Sampling for Differences in Sample Means - AP Statistics

1. A nutritionist compares two diets by repeatedly selecting independent random samples: $n_1=30$ people on Diet A and $n_2=50$ people on Diet B, then computing the difference in sample mean weight loss, $\bar{x}_A-\bar{x}_B$. Which statement is correct about the mean of the sampling distribution of $\bar{x}_A-\bar{x}_B$?​
   Answer: It equals $\mu_A-\mu_B$ because $\bar{x}_A$ and $\bar{x}_B$ are unbiased estimators of their population means. This question tests understanding of the mean of the sampling distribution. Since $\bar{x}_A$ is an unbiased estimator of $\mu_A$ and $\bar{x}_B$ is an unbiased estimator of $\mu_B$, their difference $\bar{x}_A-\bar{x}_B$ is an unbiased estimator of $\mu_A-\mu_B$. This means the sampling distribution is centered at $\mu_A-\mu_B$. Option A confuses a single sample statistic with the mean of the sampling distribution. Option C is incorrect - independence doesn't make the mean zero. Option D gives an incorrect formula. Option E is wrong - the mean doesn't depend on equal sample sizes.
2. A researcher compares average reaction time under two conditions. For each repetition, an independent random sample of $n_1=35$ people is assigned to Condition 1 and an independent random sample of $n_2=35$ people is assigned to Condition 2, and the statistic $\bar{x}_1-\bar{x}_2$ is computed. Which statement is correct about interpreting the sampling distribution of $\bar{x}_1-\bar{x}_2$?
   Answer: It describes how the difference in sample means would vary from repetition to repetition of the sampling process. This question tests interpretation of the sampling distribution concept. The sampling distribution of $\bar{x}_1-\bar{x}_2$ describes the pattern of values this statistic would take across many repetitions of the sampling process - it shows how the difference in sample means varies from sample to sample. Option A confuses it with individual differences. Option C is wrong - sampling variability exists. Option D incorrectly uses sample values instead of population parameters. Option E incorrectly identifies what the center represents.
3. Two independent random samples are taken to compare average battery life (hours) for Brand X and Brand Y. In repeated sampling, $n_X=12$ and $n_Y=48$ are used to compute $\bar{x}_X-\bar{x}_Y$. Which statement is correct?
   Answer: The sampling distribution of $\bar{x}_X-\bar{x}_Y$ describes how the difference $\bar{x}_X-\bar{x}_Y$ varies from sample to sample. This question tests the definition of a sampling distribution for the difference in sample means. The sampling distribution of x̄_X - x̄_Y describes the pattern of variation in the statistic x̄_X - x̄_Y across all possible samples of the specified sizes, making choice A correct. Choice B confuses this with the distribution of individual differences. Choice C incorrectly claims that a large $n_Y$ eliminates all variability, but both samples contribute to the spread. Choice D wrongly uses a single observed value as the mean instead of μ_X - μ_Y. Choice E incorrectly describes the shape as uniform when it's typically bell-shaped due to the CLT.
4. A hospital compares mean recovery time (days) for two treatments. An independent random sample of $n_1=45$ patients receives Treatment 1 and $n_2=60$ patients receives Treatment 2. Let $\bar{x}_1-\bar{x}_2$ be the difference in sample means. Which statement is correct about independence and the standard deviation of the sampling distribution of $\bar{x}_1-\bar{x}_2$?
   Answer: Independence allows the variances to add, giving standard deviation $\sqrt{\sigma_1^2/n_1+\sigma_2^2/n_2}$. This question explores the role of independence in determining the standard deviation of a difference in means. When samples are independent, the variances of the two sample means add, giving a standard deviation of $\sqrt{\sigma_1^2/n_1 + \sigma_2^2/n_2}$ for their difference. Choice A forgets to take the square root and incorrectly adds standard deviations instead of variances. Choice C ignores the effect of sample sizes. Choice D confuses independence's role in the spread with its (non)effect on the center. Choice E incorrectly claims independence eliminates variability. The key insight is that independence allows us to use the simple variance-addition formula; without independence, the formula would need to account for correlation between the samples.
