Example 10.11

EXAMPLE 10.11 Prediction Intervals from a Spreadsheet

In Example 10.7 (pages 511–512), we used statistical software to predict the log income of Alexander, who has $EDUC = 16$ years of education. Suppose that we have only the Excel spreadsheet. The prediction interval then requires some additional work.

Step 1. From the Excel output in Figure 10.5 (page 492), we know that $s = 1.1146$ . Excel can also find the mean and variance of the EDUC $x$ for the 100 entrepreneurs. They are $\bar{x} = 13.28$ and $s_{x}^{2} = 5.901$ .

Step 2. We need the value of $\sum {(x_{i} - \bar{x})}^{2}$ . Recalling the definition of the variance, we see that this is just

$\begin{array}{l} \sum {(x_{i} - \bar{x})}^{2} & = & (n - 1) s_{x}^{2} \\ = & (99) (5.901) = 584.2 \end{array}$

Step 3. The standard error for predicting Alexander’s log income from his years of education, $x^{*} = 16$ , is

$\begin{array}{l} SE \hat{y} & = & s \sqrt{1 + \frac{1}{n} + \frac{{(x^{*} - \bar{x})}^{2}}{\sum {(x_{i} - \bar{x})}^{2}}} \\ = & 1.1146 \sqrt{1 + \frac{1}{100} + \frac{{(16 - 13.28)}^{2}}{584.2}} \\ = & 1.1146 \sqrt{1 + \frac{1}{100} + \frac{7.3984}{584.2}} \\ = & (1.1146) (1.01127) = 1.12716 \end{array}$

Step 4. We predict Alexander’s log income from the least-squares line (Figure 10.5 again):

$\hat{y} = 8.2546 + (0.1126) (16) = 10.0562$

This agrees with the “Fit” from software in Example 10.8. The 95% prediction interval requires the 95% critical value for $t (98)$ . For hand calculation we use $t^{*} = 1.990$ from Table D with $df = 80$ . The interval is

$\begin{array}{l} \hat{y} \pm t^{*} SE \hat{y} & = & 10.0562 & \pm & (1.990) (1.12716) \\ = & 10.0562 & \pm & 2.2430 \\ = & 7.8132 & to & 12.2992 \end{array}$

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This agrees with the software result in Example 10.8, with a small difference due to roundoff and especially to not having the exact $t^{*}$ .