Sample size at 99% confidence

Question

You want to estimate a population mean to within a margin of error of E = 1 with 99\% confidence. The population standard deviation is \sigma = 10. Use z = 2.576. How many observations do you need?

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The only change from the usual setup is the confidence level: 99\% confidence uses a larger multiplier, z = 2.576, because you are demanding a wider net that catches the true mean more often. You still want the half-width at most 1: E = z \cdot \frac{\sigma}{\sqrt{n}}. Rearrange for n: n = \left(\frac{z\,\sigma}{E}ight)^2 = \left(\frac{2.576 	imes 10}{1}ight)^2 = (25.76)^2 \approx 663.6. Round up: n = 664. Notice the same problem at 95\% confidence (z = 1.96) would need only about 384 observations. Buying that extra confidence nearly doubles the sample. Takeaway. Higher confidence means …

Sample size at 99% confidence

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