both sample sizes. Choice B incorrectly subtracts standard deviations, but variances (not standard deviations) combine additively for independent variables. Choice C confuses parameters with statistics. Choice E forgets to divide by sample sizes, giving the formula for the difference of two individual observations. The crucial concept is that when finding the variability of a difference in sample means, we must account for the variability in each group and the precision gained from each sample size.

Question 16

1. If both populations are approximately normal, then $\bar{x}_1-\bar{x}_2$ is approximately normal. (correct answer)
2. Regardless of population shape, $\bar{x}_1-\bar{x}_2$ is exactly normal because the samples are the same size.
3. The mean of $\bar{x}_1-\bar{x}_2$ is $0$ because differences cancel out in repeated sampling.
4. The standard deviation of $\bar{x}_1-\bar{x}_2$ is $\sigma_1+\sigma_2$.
5. The variability of $\bar{x}_1-\bar{x}_2$ is the same as the variability of individual differences $x_1-x_2$.

Explanation: This question tests understanding of when the sampling distribution of x̄₁ - x̄₂ is approximately normal. With small samples (n₁ = n₂ = 15), the Central Limit Theorem doesn't guarantee normality, so we need the populations themselves to be approximately normal for the difference in sample means to be approximately normal. This makes choice A correct. Choice B is wrong because equal sample sizes don't ensure normality with small samples. Choice C incorrectly claims the mean is 0; it's actually μ₁ - μ₂. Choice D gives the wrong formula for standard deviation - it should involve √(σ₁²/n₁ + σ₂²/n₂), not σ₁ + σ₂. Choice E confuses the variability of sample mean differences with individual differences.

Question 17

1. It is centered at 0, but sample-to-sample variability means $\bar{x}_A-\bar{x}_B$ will not always equal 0. (correct answer)
2. It is centered at $\bar{x}_A-\bar{x}_B$ from the one set of samples collected.
3. It has no spread because the true means are equal.
4. It describes the distribution of height differences for each matched pair of plants.
5. It must be skewed because it is a difference of two sample means.

Explanation: We're testing the center and variability of the sampling distribution when population means are equal. It's centered at 0 ($\mu_A - \mu_B = 0$), but sampling variability ensures $\bar{x}_A - \bar{x}_B$ fluctuates around 0, not always equaling it. Choice C distracts by suggesting no spread when means are equal, overlooking random variation. Mini-lesson: the difference distribution describes variation in $\bar{x}_A - \bar{x}_B$ over repetitions; even with equal $\mu$, spread is $\sqrt{\sigma_A^2/n_A + \sigma_B^2/n_B} > 0$. It's for independent samples, not matched pairs. Choice A captures this accurately.

Question 18

1. The variability is determined only by $n_1$ because it is the larger sample.
2. The variability is larger than it would be if both sample sizes were 80, because the smaller sample size $n_2=20$ contributes more variability. (correct answer)
3. The variability is zero if both groups are sampled randomly.
4. The sampling distribution is the distribution of $x_1-x_2$ for randomly paired patients.
5. The sampling distribution must be centered at 0 because blood pressure is measured on the same scale in both groups.

Explanation: Here, we're examining the variability in the sampling distribution of the difference in means for unequal sample sizes. The center remains $\mu_1 - \mu_2$, but the spread $\sqrt{\sigma^2/n_1 + \sigma^2/n_2}$ is larger when one sample is small, as the smaller n contributes more to the variance. Choice A distracts by suggesting variability depends only on the larger sample, ignoring the additive nature of variances. Mini-lesson: for independent samples, the difference distribution's standard deviation combines the individual sampling variances; with n1=80 and n2=20, the term 1/n2=0.05 dominates over 1/n1=0.0125, increasing overall spread compared to equal larger samples. This makes the distribution more variable, as in choice B. Assuming equal population SDs simplifies but doesn't change the principle.

Question 19

1. It is approximately normal because each sample size is moderately large and the samples are independent. (correct answer)
2. It must have the same shape as the population distribution of GPA.
3. It is uniform because GPAs are bounded between 0 and 4.
4. It cannot be approximately normal unless $n_1$ and $n_2$ are different.
5. It has no reason to be approximately normal because it is based on two samples rather than one.

Explanation: The focus is on the shape of the sampling distribution for the difference in means. With moderately large samples (n=45) and non-extremely skewed populations, it's approximately normal due to the central limit theorem for independent samples. Choice B distracts by equating it to the population shape, which applies to small samples but not here. Mini-lesson: the difference distribution combines two sampling distributions, becoming normal for large n regardless of population shape (if not too skewed), with center $\mu_{on} - \mu_{off}$ and spread from added variances. Equal or unequal n doesn't affect normality eligibility. Choice A is correct.

Question 20

1. It represents the distribution of $\bar{x}_1-\bar{x}_2$ values from repeating the sampling process many times. (correct answer)
2. It represents the distribution of all satisfaction scores from Store 1 minus all satisfaction scores from Store 2.
3. It represents the distribution of differences in satisfaction for the same customers surveyed at both stores.
4. It represents the distribution of $\mu_1-\mu_2$ across different samples.
5. It represents a distribution with zero spread because both samples are the same size.

Explanation: This question clarifies what the sampling distribution of the difference in means represents. It's the distribution of $\bar{x}_1 - \bar{x}_2$ values from many independent samplings, centered at $\mu_1 - \mu_2$ with positive spread. Choice B is a distractor, mixing it up with all possible individual differences rather than sample means. Mini-lesson: unlike paired data, independent samples yield a difference distribution focused on means, not individual pairings or combined populations. Equal sample sizes don't eliminate variability. Choice A correctly defines it as the long-run distribution from repetitions.