Exhibit 8.1.The ages of MBA students at a university are normally distributed with a known population variance of 10.24.Suppose you are asked to construct a 95% confidence interval for the population mean age if the mean of a sample of 36 students is 26.5 years. Refer to Exhibit 8.1.What is the margin of error for a 95% confidence interval for the population mean?
A)1.645(3.20/6)
B)1.645(10.24/6)
C)1.96(3.20/6)
D)1.96(10.24/6)

Question

Quizplus · Accepted Answer

The margin of error for a 95% confidence interval is given by:

Margin of Error = Z* (Standard Error)

Where, Z* is the z-score for a 95% confidence interval (which equals 1.96), and the standard error is given by:

Standard Error = σ/√n

Where, σ is the population standard deviation (which equals sqrt(10.24)), and n is the sample size (which equals 36).

Substituting the given values, we get:

Margin of Error = 1.96 * [sqrt(10.24)/sqrt(36)] = 1.96 * (3.20/6)

Thus, the margin of error for a 95% confidence interval is 1.96 * (3.20/6) = 1.04, which is closest to option C: 1.96(3.20/6).

Exhibit 81