Solution. (a) \(y_{ij} = \mu + \tau _i + \varepsilon _{ij}\), \(i = 1, \dots , 4\), \(j = 1, \dots , 4\), where \(y_{ij}\) is the organ damage of the \(j\)th animal on ration \(i\), \(\mu \) the overall mean, \(\tau _i\) the fixed effect of ration \(i\) with \(\sum \tau _i = 0\), and \(\varepsilon _{ij}\) the random error; the errors are independent \(N(0, \sigma ^2)\) with a common variance.

(b) The ration means are \(6.575\), \(5.350\), \(9.200\) and \(6.700\), and \(\bar y_{\cdot \cdot } = 111.3/16 = 6.95625\). So \(\hat \mu = 6.956\) and \[ \hat \tau _1 = -0.381 , \quad \hat \tau _2 = -1.606 , \quad \hat \tau _3 = 2.244 , \quad \hat \tau _4 = -0.256 . \] \(\hat \sigma ^2 = MSE\) is found in (c).

(c) The correction factor is \(111.3^2/16 = 774.2306\), so \(SST = 811.39 - 774.2306 = 37.1594\), \begin {align*} SS_{\mathrm {Rations}} &= \frac {26.3^2 + 21.4^2 + 36.8^2 + 26.8^2}{4} - 774.2306 \\ &= 805.5325 - 774.2306 = 31.3019 , \end {align*}

and \(SSE = 5.8575\).

\(F_0 = 21.38 > F_{0.05,3,12} = 3.49\) (\(p = 4 \times 10^{-5}\)), so the mean organ damage differs among the rations, and \(\hat \sigma ^2 = 0.488\). Which rations differ, and whether it is the type or the proportion of fat that matters, are questions about contrasts, taken up in a later section.

Solution. With \(X\) as in the text, \[ X^\top X = \begin {pmatrix} 6 & 2 & 2 & 2 \\ 2 & 2 & 0 & 0 \\ 2 & 0 & 2 & 0 \\ 2 & 0 & 0 & 2 \end {pmatrix} , \] whose first column is the sum of the other three, so the rank is \(3\). The normal equations give \(\hat \mu + \hat \tau _i = \bar y_{i\cdot }\). With \(\hat \tau _3 = 0\): \(\hat \mu = \bar y_{3\cdot }\), \(\hat \tau _1 = \bar y_{1\cdot } - \bar y_{3\cdot }\) and \(\hat \tau _2 = \bar y_{2\cdot } - \bar y_{3\cdot }\), so \(\hat \tau _1 - \hat \tau _2 = \bar y_{1\cdot } - \bar y_{2\cdot }\). With \(\sum \hat \tau _i = 0\): \(\hat \tau _i = \bar y_{i\cdot } - \bar y_{\cdot \cdot }\), and again \(\hat \tau _1 - \hat \tau _2 = \bar y_{1\cdot } - \bar y_{2\cdot }\), as Corollary 3.2.4 requires.

\(\tau _1 + \tau _2\) is not estimable: its coefficients \((1, 1)\) do not sum to zero, and it is not of the form \(\sum a_i(\mu + \tau _i)\), since that would need \(\sum a_i = 0\) (no \(\mu \)) with \(a_1 = a_2 = 1\) and \(a_3 = 0\). Indeed the two side conditions give different values, \(\bar y_{1\cdot } + \bar y_{2\cdot } - 2\bar y_{3\cdot }\) and \(\bar y_{1\cdot } + \bar y_{2\cdot } - 2\bar y_{\cdot \cdot }\).

Solution. Expanding, \begin {align*} \sum _in_i(\bar y_{i\cdot } - \bar y_{\cdot \cdot })^2 &= \sum _in_i\bar y_{i\cdot }^2 - 2\bar y_{\cdot \cdot }\sum _in_i\bar y_{i\cdot } + N\bar y_{\cdot \cdot }^2 \\ &= \sum _i\frac {y_{i\cdot }^2}{n_i} - 2\bar y_{\cdot \cdot }y_{\cdot \cdot } + \frac {y_{\cdot \cdot }^2}{N} , \end {align*}

and \(\bar y_{\cdot \cdot }y_{\cdot \cdot } = y_{\cdot \cdot }^2/N\), which gives the formula. The same expansion gives \(SST = \sum \sum y_{ij}^2 - y_{\cdot \cdot }^2/N\), and subtracting, \(SSE = \sum \sum y_{ij}^2 - \sum _iy_{i\cdot }^2/n_i\).

Solution. If \(A\mathbf v = \lambda \mathbf v\) with \(\mathbf v \ne \mathbf 0\), then \(\lambda \mathbf v = A\mathbf v = A^2\mathbf v = \lambda ^2\mathbf v\), so \(\lambda ^2 = \lambda \) and \(\lambda \in \{0, 1\}\). A symmetric matrix is diagonalisable, so its rank is the number of non-zero eigenvalues, which here is the number of eigenvalues equal to \(1\), which is their sum, the trace.

