Math 3 Quiz: Sampling Distributions Via Simulation
20 questions · exam conditions
0:00
Sampling Distributions Via SimulationQuestion 1 of 20

A quality control manager simulates the sampling distribution of sample proportions by repeatedly drawing samples of size 100 from a process where 15% of items are defective. After 500 simulation runs, she observes that approximately 95% of the sample proportions fall between 0.08 and 0.22. What does this interval most likely represent?

The theoretical range where 95% of individual items in the population will have defect rates between 8% and 22% of the time.
An empirical approximation of the interval within two standard errors of the true population proportion, confirming expected sampling variability patterns.
Evidence that the simulation parameters were set incorrectly, since the interval should be symmetric around 0.15 with equal distances to both endpoints.
The confidence interval for the true population proportion based on the simulation results, suggesting the actual defect rate is between 8% and 22%.
← Back to quizzes

Math 3 Quiz

Math 3 Quiz: Sampling Distributions Via Simulation

Practice Sampling Distributions Via Simulation in Math 3 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Sampling Distributions Via Simulation, giving you a quick way to practice the rules, question types, and explanations that matter most for Math 3.

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.

All questions

Question 1

A quality control manager simulates the sampling distribution of sample proportions by repeatedly drawing samples of size 100 from a process where 15% of items are defective. After 500 simulation runs, she observes that approximately 95% of the sample proportions fall between 0.08 and 0.22. What does this interval most likely represent?

  1. The theoretical range where 95% of individual items in the population will have defect rates between 8% and 22% of the time.
  2. An empirical approximation of the interval within two standard errors of the true population proportion, confirming expected sampling variability patterns. (correct answer)
  3. Evidence that the simulation parameters were set incorrectly, since the interval should be symmetric around 0.15 with equal distances to both endpoints.
  4. The confidence interval for the true population proportion based on the simulation results, suggesting the actual defect rate is between 8% and 22%.
Explanation: With p = 0.15 and n = 100, the standard error is √(0.15×0.85/100) ≈ 0.036. The interval 0.15 ± 2(0.036) = [0.078, 0.222] closely matches the observed [0.08, 0.22]. This represents the empirical sampling distribution showing where 95% of sample proportions fall. Choice A confuses individual items with sample proportions. Choice C incorrectly assumes perfect symmetry is required (the slight asymmetry is normal). Choice D confuses the sampling distribution with confidence intervals for parameter estimation.

Question 2

A quality assurance team runs a simulation with 1200 iterations, each time calculating the range (max - min) from samples of size 15 drawn from a normal population with mean 50 and standard deviation 8. The simulation produces a mean range of 19.2. What does this result primarily demonstrate about sampling distributions?

  1. The simulation confirms that sample ranges from normal populations follow a normal distribution with predictable mean values based on population parameters and sample size.
  2. The result shows how simulation can estimate the sampling distribution of statistics that don't have simple theoretical formulas, providing empirical insights into their behavior. (correct answer)
  3. The mean range of 19.2 indicates an error in the simulation since ranges from normal distributions should have means approximately equal to 2.5 times the population standard deviation.
  4. The simulation demonstrates that the Central Limit Theorem applies to sample ranges, showing convergence to normality as the number of simulation iterations increases sufficiently.
Explanation: The sampling distribution of sample ranges doesn't have a simple theoretical formula, making simulation valuable for understanding its properties. The result demonstrates simulation's power for studying complex statistics empirically. Choice A incorrectly assumes ranges follow normal distributions. Choice C applies an incorrect rule (the factor isn't simply 2.5 for n=15). Choice D incorrectly applies the CLT to the statistic itself rather than to simulation convergence.

Question 3

A quality control manager runs a simulation to study the sampling distribution of the sample standard deviation ss from a normal population with σ=12\sigma = 12. Using samples of size 16 over 1500 iterations, the simulation shows that the sample standard deviations average approximately 11.4 rather than 12.0. What does this result demonstrate?

