Finite-population correction for town spending

Question

A town has N = 1000 households. You sample n = 100 of them without replacement and find the sample standard deviation of monthly spending is s = \50. What is the standard error of the average spending, using the finite-population correction?

Accepted Answer

The usual formula s/\sqrt{n} assumes you draw from an effectively infinite pool. But here you have sampled a real chunk of a finite town, and once you have measured a household you cannot draw it again. Sampling a large fraction of a finite population actually makes your average more precise than the plain formula suggests, because you have covered more of the whole. Start with the ordinary standard error: \frac{s}{\sqrt{n}} = \frac{50}{\sqrt{100}} = \frac{50}{10} = 5. Then multiply by the finite-population correction (FPC), which shaves off a bit because you sampled a real slice of the popul…

Finite-population correction for town spending

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