A researcher is interested in determining the average number of years employees of a company stay with the company. If past information shows a standard deviation of 7 months, what size sample should be selected so that at 95% confidence the margin of error will be 2 months or less?

48

Step-by-step explanation:

The significance level is 95%

Z-alpha/2 = 1.96 at 0.05 from the standard normal distribution table.

The standard deviation has been given to be 7

Margin of error is 2

We solve further using the formula:

n = (z-alpha/2 * sd/ME)²

n = (1.96 * 7/2)²

n = (1.96x3.5)²

= 6.86²

= 47.0596

This is approximately

n = 48

So the sample size they should be selected should be 48 for at 95% confidence level.

use distributive property on all like terms and change your signs if you have a double negative or if you have positive/negative