5. A sports analyst compares average free-throw percentages for two independent teams. In each repetition, the analyst takes an independent random sample of $n_1=20$ players from Team 1 and $n_2=80$ players from Team 2 and computes $\bar{x}_1-\bar{x}_2$. Assume the population distributions of player percentages are roughly symmetric with standard deviations $\sigma_1$ and $\sigma_2$. Which statement is correct about the shape of the sampling distribution of $\bar{x}_1-\bar{x}_2$?
   Answer: It will be approximately normal because both population distributions are roughly symmetric and the sample sizes are moderate to large. 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.
6. Two independent random samples are taken to compare mean daily screen time (hours) for teens in two regions. Region 1 has sample size $n_1=100$ and Region 2 has sample size $n_2=100$. Let $\bar{x}_1-\bar{x}_2$ be the difference in sample means. Which statement is correct about the variability of $\bar{x}_1-\bar{x}_2$ compared with the variability of individual screen times?
   Answer: The difference of sample means typically varies less than individual screen times because averaging reduces variability. This question compares the variability of sample means with the variability of individual observations. Sample means vary less than individual observations because averaging reduces variability by a factor of $\sqrt{n}$. The difference of sample means inherits this reduced variability from both samples. Choice A reverses this relationship. Choice C confuses the sampling distribution with individual-level comparisons. Choice D incorrectly claims no variability exists. The fundamental concept is that sample means are more stable than individual values, and this stability carries over to differences in sample means, making them useful for comparing populations despite individual-level variation.
7. A school compares two study programs by repeatedly taking random samples. Each repetition selects an independent random sample of $n_1=40$ students from Program A and $n_2=40$ students from Program B and records each group's mean test score, $\bar{x}_A$ and $\bar{x}_B$. The population standard deviations are known to be $\sigma_A=12$ and $\sigma_B=12$, and the population distributions are roughly symmetric with no extreme outliers. The statistic of interest is the difference in sample means $\bar{x}_A-\bar{x}_B$. Which statement about the sampling distribution of $\bar{x}_A-\bar{x}_B$ is correct?
   Answer: Its mean is $\mu_A-\mu_B$, and its standard deviation is $\sqrt{\sigma_A^2/n_1+\sigma_B^2/n_2}$. This question tests understanding of the sampling distribution of differences in sample means. When we repeatedly take independent samples and compute $\bar{x}_A - \bar{x}_B$, the sampling distribution has mean $\mu_A - \mu_B$ (the difference in population means) and standard deviation $\sqrt{\sigma_A^2/n_1 + \sigma_B^2/n_2}$. Choice C incorrectly claims that standard deviations subtract, but variances actually add when finding the variance of a difference. Choice D wrongly suggests equal sample sizes eliminate variability, but sampling variability always exists. Choice E confuses the center of the sampling distribution (which is $\mu_A - \mu_B$) with individual sample statistics. Since both populations are roughly symmetric and sample sizes are 40 each, the Central Limit Theorem ensures the sampling distribution is approximately normal.
8. A manufacturer compares the mean lifetime of batteries from two production lines. In repeated sampling, an independent random sample of $n_1=16$ batteries from Line 1 and $n_2=16$ batteries from Line 2 is tested, and $\bar{x}_1-\bar{x}_2$ is recorded. Which statement is correct about how the standard deviation of the sampling distribution changes if both sample sizes are quadrupled (to $n_1=n_2=64$)?​
   Answer: It becomes about half as large because each standard error term involves $\sqrt{n}$. 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.
9. A scientist compares mean plant growth (cm) under two fertilizers. She takes an independent random sample of $n_F=20$ plants with Fertilizer F and $n_G=80$ plants with Fertilizer G, then considers the sampling distribution of $\bar{x}_F-\bar{x}_G$. Which statement is correct about the standard deviation of $\bar{x}_F-\bar{x}_G$?
   Answer: It depends on $n_F$, $n_G$, $\sigma_F$, and $\sigma_G$ through $\sqrt{\sigma_F^2/n_F+\sigma_G^2/n_G}$. This question tests understanding of what determines the standard deviation of a difference in sample means. The correct formula is $\sqrt{\sigma_F^2/n_F + \sigma_G^2/n_G}$, which depends on both population standard deviations and both sample sizes. Choice A incorrectly focuses only on the first group. Choice C treats standard deviations as a ratio rather than using the variance-addition formula. Choice D incorrectly multiplies by sample sizes instead of dividing. Choice E confuses the standard deviation (which measures spread) with the mean (which measures center). The key principle is that the variability in the difference depends on the variability in each group, reduced by the respective sample sizes, with variances combining additively due to independence.
