Example 10.5

EXAMPLE 10.5 Does Log Income Increase with Education?

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CASE 10.1 The Excel regression output in Figure 10.5 (page 492) for the entrepreneur problem contains the information needed for inference about the regression coefficients. You can see that the slope of the least-squares line is $b_{1} = 0.1126$ and the standard error of this statistic is ${SE}_{b_{1}} = 0.046116$ .

Given that the response $y$ is on the log scale, this slope approximates the percent change in $y$ for a unit change in $x$ (see Example 13.10 [pages 661–662] for more details). In this case, one extra year of education is associated with an approximate 11.3% increase in income.

The $t$ statistic and $P$ -value for the test of $H_{0} : β_{1} = 0$ against the two-sided alternative $H_{a} : β_{1} \neq 0$ appear in the columns labeled “ $t$ Stat” and “ $P$ -value.” The $t$ statistic for the significance of the regression is

$t = \frac{b_{1}}{{SE}_{b_{1}}} = \frac{0.1126}{0.046116} = 2.44$

and the $P$ -value for the two-sided alternative is 0.0164. If we expected beforehand that income rises with education, our alternative hypothesis would be one-sided, $H_{a} : β_{1} > 0$ . The $P$ -value for this $H_{a}$ is one-half the two-sided value given by Excel; that is, $P = 0.0082$ . In both cases, there is strong evidence that the mean log income level increases as education increases.

A 95% confidence interval for the slope $β_{1}$ of the regression line in the population of all entrepreneurs in the United States is

$\begin{array}{l} b_{1} \pm t^{*} {SE}_{b_{1}} & = 0.1126 \pm (1.990) (0.046116) \\ = 0.1126 \pm 0.09177 \\ = 0.0208 to 0.2044 \end{array}$

This interval contains only positive values, suggesting an increase in log income for an additional year of schooling. We’re 95% confident that the average increase in income for one additional year of education is between 2.1% and 20.4%.

The $t$ distribution for this problem has $n - 2 = 98$ degrees of freedom. Table D has no entry for 98 degrees of freedom, so we use the table entry $t^{*} = 1.990$ for 80 degrees of freedom. As a result, our confidence interval agrees only approximately with the more accurate software result. Note that using the next lower degrees of freedom in Table D makes our interval a bit wider than we actually need for 95% confidence. Use this conservative approach when you don’t know $t^{*}$ for the exact degrees of freedom.