Question

find the sample size to estimate the mean salary for CIS graduated at Arkansas with 90% confidence. required precision is E=5000 dollars

standard deviation from last year= 8000 dollars

Answer 1

Solution :

Given that,

standard deviation = = 8000 dollars

margin of error = E = 5000 dollars

At 90% confidence level the z is ,

   = 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

Z_/2 = Z_0.05 = 1.645

Sample size = n = ((Z_/2 * ) / E)²

= ((1.645 *8000 ) / 5000)²

= 7

Sample size =n = 7