When the finite-population correction barely matters

Question

A national registry has N = 100{,}000 members. You sample n = 900 of them without replacement and find the sample standard deviation of an outcome is s = 15. What is the standard error of the average, using the finite-population correction?

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This is the case where the finite-population correction is barely worth the trouble, and knowing that is the point. When your sample is a tiny slice of a huge population, sampling without replacement is almost the same as sampling with replacement. Ordinary standard error: \frac{s}{\sqrt{n}} = \frac{15}{\sqrt{900}} = \frac{15}{30} = 0.5. The finite-population correction (FPC): 	ext{FPC} = \sqrt{\frac{N-n}{N-1}} = \sqrt{\frac{100{,}000-900}{100{,}000-1}} = \sqrt{\frac{99{,}100}{99{,}999}} \approx 0.9955. Corrected standard error: 	ext{SE} = 0.5 	imes 0.9955 \approx 0.498. The correction cha…

When the finite-population correction barely matters

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