A simple random sample of size n is drawn from a normally distributed population, and the mean of the sample is x Overbar, while the standard deviation is s. What is the 99% confidence interval for the population mean? Use the table below to help you answer the question.

CI = x ± 2.58 (s / √n)

Step-by-step explanation:

The confidence interval is the mean plus or minus the margin of error.  The margin of error is the critical value times the standard error.  For a normal distribution, the critical value is the z-score at the given confidence level.  The standard error is the standard deviation divided by the square root of the sample size.

CI = x ± z (s / √n)

At 99% confidence, z = 2.58.

CI = x ± 2.58 (s / √n)

is there a question here?

step-by-step explanation: