Quiz: Sampling Distributions and the Central Limit Theorem
Test your understanding of sampling distributions, the CLT, and standard error. Try to answer each question before revealing the answer.
1. A sampling distribution is the distribution of:
(a) All values in a single sample (b) All values in the population (c) A statistic computed from all possible samples of a given size (d) The sample sizes used in a study
Answer
**(c) A statistic computed from all possible samples of a given size.** A sampling distribution shows how a statistic (like $\bar{x}$ or $\hat{p}$) varies across all possible random samples of the same size from the same population. It's not about individual values — it's about the behavior of a computed statistic across many hypothetical samples.2. Which of the following best describes the Central Limit Theorem?
(a) Large samples are always normally distributed (b) The sampling distribution of the mean is approximately normal for large samples, regardless of the population shape (c) The population must be normal for statistical inference to work (d) As sample size increases, the sample becomes more representative of the population
Answer
**(b) The sampling distribution of the mean is approximately normal for large samples, regardless of the population shape.** The CLT is about the *sampling distribution of the mean*, not the sample data itself. It does not require the population to be normal, and it's specifically about the distribution of $\bar{x}$ across many samples, not about a single sample being representative.3. A population has $\mu = 100$ and $\sigma = 20$. For samples of size $n = 25$, the standard error of the mean is:
(a) 0.8 (b) 4 (c) 5 (d) 20
Answer
**(b) 4.** $\text{SE} = \sigma / \sqrt{n} = 20 / \sqrt{25} = 20 / 5 = 4$.4. According to the CLT, the mean of the sampling distribution of $\bar{x}$ equals:
(a) The sample mean $\bar{x}$ (b) The population mean $\mu$ (c) $\mu / \sqrt{n}$ (d) $\sigma / \sqrt{n}$
Answer
**(b) The population mean $\mu$.** The CLT guarantees that the center of the sampling distribution equals the population mean: $\mu_{\bar{x}} = \mu$. This means $\bar{x}$ is an unbiased estimator of $\mu$ — on average, it hits the target.5. If you quadruple the sample size, the standard error:
(a) Is divided by 4 (b) Is divided by 2 (c) Stays the same (d) Is multiplied by 2
Answer
**(b) Is divided by 2.** $\text{SE} = \sigma / \sqrt{n}$. Quadrupling $n$ means dividing by $\sqrt{4n}$ instead of $\sqrt{n}$. Since $\sqrt{4n} = 2\sqrt{n}$, the SE is cut in half. This is the "diminishing returns" property: to halve precision, you must quadruple effort.6. A population is strongly right-skewed. For which sample size is the sampling distribution of $\bar{x}$ most likely to be approximately normal?
(a) $n = 3$ (b) $n = 10$ (c) $n = 50$ (d) The sampling distribution can never be normal if the population is skewed
Answer
**(c) $n = 50$.** For strongly skewed populations, larger sample sizes are needed for the CLT to produce an approximately normal sampling distribution. The rule of thumb is $n \geq 30$ for moderate skewness and potentially more for extreme skewness. Of the given options, $n = 50$ is the most likely to yield approximate normality. Option (d) is wrong — the whole point of the CLT is that normality emerges regardless of population shape.7. Which of the following statements about the sampling distribution is FALSE?
(a) Its spread decreases as $n$ increases (b) Its center equals the population mean (c) Its spread equals the population standard deviation (d) Its shape approaches normal as $n$ increases
Answer
**(c) Its spread equals the population standard deviation.** This is false. The spread of the sampling distribution is the *standard error*, $\sigma / \sqrt{n}$, which is *smaller* than the population standard deviation $\sigma$. The SE is always less than $\sigma$ (for $n > 1$) because averaging reduces variability.8. The standard error measures:
(a) How much individual values vary in the population (b) How much sample means vary from sample to sample (c) How large the sample should be (d) How normal the population is
Answer
**(b) How much sample means vary from sample to sample.** The standard error is the standard deviation of the sampling distribution. It quantifies sampling variability — how much the sample mean (or other statistic) typically differs from the true population value across different random samples.9. The population of household incomes is right-skewed with $\mu = \$55{,}000$ and $\sigma = \$25{,}000$. For a random sample of 100 households, the sampling distribution of $\bar{x}$ is approximately:
(a) Right-skewed with mean $\$55{,}000$ and SD $\$25{,}000$ (b) Right-skewed with mean $\$55{,}000$ and SD $\$2{,}500$ (c) Normal with mean $\$55{,}000$ and SD $\$25{,}000$ (d) Normal with mean $\$55{,}000$ and SD $\$2{,}500$
Answer
**(d) Normal with mean $\$55{,}000$ and SD $\$2{,}500$.** By the CLT, with $n = 100$, the sampling distribution is approximately normal (even though the population is skewed). The mean is $\mu = \$55{,}000$, and the SE is $\sigma / \sqrt{n} = 25{,}000 / 10 = \$2{,}500$.10. A researcher takes a random sample of 36 observations from a population with $\mu = 72$ and $\sigma = 18$. What is $P(\bar{x} > 75)$?
