Question

1. determine whether the statements about the central limit theorem are true or false. state your reasoning.

a. as the sample size increases, the mean of the distribution of sample means increases.
b. as the sample size increases, the standard deviation of the distribution of sample means decreases.
c. if the sample size is at least 30 or the population is normally distributed, then the sampling distribution is approximately normally distributed by the central limit theorem.

Explanation:

• For part a:

The mean of the distribution of sample means (also known as the sampling distribution of the mean) is equal to the population mean, i.e., \(\mu_{\bar{x}}=\mu\). It does not depend on the sample size \(n\). So, as the sample size increases, the mean of the distribution of sample means remains the same.

• For part b:

The standard deviation of the distribution of sample means (also called the standard error) is given by \(\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}\), where \(\sigma\) is the population standard deviation and \(n\) is the sample size. As \(n\) increases (since \(n\) is in the denominator of the formula \(\frac{\sigma}{\sqrt{n}}\)), \(\sigma_{\bar{x}}\) decreases.

• For part c:

The Central Limit Theorem (CLT) states that if we have a population with mean \(\mu\) and standard deviation \(\sigma\), and we take sufficiently large random samples (\(n\geq30\)) from the population with replacement, the sampling distribution of the sample mean \(\bar{X}\) will be approximately normally distributed. Also, if the population is normally distributed, then for any sample size \(n\) (not just \(n\geq30\)), the sampling distribution of the sample mean is normally distributed.

Answer:

• a. False. The mean of the sampling distribution of the sample mean is equal to the population mean and is not affected by sample size.
• b. True. Since \(\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}\), as \(n\) increases, \(\sigma_{\bar{x}}\) decreases.
• c. True. This is in line with the statements of the Central Limit Theorem.