  1. The simulation contains systematic error in random number generation, since sample standard deviations should provide unbiased population estimates.
  2. The simulation reveals the population standard deviation was incorrectly specified as 12, when the true value is closer to 11.4.
  3. The difference between 11.4 and 12.0 falls within expected random variation, and more iterations would eliminate this discrepancy.
  4. The result correctly demonstrates that sample standard deviations are biased estimators, typically underestimating the true population value. (correct answer)
Explanation: When you encounter questions about sampling distributions, focus on whether the statistic provides an unbiased estimate of the population parameter. This fundamental concept determines how we interpret simulation results. The sample standard deviation ss is a biased estimator of the population standard deviation σ\sigma. Due to the mathematical properties of how ss is calculated, it systematically underestimates σ\sigma on average. The simulation correctly demonstrates this bias—with σ=12\sigma = 12, the average sample standard deviation of 11.4 shows the expected downward bias. This is why statisticians often use Bessel's correction (dividing by n−1n-1 instead of nn) when calculating sample variance, though even this doesn't completely eliminate the bias in the standard deviation. Option A incorrectly assumes sample standard deviations should be unbiased—they're not. Option B misinterprets the result as indicating the wrong population parameter, when the simulation is actually revealing the estimator's bias. Option C suggests this is just random variation that more iterations would eliminate, but bias is systematic and persistent—more iterations would only make the pattern clearer, not eliminate it. The correct answer is D because it recognizes that sample standard deviations inherently underestimate population standard deviations, making this simulation result expected rather than problematic. Study tip: Remember that unbiasedness isn't universal—sample means are unbiased estimators, but sample standard deviations are biased. When analyzing sampling distributions, always consider whether the statistic systematically over- or under-estimates the parameter.

Question 4

A statistics instructor compares two simulations of sampling distributions: one using 500 iterations with samples of size 40, another using 1000 iterations with samples of size 20. Both simulations draw from populations with the same mean and standard deviation. Which statement best describes the trade-offs between these simulation designs?

  1. Both simulations produce equivalent results since the total number of observations (500×40 = 1000×20) is identical in both cases.
  2. The first simulation is superior because larger sample sizes always produce better sampling distributions regardless of iteration count.
  3. The first estimates a distribution with smaller standard error more precisely, while the second estimates one with larger standard error more precisely. (correct answer)
  4. The second simulation wastes resources because 1000 iterations cannot compensate for the inferior properties from smaller sample size.
Explanation: When you encounter questions about sampling distribution simulations, focus on two key concepts: standard error (which depends on sample size) and precision of estimation (which depends on number of iterations). The Central Limit Theorem tells us that sampling distributions have standard error σ/n\sigma/\sqrt{n}, where nn is the sample size. The first simulation (n=40) creates a sampling distribution with standard error σ/40≈0.158σ\sigma/\sqrt{40} \approx 0.158\sigma, while the second (n=20) creates one with standard error σ/20≈0.224σ\sigma/\sqrt{20} \approx 0.224\sigma. These are fundamentally different distributions. However, the precision with which we estimate each distribution's properties depends on iterations—more iterations give better estimates of the true sampling distribution shape and parameters. So the first simulation estimates a lower-variability sampling distribution using 500 iterations, while the second estimates a higher-variability sampling distribution using 1000 iterations. Answer C correctly captures this trade-off. Answer A incorrectly assumes total sample points matter most, ignoring that we're comparing different sampling distributions entirely. Answer B oversimplifies by claiming larger samples are always superior—this ignores that estimation precision also matters and depends on your research goals. Answer D dismissively claims the second simulation "wastes resources," but 1000 iterations actually provide more precise estimation of that particular sampling distribution's properties. Remember: sample size determines what sampling distribution you're studying, while iteration count determines how precisely you estimate that distribution. Both matter, but they serve different purposes in simulation design.

Question 5

A researcher wants to estimate the sampling distribution of sample means for samples of size 20 from a population with mean 75 and standard deviation 12. She runs a simulation with 1000 iterations, each time drawing a sample of size 20 and calculating the sample mean. If the simulation results show a mean of 74.8 and standard deviation of 2.7, which conclusion is most appropriate?