(a) 0.0228 (b) 0.1587 (c) 0.4332 (d) 0.8413
Answer
**(b) 0.1587.** $\text{SE} = 18 / \sqrt{36} = 3$. $z = (75 - 72)/3 = 1.0$. $P(Z > 1) = 1 - 0.8413 = 0.1587$.11. For the standard error of a proportion, $\text{SE}_{\hat{p}} = \sqrt{p(1-p)/n}$, the SE is largest when:
(a) $p = 0.01$ (b) $p = 0.25$ (c) $p = 0.50$ (d) $p = 0.99$
Answer
**(c) $p = 0.50$.** The product $p(1-p)$ is maximized when $p = 0.5$ (giving $0.5 \times 0.5 = 0.25$). As $p$ moves toward 0 or 1, the product decreases. This means proportions near 50/50 are hardest to estimate precisely — there's maximum uncertainty when the population is evenly split.12. The CLT conditions for the sampling distribution of a proportion to be approximately normal are:
(a) $n \geq 30$ (b) $np \geq 10$ and $n(1-p) \geq 10$ (c) The population must be normal (d) $\sigma$ must be known
Answer
**(b) $np \geq 10$ and $n(1-p) \geq 10$.** These conditions ensure enough expected successes and failures for the normal approximation to be reasonable. For example, if $p = 0.02$ and $n = 100$, then $np = 2 < 10$, and the normal approximation would be poor (too few expected successes).13. Which of the following does the CLT NOT require?
(a) A random sample (b) A sufficiently large sample size (c) A normally distributed population (d) Independent observations
Answer
**(c) A normally distributed population.** This is the most remarkable feature of the CLT — it works regardless of the population shape. However, it does require random sampling, adequate sample size, and independence (or approximate independence via the 10% condition).14. A sample of $n = 49$ observations has $\bar{x} = 84$ and $s = 14$. The estimated standard error is:
(a) 0.29 (b) 2 (c) 6 (d) 14
Answer
**(b) 2.** $\widehat{\text{SE}} = s / \sqrt{n} = 14 / \sqrt{49} = 14 / 7 = 2$.15. How does the law of large numbers differ from the Central Limit Theorem?
(a) They are the same thing with different names (b) The LLN says $\bar{x}$ converges to $\mu$; the CLT describes the distribution of $\bar{x}$ (c) The LLN applies to proportions; the CLT applies to means (d) The LLN requires normality; the CLT does not
Answer
**(b) The LLN says $\bar{x}$ converges to $\mu$; the CLT describes the *distribution* of $\bar{x}$.** The law of large numbers tells us that $\bar{x}$ gets closer to $\mu$ as $n$ increases (the center). The CLT tells us the *shape* and *spread* of the distribution of $\bar{x}$ values across many samples — specifically, that it's approximately normal with standard deviation $\sigma / \sqrt{n}$.16. A survey of 400 randomly selected adults finds that 220 (55%) support a new policy. What is the standard error of $\hat{p}$?
(a) 0.0124 (b) 0.0249 (c) 0.0498 (d) 0.5500
Answer
**(b) 0.0249.** $\text{SE}_{\hat{p}} = \sqrt{\hat{p}(1-\hat{p})/n} = \sqrt{0.55 \times 0.45 / 400} = \sqrt{0.2475/400} = \sqrt{0.000619} \approx 0.0249$.17. The weight of apples from an orchard has $\mu = 150$ grams and $\sigma = 30$ grams. A grocer buys a crate of 36 randomly selected apples. What is the probability that the mean weight is less than 145 grams?
(a) 0.0228 (b) 0.1587 (c) 0.4332 (d) 0.8413
Answer
**(b) 0.1587.** $\text{SE} = 30 / \sqrt{36} = 5$. $z = (145 - 150)/5 = -1.0$. $P(Z < -1) = 0.1587$.18. To reduce the standard error from 4 to 1, you would need to multiply the sample size by:
(a) 2 (b) 4 (c) 8 (d) 16
Answer
**(d) 16.** SE is proportional to $1/\sqrt{n}$. To reduce SE by a factor of 4 (from 4 to 1), you need $\sqrt{n}$ to increase by a factor of 4, which means $n$ must increase by a factor of $4^2 = 16$.19. A sampling distribution is created by taking 5,000 samples of size 25 from a skewed population and computing the mean of each. Which graph would you expect?
(a) A histogram of 5,000 values that's strongly skewed (b) A histogram of 5,000 values that's approximately normal (c) A histogram of 25 values that's approximately normal (d) A histogram of 25 values that's strongly skewed
Answer
**(b) A histogram of 5,000 values that's approximately normal.** We're plotting 5,000 sample means (one per sample). By the CLT, with $n = 25$, the sampling distribution of $\bar{x}$ will be approximately normal (or at least substantially less skewed than the population), even though the population is skewed. The histogram shows the sampling distribution, which has 5,000 data points.20. Which of the following scenarios would make you MOST concerned about applying the CLT?
(a) A skewed population with $n = 100$ (b) A normal population with $n = 5$ (c) A population with extreme outliers and $n = 12$ (d) A symmetric population with $n = 20$