Solution. (a) Here \(p = 3\), \(n = 6\), \(n - p = 3\), and from Example 4.2.6 we have \(r_i\), \(t_i\), \(S^2_{(i)}\) and \(h_{ii}\). For plot 2, exactly, \(r_2^2 = 189/104\) and \(h_{22}/(1 - h_{22}) = 37/26\), so \[ D_2 = \frac {189}{3 \times 104}\cdot \frac {37}{26} = \frac {63}{104}\cdot \frac {37}{26} = \frac {2331}{2704} = 0.862 , \] \[ \mathrm {DFFITS}_2 = 1.753\sqrt {37/26} = 1.753 \times 1.193 = 2.091 , \] \[ \mathrm {COVRATIO}_2 = \left (\frac {0.788}{1.333}\right )^3\frac {63}{26} = 0.2068 \times 2.423 = 0.501 . \] In full:

For plots 5 and 6, \(e_i = 0\), so \(S^2_{(i)} = 4/2 = 2\) and \(\mathrm {COVRATIO}_6 = (2/1.333)^3/(2/9) = 3.375 \times 4.5 = 15.19\).

(b) The leverage cut-off is \(2p/n = 1\), which no point can exceed: with \(p = 3\) and only \(n = 6\), every point carries a large share of the fit, and the rule is useless. Relative to the average \(p/n = 0.5\), plots 2, 5 and 6 stand out. No \(D_i\) exceeds \(1\), but plot 2 comes close, and \(\mathrm {DFFITS}_2 = 2.09\) exceeds \(2\sqrt {3/6} = 1.41\). Plot 2 is the most influential observation.

Plot 6 has the highest leverage of all but no influence on the estimates: the plane passes exactly through it, so deleting it changes nothing (\(D_6 = 0\)). Its COVRATIO of \(15.2\) shows what it does do: without it the generalised variance of the estimates would be fifteen times larger. It is a high-leverage point that lies on the trend, panel 2 of Figure 10, and it is the most valuable observation in the data.

Solution. By Theorem 4.2.3(ii), \(\hat {\boldsymbol \beta } - \hat {\boldsymbol \beta }_{(i)} = \mathbf {C}\mathbf {x}_i e_i/(1 - h_{ii})\). From Example 3.3.2, with \(\sum x_j^2 = S_{xx} + n\bar x^2\), \[ \mathbf {C}\mathbf {x}_i = \frac {1}{nS_{xx}}\begin {bmatrix} \sum x_j^2 - n\bar x x_i \\ -n\bar x + n x_i \end {bmatrix} = \begin {bmatrix} \dfrac {1}{n} - \dfrac {\bar x(x_i - \bar x)}{S_{xx}} \\[2mm] \dfrac {x_i - \bar x}{S_{xx}} \end {bmatrix} , \] since \(\sum x_j^2 - n\bar x x_i = S_{xx} - n\bar x(x_i - \bar x)\). Multiply by \(e_i/(1 - h_{ii})\). At \(x_i = \bar x\) the slope change is \(0\); the point moves the line up or down, parallel to itself, by \(e_i/(n(1 - h_{ii})) = e_i/(n - 1)\), since \(h_{ii} = 1/n\) there. This is panel 1 of Figure 10. The further \(x_i\) is from \(\bar x\), the larger both factors \(x_i - \bar x\) and \(1/(1 - h_{ii})\) become.

Solution. By Theorems 5.2.3 and 5.3.2, \[ \frac {\mathrm {DFFITS}_i^2}{p\,D_i} = \frac {t_i^2\,h_{ii}/(1 - h_{ii})}{r_i^2\,h_{ii}/(1 - h_{ii})} = \frac {t_i^2}{r_i^2} = \frac {e_i^2/\bigl (S^2_{(i)}(1 - h_{ii})\bigr )}{e_i^2/\bigl (MS_{\mathrm {Res}}(1 - h_{ii})\bigr )} = \frac {MS_{\mathrm {Res}}}{S^2_{(i)}} . \] For observation 4 of Example 4.4.1, \(r_4 = 1.989\), \(t_4 = 16.39\) and \(h_{44} = 0.181\), so \(h_{44}/(1 - h_{44}) = 0.221\), and with \(p = 2\), \[ D_4 = \frac {1.989^2}{2} \times 0.221 = 0.437, \qquad \mathrm {DFFITS}_4 = 16.39 \times \sqrt {0.221} = 7.70 . \] Cook’s distance is well below \(1\); DFFITS is nearly seven times its cut-off \(2\sqrt {2/6} = 1.15\). They disagree because \(D_i\) is built on the studentized residual, which here cannot exceed \(\sqrt {n - p} = 2\), while DFFITS is built on R-student, which has no such bound. The ratio \(MS_{\mathrm {Res}}/S^2_{(4)} = 25/0.368 = 68\) is exactly the factor by which the bad value has inflated the residual mean square. Observation 4 is near the middle of the \(x\)-range, so it has little leverage and moves the slope only moderately, which is what \(D_4\) reports; but it is a gross outlier, which is what DFFITS reports.

Solution. (a) With \(n = 9\), \(p = 2\): \(2p/n = 0.444\), so both have (just) high leverage; but \(D_i = 0.144 < 1\) and \(|\mathrm {DFFITS}_i| = 0.508 < 2\sqrt {2/9} = 0.943\). On the deletion measures neither is influential.

(b) Deleting one large store leaves the other at the same point, and it holds the line almost exactly where it was: the fit to the remaining eight is the fit of Example 5.5.1, \(\hat y = 5.75 + 0.25x\), little different from \(5.94 + 0.199x\). Each store masks the other. Deleting both together gives \(\hat y = 2 + 1.25x\), the dramatic change of the example. The problem is plain in a scatter plot, and in the leverages: two points share a leverage of almost \(0.95\) out of a total of \(p = 2\). Deleting the pair, and computing the analogue of Cook’s distance for the pair, would reveal it.

Solution. (a) By Example 4.1.2, \(h_{ii} - 1/n = (x_i - \bar x)^2/S_{xx}\), and these add to \(S_{xx}/S_{xx} = 1\). They are all non-negative, so two of them cannot both be at least \(1/2\) unless they are exactly \(1/2\) each and the rest are zero.

(b) For \(0, 1, 2, 3, 4\): \(\bar x = 2\), \(S_{xx} = 10\), and \(h_{ii} = 0.2 + (x_i - 2)^2/10\), which is \(0.6, 0.3, 0.2, 0.3, 0.6\). For \(0, 0, 0, 0, 4\): \(\bar x = 0.8\), \(S_{xx} = 4(0.64) + 3.2^2 = 12.8\), and \(h_{ii} = 0.2 + 0.64/12.8 = 0.25\) for the four points at \(0\), but \(h_{55} = 0.2 + 10.24/12.8 = 1\) for the point at \(4\). In the second design the single observation at \(x = 4\) alone determines the slope: the line must pass through it (Problem 4.7.5), its residual is always zero, and an error in it cannot be detected at all. The first design spreads the leverage, so every observation is checked by the others. (The second design does have a smaller variance for \(\hat \beta _1\), \(\sigma ^2/12.8\) against \(\sigma ^2/10\), but only by trusting one observation completely; it also gives no way of seeing whether the relationship is a straight line.)