In: Statistics and Probability
A simple random sample of 81 is selected from a population with a standard deviation of 17. The degree of confidence is 90%. What is the margin of error for the mean?
Z score for 90% confidence interval = Z0.05 = 1.645
Margin of error = Z0.05 * sd / sqrt(n) = 1.645 * 17 / sqrt(81) = 3.107