Finite-population correction for a warehouse audit

Question

A warehouse holds N = 800 pallets. An auditor samples n = 100 of them without replacement and finds the sample standard deviation of pallet weight is s = 30 kg. What is the standard error of the average pallet weight, using the finite-population correction?

Accepted Answer

Because the auditor sampled one-eighth of a finite stock without replacement, the plain s/\sqrt{n} slightly overstates the uncertainty. The finite-population correction fixes this. First the ordinary standard error: \frac{s}{\sqrt{n}} = \frac{30}{\sqrt{100}} = \frac{30}{10} = 3. Now apply the finite-population correction (FPC): 	ext{FPC} = \sqrt{\frac{N-n}{N-1}} = \sqrt{\frac{800-100}{800-1}} = \sqrt{\frac{700}{799}} \approx 0.936. Corrected standard error: 	ext{SE} = 3 	imes 0.936 \approx 2.81 	ext{ kg}. Takeaway. Multiply s/\sqrt{n} by \sqrt{(N-n)/(N-1)} when sampling without replacemen…

Finite-population correction for a warehouse audit

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