By Lemma 3.1.2 with \(A = H - H_0\), \(\Sigma = \sigma ^2I\) and \(\boldsymbol \mu \) the vector of means, \begin {align*} E(SS_{\mathrm {Treatments}}) &= \sigma ^2\operatorname {tr}(H - H_0) + \boldsymbol \mu ^\top (H - H_0)\boldsymbol \mu \\ &= (t - 1)\sigma ^2 + \sum _in_i(\mu _i - \bar \mu )^2 , \end {align*}

since \(\operatorname {tr}(H) = t\), \(\operatorname {tr}(H_0) = 1\), and \((H - H_0)\boldsymbol \mu \) has entries \(\mu _i - \bar \mu = \tau _i - \bar \tau \).

Solution. \(\lambda = \sum n_i\tau _i^2/\sigma ^2 = 4(10.90^2 + 5.15^2 + 0.10^2 + 6.85^2 + 9.10^2)/8.15 = 1100.3/8.15 = 135.0\). Under \(H_1\), \(E(MS_{\mathrm {Treatments}}) = \sigma ^2(1 + \lambda /4) \approx 34.8\sigma ^2\), so \(F_0\) is expected to be about \(35\), far beyond the critical value \(3.06\). The coating means differ by up to \(20\) units while the error standard deviation is under \(3\); with such large effects even four replicates give overwhelming evidence.

Solution. Here \(S_p^2 = MSE = 0.4881\), \(N - t = 12\) and \(n_i - 1 = 3\). Then \begin {align*} q &= 12\log _{10}0.4881 \\ &\quad - 3\bigl (\log _{10}0.2825 + \log _{10}0.8700 + \log _{10}0.5133 + \log _{10}0.2867\bigr ) \\ &= 0.5874 , \end {align*}

\(c = 1 + \frac 19\bigl (\frac 43 - \frac {1}{12}\bigr ) = 1.1389\), and \(\chi _0^2 = 2.3026(0.5874)/1.1389 = 1.188 < \chi ^2_{0.05,3} = 7.81\) (\(p = 0.76\)). There is no evidence against equal variances, although with four observations per ration the test has little power: variances differing by a factor of three are not unusual in samples this small.

Solution. With \(g(p) = \arcsin \sqrt p\), \[ g'(p) = \frac {1}{\sqrt {1 - p}}\cdot \frac {1}{2\sqrt p} = \frac {1}{2\sqrt {p(1 - p)}} . \] Since \(\Var (\hat p) = p(1 - p)/m\), the delta method gives \(\Var \bigl (g(\hat p)\bigr ) \approx \dfrac {1}{4p(1 - p)}\cdot \dfrac {p(1 - p)}{m} = \dfrac {1}{4m}\), free of \(p\).

Solution. The ratio \(s_i/\bar y_{i\cdot }\) is about \(0.2\) for every treatment, so \(\sigma \propto \mu \) and \(\alpha = 1\). (Equivalently, the slope of \(\log s_i\) on \(\log \bar y_{i\cdot }\) is close to \(1\): each doubling of the mean doubles the standard deviation.) The logarithm is indicated, and the analysis of \(\log y\) compares treatments by the ratios of their typical responses.

Solution. The level means are \(3.3\), \(4.4\), \(5.6\), \(6.6\) and \(7.5\), and \(SST = 22.976\), \(SS_{\mathrm {Treatments}} = 22.536\), so \(SS_{\mathrm {PE}} = 0.440\) on \(10 - 5 = 5\) degrees of freedom. The least squares line is \(\hat y = 2.30 + 1.06x\), with regression sum of squares \(22.472\), so \(SS_{\mathrm {Res}} = 0.504\) and \(SS_{\mathrm {LOF}} = 0.504 - 0.440 = 0.064\) on \(5 - 2 = 3\) degrees of freedom. Then \[ F_0 = \frac {0.064/3}{0.440/5} = 0.24 < F_{0.05,3,5} = 5.41 \qquad (p = 0.86) , \] so there is no evidence of lack of fit, and the straight line, whose slope is highly significant (\(F_0 = 22.472/(0.504/8) = 357\)), describes the data.

Solution. The least favourable non-centrality is \(\lambda = r(1.5\sigma )^2/(2\sigma ^2) = 1.125r\), and the power is computed from the non-central \(F_{4,\,5(r-1)}(\lambda )\) distribution. It is \(0.773\) at \(r = 11\) (\(\lambda = 12.375\)) and \(0.817\) at \(r = 12\) (\(\lambda = 13.5\)), so \(12\) replicates per treatment, \(60\) runs in all, are needed.