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.

no i do not : ( what is it?

if gloria has "carry over" minutes, (that is, any minutes paid for but not used in a month can be applied to her payment for minutes in the next month,) her total usage for the three months is: total minutes = 320 + 243 + 489 = 1052 minutes number of 200 minute blocks = 1052/200 = 5 r52 (that is 5 with a remainder of 52) rounding up the number of 200 minute blocks gives her usage for the three months as 6 blocks. the cost for six blocks is 6 x $25 = $150 she also has 200 - 52 or 148 carry over minutes if gloria pays each month and unused minutes are lost, applying the rounding up procedure to each month gives: first month usage of 320 minutes rounds up to 2 blocks of 200 minutes for a payment of 2 x $25 = $50 second month usage of 243 minutes rounds up to 2 blocks of 200 minutes for a payment of 2 x $25 = $50 third month usage of 489 minutes rounds up to 3 blocks of 200 minutes for a payment of 3 x $25 = $75 total payments for the three months are $50 + $50 + $75 = $175 get the plan with carry over minutes!