Sample size for a 0.5-unit margin of error

Question

You want to estimate a population mean to within a margin of error of E = 0.5 with 95\% confidence. The population standard deviation is \sigma = 5. Use z = 1.96. How many observations do you need?

Accepted Answer

A 95\% confidence interval extends about 1.96 standard errors on each side of the estimate. You want that half-width to be at most 0.5: E = z \cdot \frac{\sigma}{\sqrt{n}}. Rearrange for n: n = \left(\frac{z\,\sigma}{E}ight)^2 = \left(\frac{1.96 	imes 5}{0.5}ight)^2 = \left(\frac{9.8}{0.5}ight)^2 = (19.6)^2 \approx 384.2. Round up so you meet (not just miss) the target: n = 385. Takeaway. Use n = (z\sigma/E)^2 and round up: demanding a small margin E is costly because it enters squared, so halving E quadruples the sample.

Sample size for a 0.5-unit margin of error

Your answer

More Statistics questions