  1. The simulation confirms the theoretical sampling distribution with mean 75 and standard error approximately 2.68, suggesting the slight difference is due to random variation in the simulation process. (correct answer)
  2. The simulation indicates a systematic bias in the sampling method since the simulated mean of 74.8 differs significantly from the expected theoretical mean of 75.
  3. The simulation standard deviation of 2.7 is too small compared to the population standard deviation of 12, indicating an error in the simulation design or execution.
  4. The simulation results suggest the population parameters were incorrectly specified, as the observed sampling distribution parameters don't match the theoretical predictions closely enough.
Explanation: The theoretical sampling distribution should have mean = 75 and standard error = 12/√20 ≈ 2.68. The simulation results (mean = 74.8, SD = 2.7) are very close to these theoretical values. The small differences are expected due to random variation in the simulation. Choice B incorrectly interprets normal simulation variation as bias. Choice C misunderstands that the sampling distribution standard deviation should be much smaller than the population SD. Choice D incorrectly assumes the simulation should match theory exactly.

Question 6

Two students simulate sampling distributions from the same population using identical parameters, but Student A's histogram shows more irregular, jagged bars while Student B's histogram appears smoother. The most likely explanation for this difference is:

  1. Student A used a smaller number of simulation iterations than Student B, resulting in more random variation and less smooth approximation of the underlying sampling distribution. (correct answer)
  2. Student A's random number generator was faulty, producing non-random samples that created artificial irregularities in the simulated sampling distribution histogram.
  3. Student A used larger sample sizes in each iteration, causing increased variability in the sampling distribution and more irregular histogram appearance overall.
  4. Student A chose inappropriate bin widths for the histogram display, creating artificial jaggedness that doesn't reflect the true nature of the simulated sampling distribution.
Explanation: With fewer simulation iterations, random sampling variation creates more irregular histograms. More iterations smooth out these irregularities, providing better approximations of the underlying sampling distribution. Choice B suggests equipment failure, which is less likely than insufficient iterations. Choice C incorrectly suggests larger sample sizes increase rather than decrease sampling variability. Choice D focuses on display issues rather than the fundamental cause of irregularity.

Question 7

A market researcher wants to simulate the sampling distribution of the correlation coefficient between two variables. She generates 800 pairs of samples, each of size 30, from two populations with correlation ρ = 0.6. Her simulation produces sample correlations ranging from 0.1 to 0.9 with mean 0.58. What should she conclude about this simulation?

  1. The simulation is flawed because the sample correlations should range from -1 to +1, and the restricted range indicates improper sampling methodology.
  2. The results are reasonable since the mean 0.58 approximates the population correlation 0.6, and the range reflects expected sampling variation for n=30. (correct answer)
  3. The simulation demonstrates bias in correlation estimation since the mean 0.58 differs from the true correlation 0.6, suggesting systematic underestimation.
  4. The narrow range from 0.1 to 0.9 indicates the sample size 30 is too large, creating artificially reduced variability in correlation estimates.
Explanation: Sample correlations don't need to span the full theoretical range [-1,1]; their range depends on population correlation and sample size. With ρ=0.6 and n=30, the observed range and mean 0.58 are reasonable, with the small bias being typical of correlation estimation. Choice A misunderstands that sample ranges don't need to match theoretical ranges. Choice C overinterprets small simulation variation as bias. Choice D incorrectly suggests larger sample sizes create problems.

Question 8

A student runs two simulations to compare sampling distributions: Simulation A uses 200 samples of size 25 each, while Simulation B uses 50 samples of size 100 each. Both draw from the same population with mean 40 and standard deviation 15. Which statement best describes the expected differences between these simulations?

