How much wider is a 99% interval than a 95% one?

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From a sample of n = 100 with mean \bar{x} = 50 and known population standard deviation \sigma = 10, you build confidence intervals for the mean. Construct the 95% and 99% intervals, then report the ratio of the 99% width to the 95% width.

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Setup. Both intervals share the same standard error, \sigma/\sqrt{n} = 10/10 = 1. Only the critical value changes: z^{} = 1.96 at 95% and z^{} = 2.576 at 99%. Compute the intervals. 95\%:\ 50 \pm 1.96(1) = [48.04,\ 51.96], \quad 	ext{width } 3.92. 99\%:\ 50 \pm 2.576(1) = [47.424,\ 52.576], \quad 	ext{width } 5.152. The ratio. Width is 2 z^{} 	imes 	ext{SE}, so the standard errors cancel and the ratio is just the ratio of critical values: \frac{5.152}{3.92} = \frac{2.576}{1.96} \approx 1.31. Interpret. The 99% interval is about 31% wider than the 95% interval, and this factor does not dep…

How much wider is a 99% interval than a 95% one?

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