Question

If we want a margin of error of 1, how large must the sample be? (assuming we are still taking a sample from a normal population with standard deviation 8 and using the 98% confidence level)

Answer 1

Solution :

Given that,

standard deviation =s =   =8

Margin of error = E = 1

At 98% confidence level the z is ,

= 1 - 98% = 1 - 0.98 = 0.02

/ 2 = 0.02/ 2 = 0.01

Z/2 = Z0.01 = 2.326 ( Using z table    )

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

n = ( 2.326* 8/ 1 )2

n =346.25

Sample size = n =346