Proof. Let \(\mathbf {H}_1 = \mathbf {X}_1(\mathbf {X}_1^\top \mathbf {X}_1)^{-1}\mathbf {X}_1^\top \) be the hat matrix of the reduced model. The correction term \(\left (\sum y_i\right )^2/n\) is the same in both regression sums of squares and cancels, so \(SS_R(\boldsymbol \beta _2 \mid \boldsymbol \beta _1) = \mathbf {y}^\top (\mathbf {H} - \mathbf {H}_1)\mathbf {y}\).

The columns of \(\mathbf {X}_1\) are columns of \(\mathbf {X}\), so \(\mathbf {H}\mathbf {X}_1 = \mathbf {X}_1\) by Theorem 3.4.2(ii), hence \(\mathbf {H}\mathbf {H}_1 = \mathbf {H}_1\) and, transposing, \(\mathbf {H}_1\mathbf {H} = \mathbf {H}_1\). Exactly as for \(\mathbf {H} - \bar {\mathbf {J}}\) in Theorem 3.8.1, it follows that \(\mathbf {H} - \mathbf {H}_1\) is symmetric and idempotent, that \((\mathbf {H} - \mathbf {H}_1)(\mathbf {I} - \mathbf {H}) = \mathbf {0}\), and that its rank is \(\operatorname {tr}(\mathbf {H}) - \operatorname {tr}(\mathbf {H}_1) = p - (p - r) = r\).

Under \(H_0\), \(\mathbf {X}\boldsymbol \beta = \mathbf {X}_1\boldsymbol \beta _1\), and \((\mathbf {H} - \mathbf {H}_1)\mathbf {X}_1 = \mathbf {X}_1 - \mathbf {X}_1 = \mathbf {0}\). So \(SS_R(\boldsymbol \beta _2 \mid \boldsymbol \beta _1) = \boldsymbol \varepsilon ^\top (\mathbf {H} - \mathbf {H}_1)\boldsymbol \varepsilon \), and Lemma 3.7.1 and (F3) finish the proof as before. □

Solution. Reduced model: \(y = \beta _0 + \beta _1 x_1 + \varepsilon \), a simple regression. From the table of Example 3.3.3, \[ S_{x_1x_1} = 70 - \frac {18^2}{6} = 16, \qquad S_{x_1y} = 244 - \frac {18 \times 69}{6} = 37, \] so \(SS_R(\beta _1) = S_{x_1y}^2/S_{x_1x_1} = 37^2/16 = 85.5625\).

Extra sum of squares: \(SS_R(\beta _2 \mid \beta _1) = 89.5 - 85.5625 = 3.9375\), on \(r = 1\) degree of freedom.

Test: \[ F_0 = \frac {3.9375/1}{1.3333} = 2.953 < F_{0.05,1,3} = 10.13 , \] so do not reject \(H_0\colon \beta _2 = 0\). This agrees with the \(t\)-test of Example 3.10.2, and \(t_0^2 = 1.7185^2 = 2.953 = F_0\): when \(r = 1\), the partial \(F\)-test and the \(t\)-test are the same test (Theorem 3.11.4).

The order in which regressors are added matters. Fitting rainfall alone gives \(SS_R(\beta _2) = 43.68\), so \(SS_R(\beta _1 \mid \beta _2) = 89.5 - 43.68 = 45.82\): fertiliser added to rainfall explains far more than rainfall added to fertiliser.