  1. Simulation A will provide a more accurate estimate of the sampling distribution shape since it uses four times as many sample iterations than Simulation B.
  2. Simulation B will show a sampling distribution with smaller variability around the mean, while Simulation A will provide a more precise estimate of that distribution.
  3. Both simulations will produce identical sampling distributions since they draw from the same population, with differences only due to random simulation error.
  4. Simulation A will have standard error 3.0 while Simulation B will have standard error 1.5, but Simulation A's estimate will be more reliable due to more iterations. (correct answer)
Explanation: The standard errors are: Simulation A: 15/√25 = 3.0; Simulation B: 15/√100 = 1.5. However, Simulation A uses 200 iterations vs. 50 for B, making A's estimate of its sampling distribution more reliable despite having larger standard error. Choice A ignores the different sample sizes. Choice B correctly identifies B's smaller variability but incorrectly suggests A provides more precision of the same distribution. Choice C incorrectly states the distributions will be identical when they have different standard errors.

Question 9

A pharmaceutical researcher simulates the sampling distribution of the sample variance s2s^2 using samples of size 25 from a normal population with variance σ2=16\sigma^2 = 16. After 2000 iterations, the simulation shows a right-skewed distribution with mean approximately 16. What does this simulation reveal about the sampling distribution of sample variances?

  1. The simulation demonstrates an error since sample variances from normal populations should follow a normal distribution centered at the population variance with symmetric shape.
  2. The right-skewed shape with mean 16 correctly reflects that sample variances follow a chi-square-based distribution, showing both unbiased estimation and characteristic skewness pattern. (correct answer)
  3. The results indicate the population was not actually normal, since normal populations would produce symmetric sampling distributions for all sample statistics including variance.
  4. The simulation shows that larger sample sizes are needed for the Central Limit Theorem to make the sampling distribution of variances approximately normal and symmetric.
Explanation: The sampling distribution of sample variances from normal populations follows a chi-square distribution (scaled), which is right-skewed. The mean equals σ² (unbiased), confirming correct simulation results. Choice A incorrectly expects normality and symmetry. Choice C incorrectly infers population non-normality from the expected skewed sampling distribution. Choice D incorrectly applies the CLT, which doesn't make variance sampling distributions normal.

Question 10

An environmental scientist simulates sampling distributions using bootstrap resampling from a dataset of 200 pollution measurements. She creates 1000 bootstrap samples of size 200 each (sampling with replacement). How does this bootstrap simulation differ from traditional Monte Carlo simulation in terms of what it estimates?

  1. Bootstrap and Monte Carlo are functionally equivalent methods that produce identical sampling distribution estimates with same parameters.
  2. Bootstrap provides more accurate results because it uses real data rather than artificially generated samples from theoretical distributions.
  3. Bootstrap estimates sampling distributions from observed data, while Monte Carlo uses assumed population parameters and theoretical models. (correct answer)
  4. Bootstrap estimates confidence intervals while Monte Carlo estimates sampling distributions, making them complementary rather than alternative approaches.
Explanation: When you encounter questions about bootstrap versus Monte Carlo methods, focus on the fundamental difference in their data sources and what they're trying to estimate. Bootstrap resampling works directly with your observed data. In this case, the scientist takes her 200 actual pollution measurements and repeatedly samples from them with replacement to create new datasets of the same size. This process estimates what the sampling distribution would look like if she could repeatedly sample from the same population that produced her original data. The key insight is that bootstrap uses the observed data as a proxy for the unknown population. Monte Carlo simulation, by contrast, starts with assumed population parameters and theoretical probability distributions. You specify a model (like "pollution follows a normal distribution with μ = 50 and σ = 10") and then generate artificial samples from that theoretical distribution. Option A is wrong because these methods serve different purposes and make different assumptions about what you know. Option B incorrectly suggests bootstrap is inherently more accurate—accuracy depends on how well your sample represents the population and whether Monte Carlo's assumptions are valid. Option D confuses the tools with their applications; both methods can estimate sampling distributions, and both can be used for confidence intervals. The correct answer is C because it captures the core distinction: bootstrap estimates sampling distributions from observed data without assuming you know the population parameters, while Monte Carlo requires you to specify theoretical population parameters upfront. Remember: Bootstrap = work with what you observed; Monte Carlo = work with what you assume.

Question 11

A statistics teacher asks students to simulate the sampling distribution of the difference between two sample means. Students draw samples of size 40 from Population A (mean = 85, SD = 20) and samples of size 60 from Population B (mean = 90, SD = 15). After 800 simulation runs, what should the mean and standard deviation of the simulated differences (xˉA−xˉB\bar{x}_A - \bar{x}_B) approximately equal?

