Finite-population correction sampling a fifth of accounts

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A firm has N = 2000 customer accounts. A reviewer samples n = 400 of them without replacement and finds the sample standard deviation of account balance is s = \100. What is the standard error of the average balance, using the finite-population correction?

Accepted Answer

Here you have sampled a full fifth of the accounts, so the finite-population correction matters more than usual. Ignoring it would exaggerate your uncertainty. Ordinary standard error: \frac{s}{\sqrt{n}} = \frac{100}{\sqrt{400}} = \frac{100}{20} = 5. Apply the finite-population correction (FPC): 	ext{FPC} = \sqrt{\frac{N-n}{N-1}} = \sqrt{\frac{2000-400}{2000-1}} = \sqrt{\frac{1600}{1999}} \approx 0.895. Corrected standard error: 	ext{SE} = 5 	imes 0.895 \approx \4.47. Sampling 20\% of the population shaves roughly 10\% off the standard error. Takeaway. The bigger your sampling fraction n/N…

Finite-population correction sampling a fifth of accounts

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