Proof. \(\mathbf {u} \neq \mathbf {0}\), since \(\mathbf {x}\) is not a combination of the columns of \(\mathbf {X}_1\) (full rank), and \(\mathbf {X}_1^\top \mathbf {u} = \mathbf {0}\) by Theorem 3.4.2(iii) applied to the regression of \(\mathbf {x}\) on \(\mathbf {X}_1\); also \(\mathbf {H}_1\mathbf {u} = \mathbf {0}\). Put \[ \mathbf {P} = \mathbf {H}_1 + \frac {\mathbf {u}\mathbf {u}^\top }{\mathbf {u}^\top \mathbf {u}} . \] We show that \(\mathbf {Py}\) is the least-squares fit \(\hat {\mathbf {y}}\). First, \(\mathbf {Py}\) lies in the column space of \(\mathbf {X}\), since \(\mathbf {H}_1\mathbf {y}\) is a combination of the columns of \(\mathbf {X}_1\) and \(\mathbf {u} = \mathbf {x} - \mathbf {H}_1\mathbf {x}\) is a combination of \(\mathbf {x}\) and those columns. Second, \(\mathbf {y} - \mathbf {Py}\) is perpendicular to every column of \(\mathbf {X}\): \begin {align*} \mathbf {X}_1^\top (\mathbf {y} - \mathbf {Py}) &= \mathbf {X}_1^\top \mathbf {y} - \mathbf {X}_1^\top \mathbf {y} - \mathbf {0} = \mathbf {0}, \\ \mathbf {u}^\top (\mathbf {y} - \mathbf {Py}) &= \mathbf {u}^\top \mathbf {y} - 0 - \mathbf {u}^\top \mathbf {y} = 0 , \end {align*}

using \(\mathbf {X}_1^\top \mathbf {H}_1 = \mathbf {X}_1^\top \), \(\mathbf {X}_1^\top \mathbf {u} = \mathbf {0}\) and \(\mathbf {u}^\top \mathbf {H}_1 = \mathbf {0}^\top \); and \(\mathbf {x} = \mathbf {H}_1\mathbf {x} + \mathbf {u}\) is a combination of these. So \(\mathbf {Py}\) satisfies the normal equations, and since the least-squares fit is unique, \(\hat {\mathbf {y}} = \mathbf {Py}\).

Now compare two ways of writing \(\hat {\mathbf {y}}\). On one hand \(\hat {\mathbf {y}} = \mathbf {X}_1\hat {\boldsymbol \beta }_1 + \mathbf {x}\hat \beta _j = (\mathbf {X}_1\hat {\boldsymbol \beta }_1 + \mathbf {H}_1\mathbf {x}\hat \beta _j) + \mathbf {u}\hat \beta _j\); on the other, \(\hat {\mathbf {y}} = \mathbf {H}_1\mathbf {y} + \mathbf {u}(\mathbf {u}^\top \mathbf {y})/(\mathbf {u}^\top \mathbf {u})\). In each, the bracketed part lies in the column space of \(\mathbf {X}_1\), which is perpendicular to \(\mathbf {u}\). Taking the inner product of both with \(\mathbf {u}\) gives \(\mathbf {u}^\top \mathbf {u}\,\hat \beta _j = \mathbf {u}^\top \mathbf {y}\), the first formula.

It is linear in \(\mathbf {y}\), so \(\operatorname {Var}(\hat \beta _j) = \sigma ^2\mathbf {u}^\top \mathbf {u}/(\mathbf {u}^\top \mathbf {u})^2 = \sigma ^2/\mathbf {u}^\top \mathbf {u}\), and comparing with \(\sigma ^2 C_{jj}\) gives the second. Finally \(\mathbf {H} = \mathbf {P}\) (both give the fit of every \(\mathbf {y}\)), so \(\mathbf {H} - \mathbf {H}_1 = \mathbf {u}\mathbf {u}^\top /\mathbf {u}^\top \mathbf {u}\) and \(SS_R(\beta _j \mid \text {others}) = \mathbf {y}^\top (\mathbf {H} - \mathbf {H}_1)\mathbf {y} = (\mathbf {u}^\top \mathbf {y})^2/\mathbf {u}^\top \mathbf {u}\). Dividing by \(MS_{\mathrm {Res}}\), \[ F_0 = \frac {\hat \beta _j^2/C_{jj}}{\hat \sigma ^2} = \left ( \frac {\hat \beta _j}{\sqrt {\hat \sigma ^2 C_{jj}}} \right )^2 = t_0^2 . \qedhere \] □