Question

Suppose the population standard deviation is believed to be $812 and the salaries are normally distributed. Find the sample size needed to construct a 98% confidence interval to
estimate the average salary of accountants, within +/- $200.

Answer 1

Solution :

Given that,

standard deviation =s =   =812

Margin of error = E = 200

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*812 / 200 )2

n =89.18

Sample size = n =90 rounded