Question

You want to estimate the proportion of students at your college or university who are employed for 10 or more hours per week while classes are in session. You plan to present your results by a 95% confidence interval. Using the guessed value p* = 0.32, find the sample size required if the interval is to have an approximate margin of error of m = 0.04.

Answer 1

Solution:

Given that,

= 0.32

1 - = 1 - 0.32 = 0.68

margin of error = E = 0.04

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 = 1.960

Sample size = n = ((Z _{/ 2}) / E)² * * (1 - )

= (1.960 / 0.04)² * 0.32 * 0.68

= 522.4576 = 522

n = sample size = 522