Question

1. A student wants to estimate the average mark on a statistics test to within 5 marks (out of 100), with 95% confidence. A preliminary sample yielded a standard deviation of 15.54. How big a sample size is necessary? Round to the nearest whole number.

Answer 1

Standard Deviation ,   σ =    15.54
sampling error ,    E =   5
Confidence Level ,   CL=   95%
      
alpha =   1-CL =   5%
Z value = Zα/2 =    1.960   [excel formula =normsinv(α/2)]
      
Sample Size,n = (Z*σ/E)² =   (1.96*15.54/5)² =    37.107
So,Sample Size needed=       38