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 99​% confident that the sample mean is correct to within ±​$60 of the population mean annual family medical expenses. A previous study indicates that the standard deviation is approximately $487.

a. How large a sample is​ necessary?

b. If management wants to be correct to within ±​$25 how many employees need to be​ selected?

Answer 1

Solution :

Given that,

standard deviation = = 487

a )  margin of error = E = 60

At 99% confidence level the z is ,

  = 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z_/2 = Z_0.005 = 2.576

Sample size = n = ((Z_/2 * ) / E)²

= ((2.576 * 487) / 60)²

= 437

Sample size = 437

b )  margin of error = E = 25

At 99% confidence level the z is ,

  = 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z_/2 = Z_0.005 = 2.576

Sample size = n = ((Z_/2 * ) / E)²

= ((2.576 * 487) / 25)²

= 2518

Sample size = 2518