Put numbers on the confidence-versus-prediction gap

Question

You fit a regression on n = 100 points with residual standard deviation s = 3. You want an interval at x0 = \bar x (the center of the data, so the extrapolation term is negligible). Compute the confidence-interval and prediction-interval half-widths (use z \approx 1.96), and give their ratio.

Accepted Answer

At the center, the two variances simplify: - Confidence interval (mean response): \operatorname{Var}(\hat y0) = s^2 \cdot \frac{1}{n}. - Prediction interval (new observation): \operatorname{Var}(y0 - \hat y0) = s^2\big(1 + \frac{1}{n}\big). Confidence half-width: 1.96 \cdot s\sqrt{	frac{1}{n}} = 1.96 \cdot 3 \cdot \sqrt{0.01} = 1.96 \cdot 3 \cdot 0.1 = 0.588. Prediction half-width: 1.96 \cdot s\sqrt{1 + 	frac{1}{n}} = 1.96 \cdot 3 \cdot \sqrt{1.01} = 1.96 \cdot 3 \cdot 1.00499 = 5.909. Ratio: \frac{	ext{prediction}}{	ext{confidence}} = \frac{\sqrt{1 + 1/n}}{\sqrt{1/n}} = \sqrt{n + 1} = \s…

Put numbers on the confidence-versus-prediction gap

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