Solution. The fit. \(\sum x_i = 48\), \(\sum y_i = 58\), so \(\bar x = 6\) and \(\bar y = 7.25\). Then \begin {align*} S_{xx} &= \sum (x_i - 6)^2 = 25 + 16 + 9 + 4 + 1 + 0 + 1 + 196 = 252, \\ S_{xy} &= \sum (x_i - 6)y_i = -15 - 20 - 15 - 16 - 8 + 0 + 11 + 126 = 63, \end {align*}

so \(\hat \beta _1 = 63/252 = 0.25\), \(\hat \beta _0 = 7.25 - 0.25 \times 6 = 5.75\), and \[ \hat y = 5.75 + 0.25x . \] The residuals are \(y_i - 5.75 - 0.25x_i\), and \(SS_{\mathrm {Res}} = 33.75\), so \(MS_{\mathrm {Res}} = 33.75/6 = 5.625\). Also \(SS_T = 49.5\) and \(R^2 = 0.318\).

Leverage and residuals, with \(h_{ii} = 1/8 + (x_i - 6)^2/252\):

For branch 8 exactly, \(h_{88} = 1/8 + 196/252 = 1/8 + 7/9 = 65/72\), so \(1 - h_{88} = 7/72\), and \begin {align*} r_8^2 &= \frac {1.75^2}{5.625 \times 7/72} = \frac {3.0625}{0.546875} = 5.6, \\ S^2_{(8)} &= \frac {33.75 - 3.0625/(7/72)}{5} = \frac {33.75 - 31.5}{5} = 0.45, \\ t_8 &= \frac {-1.75}{\sqrt {0.45 \times 7/72}} = \frac {-1.75}{0.2092} = -8.367 . \end {align*}

Branch 8 has by far the highest leverage, \(0.903\) against the cut-off \(2p/n = 0.5\). Its raw residual, \(-1.75\), is smaller than those of branches 1 and 7, and a plot of residuals against fitted values would not single it out. Its R-student residual tells a different story: \(|t_8| = 8.37\) exceeds the Bonferroni value \(t_{0.05/16,\,5} = 4.53\).

Influence measures. With \(p = 2\), \(n = 8\), the cut-offs are \(D_i > 1\), \(|\mathrm {DFFITS}_i| > 2\sqrt {2/8} = 1\), \(|\mathrm {DFBETAS}_{j,i}| > 2/\sqrt 8 = 0.707\) and COVRATIO outside \((0.25, 1.75)\). For branch 8, by Theorems 5.2.3, 5.3.2 and 5.4.3, \begin {align*} D_8 &= \frac {5.6}{2}\cdot \frac {65/72}{7/72} = 2.8 \times \frac {65}{7} = 26, \\ \mathrm {DFFITS}_8 &= -8.367\sqrt {65/7} = -8.367 \times 3.047 = -25.5, \\ \mathrm {COVRATIO}_8 &= \left (\frac {0.45}{5.625}\right )^2\frac {72}{7} = 0.0064 \times 10.29 = 0.066 . \end {align*}

For all eight branches:

Branch 8 is flagged by every measure, and overwhelmingly: \(D_8 = 26\) is twenty-six times the cut-off, and its DFBETAS for the slope says that deleting it changes \(\hat \beta _1\) by almost \(24\) standard errors. The only other flag is a mild one, branch 1’s DFBETAS for the intercept (\(-0.87\)); branch 1 is the point at the far left, and has the largest residual among the others.

The fit without branch 8. The other seven give \(\bar x = 4\), \(\bar y = 7\), \(S_{xx} = 28\), \(S_{xy} = 35\), so \[ \hat y = 2 + 1.25x , \] with \(SS_{\mathrm {Res}} = 2.25\) on \(5\) degrees of freedom (which is \(5S^2_{(8)}\), as it should be) and \(R^2 = 0.951\). The slope is five times as large. This agrees with the deletion formula: in simple regression, deleting observation \(i\) changes the slope by \((x_i - \bar x)e_i/\bigl (S_{xx}(1 - h_{ii})\bigr )\) (Problem 5.7.2), here \[ \frac {14 \times (-1.75)}{252 \times 7/72} = \frac {-24.5}{24.5} = -1 , \] and indeed \(0.25 - 1.25 = -1\).

The two lines are drawn in Figure 11. The seven ordinary branches lie close to a line of slope \(1.25\); the warehouse store, far to the right, sells much less per square metre, and on its own it flattens the fitted line until it fits none of the branches well.