Proof. (i) The intercept and the \(a - 1\) indicators span the same space as the \(a\) indicators of all the levels (the reference indicator is the ones column minus the others). That space consists of the vectors that are constant within each level, and, as in the proof of Theorem 4.6.2, projecting onto it replaces each \(y_{ij}\) by its level mean. Matching \(\beta _0 + \beta _j\) with \(\bar y_j\) and \(\beta _0\) with the reference mean gives the coefficients.

(ii) \(SS_{\mathrm {Res}} = \sum (y_{ij} - \hat y_{ij})^2\) with \(\hat y_{ij} = \bar y_i\), and \(SS_R = \sum (\hat y_{ij} - \bar y)^2 = \sum _i n_i(\bar y_i - \bar y)^2\). The regression has \(a - 1\) degrees of freedom and the residual \(n - a\), the same as the analysis of variance, so the two \(F\) statistics coincide. □

Solution. The level means are \(\bar y_A = 6\), \(\bar y_B = 9\), \(\bar y_C = 7\) and \(\bar y = 88/12 = 7.333\). By Theorem 9.3.1, \[ \hat y = 6 + 3z_1 + 1z_2 . \] Method \(B\) scores \(3\) marks more than method \(A\) on average, method \(C\) \(1\) mark more. Then \begin {align*} SS_R &= 4\bigl [(6 - 7.333)^2 + (9 - 7.333)^2 + (7 - 7.333)^2\bigr ] \\ &= 4(1.778 + 2.778 + 0.111) = 18.67 , \\ SS_{\mathrm {Res}} &= (1 + 1 + 0 + 0) + (0 + 1 + 1 + 0) + (0 + 1 + 1 + 0) = 6 , \end {align*}

on \(2\) and \(9\) degrees of freedom, so \[ F_0 = \frac {18.67/2}{6/9} = \frac {9.333}{0.667} = 14.0 > F_{0.05,2,9} = 4.26 . \] The method matters. The same table is the one-way analysis of variance of the three groups. With \(MS_{\mathrm {Res}} = 0.667\) and four learners per group, each comparison with \(A\) has standard error \(\sqrt {0.667(1/4 + 1/4)} = 0.577\), so \(\hat \beta _1 = 3\) gives \(t = 5.2\) and \(\hat \beta _2 = 1\) gives \(t = 1.73\): \(B\) is clearly better than \(A\), while the evidence for \(C\) over \(A\) is weak.