Question

There are a large number of individuals working at a certain factory. Julie wishes to obtain a​ 95% confidence interval estimate for the true mean age of all of the workers by obtaining a random sample of 114 workers. If she knows that the population standard deviation is 17.3 ​years, what will the margin of error​ be? (Round to 2 decimal​ places.)

Answer 1

Solution :

Given that,

The population standard deviation = = 17.3

sample = n = 114

At 95% confidence level the z is ,

  = 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z_/2 = Z_0.025 = 1.960

Margin of error = E = Z_/2* (/ n)

= 1.960 * (17.3 / 114 )

= 3.17

Margin of error = E = 3.17