Question

A survey is planned to determine the mean annual family medical expenses of employees of a large company. The management of the company wishes to be

90% confident that the sample mean is correct to within plus or minus $60 of the population mean annual family medical expenses. A previous study indicates that the standard deviation is approximately ​$438

a. How large a sample is​ necessary?

b. If management wants to be correct to within plus or minus $20 how many employees need to be​ selected?

Answer 1

Solution :

Given that,

Population standard deviation = = 438

a) Margin of error = E = 60

At 90% confidence level

= 1-0.90% =1-0.90 =0.10

/2 =0.10/ 2= 0.05

Z/2 = Z0.05 = 1.645

Z/2 = 1.645  

sample size = n = [Z/2* / E] 2

n = [ 1.645 * 438 / 60 ]2

n = 144

Sample size = n = 144

b)

Margin of error = E = 20

At 90% confidence level

= 1-0.90% =1-0.90 =0.10

/2 =0.10/ 2= 0.05

Z/2 = Z0.05 = 1.645

Z/2 = 1.645  

sample size = n = [Z/2* / E] 2

n = [ 1.645 * 438 / 20 ]2

n = 1298

Sample size = n = 1298

Answer = 1298 employees.