10. Two independent random samples are drawn to compare average daily screen time (hours) for teens in Region A and Region B. Each repetition takes $n_A=50$ from Region A and $n_B=50$ from Region B and computes $\bar{x}_A-\bar{x}_B$. Which statement is correct?
    Answer: The mean of the sampling distribution of $\bar{x}_A-\bar{x}_B$ is $\mu_A-\mu_B$. This question tests the fundamental property that the sampling distribution of x̄_A - x̄_B is centered at μ_A - μ_B, the difference in population means. This is true regardless of sample sizes or population distributions - the expected value of the difference in sample means equals the difference in population means. Choice B incorrectly describes the standard deviation formula, which should be √(σ_A²/50 + σ_B²/50), not σ_A/50. Choice C is wrong because equal sample sizes don't eliminate variability. Choice D incorrectly adds the means instead of subtracting. Choice E confuses the sampling distribution of the difference in sample means with the distribution of individual differences.
11. To compare two fertilizers, a researcher repeatedly takes an independent random sample of $n_1=10$ plants grown with Fertilizer A and $n_2=10$ plants grown with Fertilizer B, then records the mean height for each sample and computes $\bar{x}_A-\bar{x}_B$. Suppose the population of heights for each fertilizer is strongly right-skewed, with no extreme outliers. Which statement is correct about the sampling distribution of $\bar{x}_A-\bar{x}_B$?
    Answer: It is likely to be closer to normal if both sample sizes were increased. 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.
12. An online retailer compares the mean number of items per order for two types of customers. Each week, it takes an independent random sample of $n_1=60$ new customers and $n_2=60$ returning customers and computes $\bar{x}_{new}-\bar{x}_{ret}$. Which statement is correct about what makes the sampling distribution of $\bar{x}_{new}-\bar{x}_{ret}$ approximately normal?
    Answer: It is approximately normal if both populations are normal or if both sample sizes are large enough for the CLT to apply. This question addresses conditions for approximate normality of the sampling distribution. The sampling distribution of $\bar{x}_{new}-\bar{x}_{ret}$ is approximately normal if either both populations are normal OR both sample sizes are large enough for the CLT to apply to each sample mean individually. With n=60 for each group, the CLT likely applies. Option A incorrectly links normality to equal means. Option B is wrong - small samples need normal populations. Option D is false - differences can have normal distributions. Option E incorrectly requires equal population standard deviations.
13. A nutritionist compares mean sodium intake (mg/day) for two independent groups: people following Diet 1 and people following Diet 2. She takes an SRS of $n_1=50$ from Diet 1 and an independent SRS of $n_2=10$ from Diet 2 and computes $\bar{x}_1-\bar{x}_2$. Which statement is correct about how changing sample sizes affects the sampling distribution of $\bar{x}_1-\bar{x}_2$?
    Answer: Increasing $n_2$ (the smaller sample) would generally reduce the standard deviation of $\bar{x}_1-\bar{x}_2$. The skill involves how sample size changes impact the sampling distribution of the mean difference. The center stays at $\mu_1 - \mu_2$ regardless of n, but increasing the smaller sample size reduces the spread more effectively by targeting the larger variance term. Choice A distracts by claiming doubling both increases variability, which is the opposite of the truth. Mini-lesson: the difference distribution's standard deviation $\sqrt{\sigma_1^2/n_1 + \sigma_2^2/n_2}$ decreases with larger n; for unequal sizes like n1=50 and n2=10, boosting n2 shrinks the dominant $\sigma_2^2/n_2$ term. Unequal n doesn't prevent normality for large samples. Thus, choice B is accurate.
14. A scientist compares mean plant height (cm) for plants grown under Light A versus Light B. She takes an SRS of $n_1=36$ plants under Light A and an independent SRS of $n_2=36$ plants under Light B, then computes $\bar{x}_A-\bar{x}_B$. If the two populations have equal means ($\mu_A=\mu_B$), which statement is correct about the sampling distribution of $\bar{x}_A-\bar{x}_B$?
