Question

t is known that the population variance equals 529. With a 0.95 probability, the sample size that needs to be taken if the desired margin of error is 5 or less is ?

Answer 1

Solution :

Given that,

variance = 2 = 529

Population standard deviation = = 2 = 529 = 23

Margin of error = E = 5

At 95% confidence level the z is,

= 1 - 95%

= 1 - 0.95 = 0.05

/2 = 0.025

Z/2 = Z0.025 = 1.96  

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

n = [1.96 * 23 / 5 ]2

n = 81.28

Sample size = n = 82