Solution. \(w = 1 - z\), so \(9.1 + 2.1x + 3.0z = 9.1 + 2.1x + 3.0(1 - w) = 12.1 + 2.1x - 3.0w\). In general, the columns \(\mathbf {1}, \mathbf {x}, \mathbf {z}\) and \(\mathbf {1}, \mathbf {x}, \mathbf {1} - \mathbf {z}\) span the same space, and the least-squares fitted values are the projection of \(\mathbf {y}\) onto that space, whatever basis describes it. Only the meaning of the coefficients changes: the intercept becomes the mean for the new reference level, and the indicator coefficient changes sign.

Solution. The columns of the \(a\) indicators are non-zero and have no \(1\) in a common row, so they are orthogonal and hence linearly independent. \(\mathbf {X}^\top \mathbf {X} = \operatorname {diag}(n_1, \dots , n_a)\) and \(\mathbf {X}^\top \mathbf {y}\) has entries \(\sum _j y_{ij} = n_i\bar y_i\), so \(\hat \gamma _i = \bar y_i\). It spans the same column space as the model with an intercept and \(a - 1\) indicators (Theorem 9.3.1), so it gives the same fit, reparametrised: \(\gamma _A = \beta _0\), and \(\gamma _i = \beta _0 + \beta _i\) for the other levels. This cell-means form avoids the dummy-variable trap by dropping the intercept instead of an indicator. Note that its \(R^2\), computed by software as for a model without an intercept, is not comparable with the usual \(R^2\).

Solution. (a) Rural (\(z = 0\)): \(\hat y = 420 + 35x\). City (\(z = 1\)): \(\hat y = 570 + 55x\).

(b) The difference is \(150 + 20x\): \(250\) kwacha at \(x = 5\), and \(150\) at \(x = 0\).

(c) With an interaction, \(\beta _2 = 150\) is the difference only at \(x = 0\), for traders with no experience. The gap grows by \(20\) kwacha with each year of experience, so there is no single difference to report; it must be given as a function of \(x\), or at stated values of \(x\) that are typical of the data.

Solution. The values of \((x - 4)_+\) are \(0, 0, 0, 0, 1, 2, 3, 4\). Then \[ \mathbf {X}^\top \mathbf {X} = \begin {bmatrix} 8 & 36 & 10 \\ 36 & 204 & 70 \\ 10 & 70 & 30 \end {bmatrix} , \qquad \mathbf {X}^\top \mathbf {y} = \begin {bmatrix} 71 \\ 405 \\ 144 \end {bmatrix} , \] with \(\det (\mathbf {X}^\top \mathbf {X}) = 880\). By cofactors, as in Example 3.3.3, \[ (\mathbf {X}^\top \mathbf {X})^{-1} = \frac {1}{880}\begin {bmatrix} 1220 & -380 & 480 \\ -380 & 140 & -200 \\ 480 & -200 & 336 \end {bmatrix} , \] and \[ \hat {\boldsymbol \beta } = \frac {1}{880}\begin {bmatrix} 1220(71) - 380(405) + 480(144) \\ -380(71) + 140(405) - 200(144) \\ 480(71) - 200(405) + 336(144) \end {bmatrix} = \frac {1}{880}\begin {bmatrix} 1840 \\ 920 \\ 1464 \end {bmatrix} = \begin {bmatrix} 2.091 \\ 1.045 \\ 1.664 \end {bmatrix} . \] So the slope is \(1.045\) up to \(x = 4\) and \(1.045 + 1.664 = 2.709\) beyond it. The residual sum of squares is \(1.573\) on \(5\) degrees of freedom, and \[ F = \frac {8.821 - 1.573}{1.573/5} = \frac {7.248}{0.3145} = 23.0 > F_{0.05,1,5} = 6.61 . \] The slope does change at \(x = 4\): the response rises nearly three times as fast beyond it.

Solution. With indicators \(z_1\), \(z_2\) for two of the schools, the regressors are \(x\), \(z_1\), \(z_2\), \(xz_1\) and \(xz_2\), and with the intercept there are \(6\) parameters: an intercept and slope for each school. Equal slopes is \(H_0\colon \beta _{xz_1} = \beta _{xz_2} = 0\), a partial \(F\) comparing this model with the parallel-lines model in \(x\), \(z_1\), \(z_2\), on \(2\) and \(n - 6\) degrees of freedom.