Question

If the standard deviation of a national accounting examination is 900, how large a sample is needed to estimate the true mean score within 5 points with 99% confidence?

Answer 1

Variance = 900
Standard deviation = 30
z(alpha) = 2.575
error = 5
n = ( z(alpha) * SD /error)² = ( 2.575 * 30 / 5 )² = 238.7025
sample of 239 is needed.