Sample size to estimate mean weight

Question

You want to estimate the mean weight of a produced item to within a margin of error of $E = 0.5$ kg, with 95% confidence. The process standard deviation is $\sigma = 3$ kg.

What is the minimum sample size?

Accepted Answer

Setup. From E = z^{}\sigma/\sqrt{n}, solve n = (z^{}\sigma/E)^{2} with z^{} = 1.96. Compute. n = \left(\frac{1.96 	imes 3}{0.5}ight)^{2} = \left(\frac{5.88}{0.5}ight)^{2} = (11.76)^{2} = 138.30. Round up: n = 139. Interpret. A tight margin of half a kilogram against a 3 kg spread forces a fairly large sample. Takeaway. Tighter margins are expensive: because n depends on 1/E^{2}, demanding twice the precision costs four times the data.

Sample size to estimate mean weight

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