Sample size at 90% confidence

Question

You want to estimate a population mean to within a margin of error of E = 2 with 90\% confidence. The population standard deviation is \sigma = 20. Use z = 1.645. How many observations do you need?

Accepted Answer

At 90\% confidence the multiplier is smaller, z = 1.645, because you accept a slightly higher chance of missing the true mean in exchange for a smaller sample. You want the half-width at most 2: E = z \cdot \frac{\sigma}{\sqrt{n}}. Rearrange for n: n = \left(\frac{z\,\sigma}{E}ight)^2 = \left(\frac{1.645 	imes 20}{2}ight)^2 = \left(\frac{32.9}{2}ight)^2 = (16.45)^2 \approx 270.6. Round up: n = 271. The same margin at 95\% confidence (z = 1.96) would need about 384, so relaxing to 90\% confidence saves roughly a third of the sample. Takeaway. Lower confidence means a smaller z and a sma…

Sample size at 90% confidence

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