Question

In order to estimate the annual cost of full time-employees absences full time employees take per year. Based on previous studies, it's assumed s= 2.1 days.

a) How big a sample must be selected if the company wants to be 90% confident that their estimate is within 1 day of the true mean?

b) Repeat (a) using a 95% confidence interval.

c) Which level of confidence requires a larger sample size? Explain.

Answer 1

Solution :

Given that,

standard deviation = S = 6250

margin of error = E = 500

At 90% confidence level the z is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

Z/2 = Z0.05 = 1.645

Sample size = n = ((Z/2 * S) / E)2

= ((1.645 * 2.1) / 1)2

= 12

Sample size = 12

b) Given that,

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z/2 = Z0.025 = 1.960

Sample size = n = ((Z/2 * S) / E)2

= ((1.960 * 2.1) / 1)2

= 17

Sample size = 17

C ) Confidence level increased than sample size is larger