Question

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

Answer 1

The margin of error for a 95% confidence interval for the population mean is of 1.05 years.

Explanation:

We have that to find our
`\alpha` level, that is the subtraction of 1 by the confidence interval divided by 2. So:

`\alpha = (1-0.95)/(2) = 0.025`

Now, we have to find z in the Ztable as such z has a pvalue of
`1-\alpha`.

So it is z with a pvalue of
`1-0.025 = 0.975`, so
`z = 1.96`

Now, find the margin of error M as such

`M = z*(\sigma)/(√(n))`

In which
`\sigma` is the standard deviation of the population(square root of the variance) and n is the size of the sample.

What is the margin of error for a 95% confidence interval for the population mean

36 students, so
`n = 36`

Variance of 10.24, so
`\sigma = √(10.24) = 3.2`

`M = 1.96*(3.2)/(√(36)) = 1.05`

The margin of error for a 95% confidence interval for the population mean is of 1.05 years.