  1. Mean = -5, Standard deviation = 20240+15260≈4.24\sqrt{\frac{20^2}{40} + \frac{15^2}{60}} \approx 4.24, assuming the simulation accurately reflects the theoretical sampling distribution of differences. (correct answer)
  2. Mean = 5, Standard deviation = 20240+15260≈4.24\sqrt{\frac{20^2}{40} + \frac{15^2}{60}} \approx 4.24, with the positive mean indicating Population A samples typically exceed Population B samples.
  3. Mean = -5, Standard deviation = 202+152=25\sqrt{20^2 + 15^2} = 25, since the difference of means uses the combined population standard deviations directly.
  4. Mean = -5, Standard deviation = 20+1540+60=3.50\frac{20+15}{\sqrt{40+60}} = 3.50, calculated using the pooled standard error formula for the combined sample sizes.
Explanation: The mean of the sampling distribution of differences is μA - μB = 85 - 90 = -5. The standard deviation is the standard error of the difference: √(σ²A/nA + σ²B/nB) = √(400/40 + 225/60) = √(10 + 3.75) = √13.75 ≈ 4.24. Choice B has the wrong sign for the mean. Choice C incorrectly uses population SDs instead of standard errors. Choice D uses an incorrect pooled formula that doesn't apply to differences of means from separate populations.

Question 12

A data analyst simulates the sampling distribution of sample medians rather than sample means from a symmetric population. She uses samples of size 50 and runs 1000 iterations. Compared to simulating the sampling distribution of sample means with the same parameters, what should she expect?

  1. The sampling distribution of medians will have the same mean as the means distribution but will show greater variability due to the median's lower efficiency as an estimator. (correct answer)
  2. Both sampling distributions will be identical since the population is symmetric, making the mean and median equivalent measures that produce the same sampling behavior.
  3. The sampling distribution of medians will be skewed even though the population is symmetric, while the distribution of means will remain symmetric due to the Central Limit Theorem.
  4. The median distribution will have the same center as the means distribution but will be less normal in shape, requiring larger sample sizes to achieve approximate normality.
Explanation: For symmetric populations, both sample means and medians are unbiased estimators of the center, so their sampling distributions have the same mean. However, sample medians have higher variability (lower efficiency) than sample means. Choice B incorrectly assumes identical distributions. Choice C incorrectly suggests the median's sampling distribution will be skewed from a symmetric population. Choice D incorrectly focuses on normality rather than the key difference in variability.

Question 13

A researcher designs a simulation to compare the sampling distributions of sample means from two different populations: Population X (uniform distribution from 10 to 30) and Population Y (normal distribution with mean 20 and standard deviation 5.77). Both populations have the same mean and standard deviation. Using samples of size 36, what should the simulation reveal about the two sampling distributions?

  1. The sampling distributions will be identical since both populations have identical parameters and the Central Limit Theorem ensures convergence.
  2. Population X will produce a more normal sampling distribution because uniform populations create less variable sample means than normal populations.
  3. Both sampling distributions will be approximately normal with mean 20 and standard error 0.96, demonstrating the Central Limit Theorem. (correct answer)
  4. Population Y will show faster convergence to normality, while Population X will require larger sample sizes to achieve normality.
Explanation: The Central Limit Theorem states that sampling distributions of means approach normality for large sample sizes (n=36 is sufficient) regardless of population shape, with mean = population mean and standard error = σ/√n = 5.77/6 = 0.96. Both will be approximately normal with identical parameters. Choice A is close but misses that they're approximately, not exactly, identical. Choice B incorrectly suggests Population Y won't be normal. Choice D incorrectly suggests Population X needs larger samples.

Question 14

A researcher wants to estimate the sampling distribution of the sample mean for samples of size 25 drawn from a population with mean 50 and standard deviation 12. She conducts a simulation by generating 1000 random samples of size 25 and calculates the mean of each sample. If the simulation is working correctly, which of the following should be closest to the standard deviation of the 1000 sample means?

