Question

An estimate is needed of the mean average of farms in a certain city. A 95% confidence interval should have a margin of error of 29 acres. A study ten years ago inn this city had a sample standard deviation of 110 acres for farm size. About how large a sample of farms is needed?

Answer 1

Solution :

Given that,

sample standard deviation =s = 110

Margin of error = E = 29

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 = 1.96 ( Using z table ( see the 0.025 value in standard normal (z) table corresponding z value is 1.96 )

sample size = n = [Z_/2* S / E] ²

n = ( 1.96* 110 /29 )²

n =55.27(Rounded)

Sample size = n =55