Question

The manager of a university credit union wants to know the mean balance in savings accounts. He wants to estimate this mean with 98% confidence to within $20. When he performed this analysis last year he found a standard deviation of $43. What size sample does he need to take?

Answer 1

Solution :

Given that,

standard deviation =   =$43

Margin of error = E = $20

At 98% confidence level the z is ,

= 1 - 98% = 1 - 0.98 = 0.02

/2 = 0.02/ 2 = 0.01

Z/2 = Z_0.01 = 2.326 ( Using z table    )

sample size = n = [Z_/2* / E] ²

n = ( 2.326* 43 / 20 )²

n =25.00

Sample size = n =25