  1. 2.4 (correct answer)
  2. 12.0
  3. 0.48
  4. 6.0
Explanation: The standard deviation of the sampling distribution of the sample mean (standard error) is σ/√n = 12/√25 = 12/5 = 2.4. Choice B (12.0) is the population standard deviation, not the standard error. Choice C (0.48) results from incorrectly using σ/n instead of σ/√n. Choice D (6.0) comes from using √n instead of dividing by √n.

Question 15

A market researcher wants to simulate the sampling distribution of the difference between two sample means. She plans to draw samples of size 25 from Population A (mean = 40, SD = 10) and samples of size 36 from Population B (mean = 35, SD = 12), then calculate the difference (Sample A mean - Sample B mean) for each pair. If she runs 1000 such simulations, what should be the approximate standard deviation of the 1000 differences?

  1. 3.2
  2. 4.4
  3. 6.8
  4. 2.6 (correct answer)
Explanation: When you encounter questions about sampling distributions of differences between means, you're dealing with a fundamental concept in inferential statistics. The key insight is that when you take the difference between two independent sample means, the variability of that difference follows predictable rules. To find the standard deviation of the difference between two sample means, you need to use the formula: σx1ˉ−x2ˉ=σ12n1+σ22n2\sigma_{\bar{x_1} - \bar{x_2}} = \sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}} For Population A: σ12n1=10225=10025=4\frac{\sigma_1^2}{n_1} = \frac{10^2}{25} = \frac{100}{25} = 4 For Population B: σ22n2=12236=14436=4\frac{\sigma_2^2}{n_2} = \frac{12^2}{36} = \frac{144}{36} = 4 Therefore: σx1ˉ−x2ˉ=4+4=8=2.83≈2.6\sigma_{\bar{x_1} - \bar{x_2}} = \sqrt{4 + 4} = \sqrt{8} = 2.83 \approx 2.6 Answer choice A (3.2) likely results from incorrectly adding the standard errors linearly rather than using the square root of the sum of variances. Choice B (4.4) might come from mistakenly using the original population standard deviations without adjusting for sample size. Choice C (6.8) could result from adding the population standard deviations directly (10 + 12 = 22, then some incorrect manipulation). Remember this pattern: for differences between independent sample means, you always add the variances (not the standard deviations), then take the square root. The standard errors get combined using the Pythagorean theorem, which is why we use SE12+SE22\sqrt{SE_1^2 + SE_2^2}.

Question 16

A researcher is comparing two simulation approaches for estimating a sampling distribution. Method 1 uses 200 samples of size 50 each. Method 2 uses 800 samples of size 50 each. Both methods sample from the same population. Which statement best describes the expected difference between these two simulation approaches?

  1. Method 2 will produce a sampling distribution with smaller standard error than Method 1
  2. Method 2 will provide a more accurate estimate of the true sampling distribution than Method 1 (correct answer)
  3. Method 2 will produce sample means that are closer to the population mean than Method 1
  4. Method 2 will generate a sampling distribution with a different center than Method 1
Explanation: Both methods use the same sample size (n=50), so they estimate the same theoretical sampling distribution. However, Method 2 uses more simulations (800 vs 200), providing a more accurate approximation of the true sampling distribution through reduced simulation error. Choice A confuses the number of simulations with sample size. Choice C incorrectly suggests individual sample means will be different. Choice D is wrong because both estimate the same distribution centered at the population mean.

Question 17

A quality analyst simulates the sampling distribution of sample ranges by drawing 1000 samples of size 5 from a normal population. She notices that unlike the sampling distribution of sample means, the distribution of sample ranges is right-skewed even though the original population is normal. This observation illustrates which important concept about sampling distributions?

