What is the standard error of the sample mean for n observations from a population with known standard deviation sigma?

What is the standard error of the sample mean for n observations from a population with known standard deviation sigma?

• A. sigma / sqrt(n)
• B. sqrt(n) / sigma
• C. sigma * sqrt(n)
• D. sigma^2 / n

Answer: (A)

The standard error of the sample mean measures how spread out the sample mean is across repeated samples. For n independent observations from a population with standard deviation sigma, the distribution of the sample mean has variance equal to sigma^2/n. This comes from the fact that variances add for independent variables and the mean involves dividing by n, so the variance scales by 1/n^2, giving Var(X-bar) = sigma^2/n. Therefore the standard error, which is the standard deviation of the sample mean, is sqrt(sigma^2/n) = sigma / sqrt(n). This shows why the precision improves as you take more observations: the spread of the sample mean decreases with the square root of n. The other expressions don’t describe that spread correctly: they would imply wrong scaling (increasing with n) or refer to the variance instead of the standard deviation. With known sigma, you can state the standard error exactly as sigma divided by the square root of n.

The standard error of the sample mean measures how spread out the sample mean is across repeated samples. For n independent observations from a population with standard deviation sigma, the distribution of the sample mean has variance equal to sigma^2/n. This comes from the fact that variances add for independent variables and the mean involves dividing by n, so the variance scales by 1/n^2, giving Var(X-bar) = sigma^2/n. Therefore the standard error, which is the standard deviation of the sample mean, is sqrt(sigma^2/n) = sigma / sqrt(n). This shows why the precision improves as you take more observations: the spread of the sample mean decreases with the square root of n. The other expressions don’t describe that spread correctly: they would imply wrong scaling (increasing with n) or refer to the variance instead of the standard deviation. With known sigma, you can state the standard error exactly as sigma divided by the square root of n.