Track 6 · Sampling and the Central Limit Theorem · Lesson 7 of 8
Finite populations
Apply the finite population correction to the standard error when sampling without replacement, and decide when it matters.
An auditor checks a random sample of 90 of the 120 expense claims a department filed last quarter. The usual standard error formula treats those 90 as a drop in an endless ocean — but they are three quarters of every claim there is. Their average cannot be far from the average of all 120, and the finite population correction says by how much. (It corrects random sampling error only: if 90 of 120 staff had chosen to answer a survey, the correction would say nothing about who stayed silent.)
The formula assumes independent observations — sampling with replacement, or from a population so large that removing a few units changes nothing. Real surveys sample without replacement: once an employee is chosen, she cannot be chosen again. In a small population this matters, because each selected unit removes some of the remaining uncertainty. In the extreme case, a census () has no sampling error at all.
Included with a Formula Quest membership
It has 7 graded exercises with feedback, hints and worked solutions. After it you will be able to:
- Compute the finite population correction √((N − n)/(N − 1)) and apply it to a standard error
- Use the sampling fraction to decide when the correction matters
- Adjust a planned sample size for a small population
- Explain why a large sampling fraction reduces random error but not nonresponse bias