Width ratio of a 95% to a 90% interval

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From a sample of n = 49 with mean \bar{x} = 100 and known standard deviation \sigma = 14, you build intervals for the mean. Construct the 90% and 95% intervals, then report the ratio of the 95% width to the 90% width.

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Setup. The shared standard error is \sigma/\sqrt{n} = 14/7 = 2. Critical values: z^{} = 1.645 at 90% and z^{} = 1.96 at 95%. Compute the intervals. 90\%:\ 100 \pm 1.645(2) = [96.71,\ 103.29], \quad 	ext{width } 6.58. 95\%:\ 100 \pm 1.96(2) = [96.08,\ 103.92], \quad 	ext{width } 7.84. The ratio. The standard errors cancel: \frac{7.84}{6.58} = \frac{1.96}{1.645} \approx 1.19. Interpret. Moving from 90% to 95% confidence widens the interval by about 19%, regardless of the sample values. Takeaway. Small increases in confidence near the middle of the scale cost relatively little width; the price…

Width ratio of a 95% to a 90% interval

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