Solution. Two contrasts \(\sum c_i\mu _i\) and \(\sum d_i\mu _i\) are orthogonal if \(\sum c_id_i/n_i = 0\), which for equal \(n_i\) is \(\sum c_id_i = 0\). Each vector sums to zero, so each is a contrast. For the pairs: \(1 - 1 = 0\); \(1 - 1 = 0\); \(1 - 1 = 0\); \(1 + 1 - 2 = 0\); \(1 + 1 - 2 = 0\); \(1 + 1 + 1 - 3 = 0\). They are four mutually orthogonal contrasts among five treatments (the Helmert contrasts, each comparing a treatment with the average of those before it), so their four sums of squares add up to \(SS_{\mathrm {Treatments}}\) and are independent of one another and of \(SSE\).

Solution. With integer coefficients \((2, 2, 2, -3, -3)\), \(\sum c_iy_{i\cdot } = 2(331 + 308 + 287) - 3(260 + 251) = 1852 - 1533 = 319\) and \[ SS_C = \frac {319^2}{4(4 + 4 + 4 + 9 + 9)} = \frac {101761}{120} = 848.01 , \] so \(F_0 = 848.01/8.15 = 104.05 > 4.54\), and the contrast is significant. With coefficients \((\frac 13, \frac 13, \frac 13, -\frac 12, -\frac 12)\), \(\hat C = 77.167 - 63.875 = 13.29\) and \[ \operatorname {se}(\hat C) = \sqrt {\frac {8.15}{4}\Bigl (\frac 39 + \frac 24\Bigr )} = 1.303 , \] so the interval is \(13.29 \pm 2.131(1.303)\), that is \((10.51,\ 16.07)\).

Solution. \(HSD = 4.199\sqrt {0.4881/4} = 1.467\). The means are \(5.350\) (ration \(2\)), \(6.575\) (\(1\)), \(6.700\) (\(4\)) and \(9.200\) (\(3\)). Ration \(3\) differs from each of the others (by \(2.500\), \(2.625\) and \(3.850\), all above \(1.467\)), while the differences among rations \(1\), \(2\) and \(4\) (\(1.225\), \(1.350\), \(0.125\)) are not significant. So ration \(3\), animal fat in a large proportion, causes more damage than the rest, which form a single group. This is consistent with the factorial contrasts of Example 4.2.3: both main effects push ration \(3\) upwards.

Solution. For \(m = 10\) the Bonferroni multiplier is \(t_{0.0025,15} = 3.286\), smaller than Scheffé’s \(3.496\), so Bonferroni gives shorter intervals. For \(m = 50\) it is \(4.073\), larger than \(3.496\), and Scheffé’s intervals are shorter even though they cover every contrast. Bonferroni is preferable for a modest number of contrasts fixed in advance; Scheffé for many contrasts or for contrasts suggested by the data. (For the pairwise comparisons alone, Tukey’s \(q/\sqrt 2 = 3.088\) beats both.)

Solution. With \(r = 2\) and the coefficients for \(t = 5\):

The total is \(22.536 = SS_{\mathrm {Treatments}}\). Only the linear component is significant (\(F_{0.05,1,5} = 6.61\)), and the three higher components together give \(0.064\) on \(3\) degrees of freedom, which is exactly the lack-of-fit sum of squares of the straight line.

Solution. With \(n_i = r\), \(SS_C = r\bigl (\sum c_i\bar y_{i\cdot }\bigr )^2/\sum c_i^2 = r\bigl (\sum c_i(\bar y_{i\cdot } - \bar y_{\cdot \cdot })\bigr )^2/\sum c_i^2\), since \(\sum c_i = 0\). By the Cauchy–Schwarz inequality this is at most \(r\sum (\bar y_{i\cdot } - \bar y_{\cdot \cdot })^2 = SS_{\mathrm {Treatments}}\), with equality when \(c_i \propto \bar y_{i\cdot } - \bar y_{\cdot \cdot }\). The best single contrast compares the treatments in proportion to their observed deviations, which is a contrast chosen by the data; this is why testing it as if it had been planned would be wrong, and why Scheffé’s critical value \((t - 1)F_\alpha \), rather than \(F_\alpha \), is the right yardstick for it.