Question

What sample size is needed to estimate a population mean with a margin of error of 10 of the true mean value using a confidence level of 95%, if the true population variance (not the standard deviation!) is known to be 1600? (10 points)

Answer 1

Solution

variance 2 = 1600

standard deviation =   =40

Margin of error = E = 10

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*40 /10 )2

n =61.4656

Sample size = n =61