    Answer: It is centered at 0, but sample-to-sample variability means $\bar{x}_A-\bar{x}_B$ will not always equal 0. 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.
15. A college compares mean GPA for students living on campus versus off campus. An SRS of $n_1=45$ on-campus students and an independent SRS of $n_2=45$ off-campus students are selected, and $\bar{x}_{on}-\bar{x}_{off}$ is computed. Assume independence and that both populations are not extremely skewed. Which statement is correct about the shape of the sampling distribution of $\bar{x}_{on}-\bar{x}_{off}$?
    Answer: It is approximately normal because each sample size is moderately large and the samples are independent. 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.
16. A school compares the mean number of minutes of homework done per night by students in two different programs. A random sample of $n_1=40$ students from Program A and an independent random sample of $n_2=35$ students from Program B are taken. Let $\bar{x}_A-\bar{x}_B$ be the difference in sample means. Which statement is correct about the sampling distribution of $\bar{x}_A-\bar{x}_B$?
    Answer: It is centered at $\mu_A-\mu_B$ and has standard deviation $\sqrt{\sigma_A^2/n_1+\sigma_B^2/n_2}$. This question tests understanding of the sampling distribution of the difference between two sample means. When we have two independent random samples, the sampling distribution of $\bar{x}_A - \bar{x}_B$ is centered at the difference of the population means, $\mu_A - \mu_B$. The standard deviation follows the formula $\sqrt{\sigma_A^2/n_1 + \sigma_B^2/n_2}$, which comes from the fact that variances add for independent random variables. Choice B incorrectly uses sample statistics instead of population parameters for the center. Choice D forgets to square the population standard deviations before dividing by sample sizes. Choice E incorrectly adds the population variances without dividing by sample sizes. When comparing two groups, remember that the sampling distribution describes what happens across all possible samples, not just the one sample you collected.
17. A nutritionist compares mean sodium intake (mg) for two independent populations: people who eat breakfast daily (B) and people who do not (N). In repeated sampling, she takes an SRS of $n_B=10$ and an SRS of $n_N=10$ and calculates $\bar{x}_B-\bar{x}_N$. Which statement is correct about when the sampling distribution of $\bar{x}_B-\bar{x}_N$ will be approximately normal?
    Answer: It will be approximately normal whenever the two population distributions are approximately normal. 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.
18. To compare mean systolic blood pressure for two independent groups, a clinic takes an SRS of $n_1=80$ patients who exercise regularly and an independent SRS of $n_2=20$ patients who do not. The statistic is $\bar{x}_1-\bar{x}_2$. Assume both populations have the same standard deviation. Which statement is correct about the variability of the sampling distribution of $\bar{x}_1-\bar{x}_2$?
    Answer: 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. 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.
19. A hospital compares patient wait times in two departments. Repeatedly, it takes an independent random sample of $n_1=100$ patients from the ER and $n_2=20$ patients from Urgent Care and computes $\bar{x}_{ER}-\bar{x}_{UC}$. Which statement is correct about the sampling distribution of $\bar{x}_{ER}-\bar{x}_{UC}$?
    Answer: It is centered at $\mu_{ER}-\mu_{UC}$, and its variability depends on both $n_1$ and $n_2$. 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.
20. A consumer group repeatedly compares two brands of batteries by sampling. In each repetition, an independent random sample of $n_1=9$ batteries of Brand A and $n_2=9$ batteries of Brand B is tested for lifetime (hours), and $\bar{x}_A-\bar{x}_B$ is recorded. Assume both population lifetime distributions are normal with standard deviations $\sigma_A$ and $\sigma_B$. Which statement is correct?
    Answer: The sampling distribution of $\bar{x}_A-\bar{x}_B$ is normal because each population is normal. This question tests understanding of sampling distributions when both populations are normal. When both populations are normal, the sampling distribution of $\bar{x}_A - \bar{x}_B$ is exactly normal regardless of sample size, because linear combinations of normal random variables are normal. Choice B incorrectly requires equal population standard deviations for normality. Choice C wrongly claims small equal sample sizes eliminate variability. Choice D confuses the mean of the sampling distribution ($\mu_A - \mu_B$) with individual sample statistics. Choice E shows a fundamental misunderstanding of how standard deviations combine - they don't divide. The exact normality comes from the mathematical property that differences of normal distributions remain normal.

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