3.10 Tests on individual coefficients
Theorem 3.10.1. Under A1–A4, \[ \frac {\hat \beta _j - \beta _j}{\operatorname {se}(\hat \beta _j)} \sim t_{n-p}, \qquad \operatorname {se}(\hat \beta _j) = \sqrt {\hat \sigma ^2 C_{jj}} . \]
Proof. \((\hat \beta _j - \beta _j)/\sqrt {\sigma ^2 C_{jj}} \sim N(0,1)\) and \((n-p)\hat \sigma ^2/\sigma ^2 \sim \chi ^2_{n-p}\), independently, by Theorem 3.7.2. By (F3) their ratio, \(Z/\sqrt {V/(n-p)}\), is \(t_{n-p}\), and \(\sigma \) cancels as in Theorem 2.8.1. □
To test \(H_0\colon \beta _j = 0\), use \(t_0 = \hat \beta _j/\operatorname {se}(\hat \beta _j)\) and reject when \(|t_0| > t_{\alpha /2, n-p}\).
Remark. This is a marginal or partial test: it asks whether \(x_j\) adds anything given that all the other regressors are in the model. Its verdict can change when other regressors are added or removed.
Solution. With \(t_{0.025,3} = 3.182\) and the standard errors of Example 3.6.2:
| Coefficient | Estimate | se | \(t_0\) | Decision |
| \(\beta _1\) (fertiliser) | 2 | 0.3412 | 5.862 | reject \(H_0\) (\(p = 0.010\)) |
| \(\beta _2\) (rainfall) | 1 | 0.5819 | 1.719 | do not reject (\(p = 0.184\)) |
Fertiliser is significant. Once fertiliser is in the model, rainfall does not add a significant amount; that is not the same as saying rainfall does not affect yield.
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