2.8 Hypothesis tests on the slope and intercept
Theorem 2.8.1. \[ \frac {\hat \beta _1 - \beta _1}{\operatorname {se}(\hat \beta _1)} \sim t_{n-2} \qquad \text {and}\qquad \frac {\hat \beta _0 - \beta _0}{\operatorname {se}(\hat \beta _0)} \sim t_{n-2} . \]
Proof. For the slope, \(Z = (\hat \beta _1 - \beta _1)\big /\sqrt {\sigma ^2/S_{xx}} \sim N(0,1)\) by Theorem 2.7.1, and \(V = SS_{\mathrm {Res}}/\sigma ^2 = (n-2)\hat \sigma ^2/\sigma ^2 \sim \chi ^2_{n-2}\) independently of it, by Theorem 2.7.3. By (F3), \[ \frac {Z}{\sqrt {V/(n-2)}} = \frac {(\hat \beta _1 - \beta _1)\big /\sqrt {\sigma ^2/S_{xx}}}{\sqrt {\hat \sigma ^2/\sigma ^2}} = \frac {\hat \beta _1 - \beta _1}{\sqrt {\hat \sigma ^2/S_{xx}}} \sim t_{n-2} . \] The unknown \(\sigma \) cancels. The intercept is the same with \(\sigma ^2\left (\frac 1n + \frac {\bar x^2}{S_{xx}}\right )\) in place of \(\sigma ^2/S_{xx}\). □
To test \(H_0\colon \beta _1 = \beta _{10}\) for a given value \(\beta _{10}\), compute \[ t_0 = \frac {\hat \beta _1 - \beta _{10}}{\operatorname {se}(\hat \beta _1)} . \] Under \(H_0\), \(t_0 \sim t_{n-2}\). Writing \(t_{\alpha , \nu }\) for the point with probability \(\alpha \) above it in the \(t_\nu \) distribution, at significance level \(\alpha \):
| Alternative | Reject \(H_0\) when |
| \(H_1\colon \beta _1 \neq \beta _{10}\) | \(|t_0| > t_{\alpha /2, n-2}\) |
| \(H_1\colon \beta _1 > \beta _{10}\) | \(t_0 > t_{\alpha , n-2}\) |
| \(H_1\colon \beta _1 < \beta _{10}\) | \(t_0 < -t_{\alpha , n-2}\) |
Tests on \(\beta _0\) are the same, with \(\operatorname {se}(\hat \beta _0)\).
The most important case is \(H_0\colon \beta _1 = 0\), the test of significance of regression. If \(\beta _1 = 0\) then \(E(y) = \beta _0\) whatever \(x\) is: \(x\) is of no use in explaining \(y\), at least through a straight line. Failing to reject it can mean that there is no relationship, or that the relationship is not linear (draw the scatter plot), or simply that there are too few data to detect it.
Example 2.8.2. For the advertising data, test at the 5% level whether sales are linearly related to advertising, and whether the intercept is zero.
Solution. From Examples 2.2.6 and 2.5.5: \(\hat \beta _1 = 0.7\), \(\operatorname {se}(\hat \beta _1) = 0.1915\), \(\hat \beta _0 = -0.1\), \(\operatorname {se}(\hat \beta _0) = 0.6351\), with \(n - 2 = 3\) degrees of freedom. From tables, \(t_{0.025,3} = 3.182\).
Slope. \(H_0\colon \beta _1 = 0\) against \(H_1\colon \beta _1 \neq 0\). \[ t_0 = \frac {0.7 - 0}{0.1915} = 3.656 . \] Since \(|t_0| = 3.656 > 3.182\), reject \(H_0\) at the 5% level (the \(p\)-value is \(0.035\)). Advertising and sales are linearly related.
Intercept. \(H_0\colon \beta _0 = 0\) against \(H_1\colon \beta _0 \neq 0\). \[ t_0 = \frac {-0.1 - 0}{0.6351} = -0.157 . \] \(|t_0| < 3.182\), so do not reject (\(p = 0.885\)). The data are consistent with a line through the origin.
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