Question

The weekly earnings of students in one age group are normally distributed with a standard deviation of 92 dollars. A researcher wishes to estimate the mean weekly earnings of students in this age group. Find the sample size needed to assure with 95 percent confidence that the sample mean will not differ from the population mean by more than 4 dollars.

Answer 1

Solution

standard deviation =   =92

Margin of error = E = 4

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z/2 = Z0.025 = 1.96

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

n = ( 1.96*92 / 4 )2

n =2032.2064

Sample size = n =2033