Finite-population correction for a small class

Question

A lecture has N = 250 enrolled students. A grader samples n = 25 exams without replacement and finds the sample standard deviation of scores is s = 8 points. What is the standard error of the average score, using the finite-population correction?

Accepted Answer

You sampled a tenth of a finite class without replacement, so the finite-population correction applies, though it is modest at this sampling fraction. Ordinary standard error: \frac{s}{\sqrt{n}} = \frac{8}{\sqrt{25}} = \frac{8}{5} = 1.6. Apply the finite-population correction (FPC): 	ext{FPC} = \sqrt{\frac{N-n}{N-1}} = \sqrt{\frac{250-25}{250-1}} = \sqrt{\frac{225}{249}} \approx 0.951. Corrected standard error: 	ext{SE} = 1.6 	imes 0.951 \approx 1.52 	ext{ points}. Takeaway. At a 10\% sampling fraction the FPC \sqrt{(N-n)/(N-1)} trims only about 5\% off the SE, here from 1.6 to roughly 1.…

Finite-population correction for a small class

Your answer

More Statistics questions