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.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 can be calculated using the formula:
Margin of error = z* (standard deviation / square root of sample size)
Here, the population variance is given as 10.24, which can be used as an estimate of standard deviation. The sample size is given as 36 and the confidence level is 95%, which corresponds to a z-value of 1.96. Therefore:
Margin of error = 1.96 * (sqrt(10.24) / sqrt(36)) = 1.96 * (3.20 / 6) = 1.04
So the margin of error is 1.04 years. The only choice that matches this calculation is C.

The Ages of MBA Students at a University Are Normally