Suppose you have a random sample of size $n$ from a population with standard deviation $\sigma$. To calculate the standard error of the sample mean, which formula should you use, and under what condition can you assume the sampling distribution of the mean is approximately normal?

• A. Use $\frac{\sigma }{\sqrt{n}}$; normality can be assumed only if the population is normal, regardless of sample size.
• B. Use $\frac{\sigma }{\sqrt{n}}$; normality can be assumed if the sample size is large due to the Central Limit Theorem.
• C. Use $\frac{\sigma }{n}$; normality can be assumed if the sample size is large due to the Central Limit Theorem.
• D. Use $\frac{\sigma }{n}$; normality can be assumed only if the population is normal, regardless of sample size.

Answer: (B)

Verified step by step guidance

1. Identify the formula for the standard error of the sample mean when the population standard deviation \( \sigma \) is known. The standard error (SE) is given by the formula: \[ SE = \frac{\sigma}{\sqrt{n}} \] where \( n \) is the sample size.
2. Understand the concept of the sampling distribution of the sample mean. According to the Central Limit Theorem (CLT), the distribution of the sample mean will be approximately normal if the sample size \( n \) is sufficiently large, regardless of the shape of the population distribution.
3. Recognize the condition under which the sampling distribution of the mean is approximately normal: - If the population itself is normal, then the sampling distribution of the mean is normal for any sample size. - If the population is not normal, the sampling distribution of the mean is approximately normal only when the sample size \( n \) is large (commonly \( n \geq 30 \) is used as a rule of thumb).
4. Combine these points to conclude that the correct formula for the standard error is \( \frac{\sigma}{\sqrt{n}} \), and the sampling distribution of the mean can be assumed approximately normal if the sample size is large due to the Central Limit Theorem.
5. Summarize: Use the formula \( SE = \frac{\sigma}{\sqrt{n}} \) for the standard error, and rely on the Central Limit Theorem to justify normality of the sampling distribution when the sample size is large.