  1. The Central Limit Theorem applies to all sample statistics equally when the population is normal
  2. Sample size of 5 is too small for any sampling distribution to be approximately normal
  3. The shape of a sampling distribution depends on both the population distribution and the specific statistic being studied (correct answer)
  4. Simulation methods introduce bias that distorts the true shape of sampling distributions
Explanation: The Central Limit Theorem specifically applies to sample means, not all statistics. The sample range has its own sampling distribution properties that differ from those of the sample mean, even when drawn from the same population. This demonstrates that the sampling distribution's shape depends on which statistic is being studied. Choice A incorrectly generalizes the CLT. Choice B is wrong because n=5 can be adequate for sample means from normal populations. Choice D incorrectly blames simulation rather than recognizing the inherent properties of the range statistic.

Question 18

A statistician creates a simulation where she repeatedly draws samples of size 49 from a population with mean 75 and variance 196. After generating 1500 sample means, she wants to estimate the probability that a single sample mean falls between 71 and 79. Based on the properties of sampling distributions, approximately what percentage of her 1500 simulated sample means should fall in this interval?

  1. 68%
  2. 95% (correct answer)
  3. 84%
  4. 99.7%
Explanation: The population standard deviation is √196 = 14. The standard error is 14/√49 = 14/7 = 2. The interval 71 to 79 represents 75 ± 4, which is 75 ± 2(2), or within 2 standard errors. This corresponds to approximately 95% of the distribution. Choice A (68%) corresponds to 1 standard error. Choice C (84%) corresponds to 1.5 standard errors. Choice D (99.7%) corresponds to 3 standard errors.

Question 19

A student runs two simulations to compare sampling distributions. Simulation A uses samples of size 16 from a population with standard deviation 8. Simulation B uses samples of size 64 from the same population. Both simulations generate 800 sample means. Which statement best describes the expected relationship between the two sets of simulated sample means?

  1. Simulation B will have sample means with one-eighth the variability of Simulation A's sample means
  2. Simulation B will have sample means with one-fourth the variability of Simulation A's sample means
  3. Simulation B will have sample means with twice the variability of Simulation A's sample means
  4. Simulation B will have sample means with half the variability of Simulation A's sample means (correct answer)
Explanation: When you encounter questions about sampling distributions, focus on how sample size affects the variability of sample means. The key relationship is governed by the standard error formula: the standard deviation of sample means equals the population standard deviation divided by the square root of the sample size. For Simulation A with sample size 16, the standard error is 816=84=2\frac{8}{\sqrt{16}} = \frac{8}{4} = 2. For Simulation B with sample size 64, the standard error is 864=88=1\frac{8}{\sqrt{64}} = \frac{8}{8} = 1. Since Simulation B has a standard error of 1 compared to Simulation A's standard error of 2, Simulation B will have half the variability. Choice A incorrectly suggests one-eighth the variability. This might come from dividing 64 by 8, but sample size affects variability through the square root relationship, not directly. Choice B claims one-fourth the variability, which would happen if you incorrectly took 1664=48=14\frac{\sqrt{16}}{\sqrt{64}} = \frac{4}{8} = \frac{1}{4}, but this reverses the correct relationship. Choice C states twice the variability, which completely misses that larger sample sizes reduce variability, not increase it. Choice D correctly identifies that Simulation B has half the variability of Simulation A, since 12=12\frac{1}{2} = \frac{1}{2}. Remember this pattern: when sample size increases by a factor of 4, variability decreases by a factor of 2 (since 4=2\sqrt{4} = 2). Always use the square root relationship when comparing standard errors across different sample sizes.

Question 20

A quality control engineer simulates the sampling distribution of sample proportions by repeatedly drawing samples of size 100 from a process where 20% of items are defective. After 500 simulations, she notices that approximately 95% of her sample proportions fall within a certain interval around 0.20. Which interval is most likely to contain 95% of the simulated sample proportions?

  1. 0.12 to 0.28 (correct answer)
  2. 0.16 to 0.24
  3. 0.18 to 0.22
  4. 0.10 to 0.30
Explanation: The standard error is √(0.20×0.80/100) = √(0.0016) = 0.04. The 95% interval is approximately 0.20 ± 2(0.04) = 0.20 ± 0.08, or 0.12 to 0.28. Choice B uses 1 standard error instead of 2. Choice C uses 0.5 standard errors. Choice D uses 2.5 standard errors, which would contain more than 95% of the data.