Solution. (a) Varieties are fixed (all of interest, the same in any repeat); farms are random (a sample from the district, and inference is meant to apply to the district). Mixed model: \(H_0 : \tau _1 = \dots = \tau _5 = 0\) for varieties, \(H_0 : \sigma ^2_\beta = 0\) for farms, \(H_0 : \sigma ^2_{\tau \beta } = 0\) for the interaction.

(b) Technicians and parts are both random: nobody cares about technician \(2\) in particular; the question is how much of the measurement variation is due to technicians (reproducibility) and to repeat measurements (repeatability). All hypotheses concern variance components. This is the usual gauge study.

(c) Temperature is fixed. Suppliers are random. Batches are random too, but each batch belongs to one supplier, so batches are nested in suppliers rather than crossed with them; nested factors are treated in the next section.

Solution. Now \(y_{i\cdot } = n_i\mu + n_i\tau _i + \varepsilon _{i\cdot }\) and \(y_{\cdot \cdot } = N\mu + \sum _in_i\tau _i + \varepsilon _{\cdot \cdot }\), so \begin {align*} E(y_{i\cdot }^2) &= n_i^2\mu ^2 + n_i^2\sigma ^2_\tau + n_i\sigma ^2 , \\ E(y_{\cdot \cdot }^2) &= N^2\mu ^2 + \textstyle \sum _in_i^2\sigma ^2_\tau + N\sigma ^2 . \end {align*}

Then \begin {align*} E(SS_{\mathrm {Treatments}}) &= \sum _i\frac {E(y_{i\cdot }^2)}{n_i} - \frac {E(y_{\cdot \cdot }^2)}N \\ &= (N\mu ^2 + N\sigma ^2_\tau + t\sigma ^2) - \Bigl (N\mu ^2 + \frac {\sum _in_i^2}N\sigma ^2_\tau + \sigma ^2\Bigr ) \\ &= (t - 1)\sigma ^2 + \Bigl (N - \frac {\sum _in_i^2}N\Bigr )\sigma ^2_\tau , \end {align*}

and dividing by \(t - 1\) gives the result. \(E(SSE) = \sum _{i,j}E(y_{ij}^2) - \sum _iE(y_{i\cdot }^2)/n_i = N\sigma ^2 - t\sigma ^2\), so \(E(MSE) = \sigma ^2\). By the Cauchy–Schwarz inequality \(\sum n_i^2 \ge N^2/t\), so \[ (t - 1)n_0 \le N - N/t = (t - 1)N/t , \] with equality only when all \(n_i\) are equal. Imbalance reduces the effective number of observations per level.

Solution. \(\Cov (w_{ij}, \bar w_{\cdot j}) = \sigma ^2_{\tau \beta }/a\) and \(\Var (\bar w_{\cdot j}) = \sigma ^2_{\tau \beta }/a\), so \begin {align*} \Var \bigl ((\tau \beta )_{ij}\bigr ) &= \sigma ^2_{\tau \beta } - 2\sigma ^2_{\tau \beta }/a + \sigma ^2_{\tau \beta }/a = \tfrac {a - 1}a\sigma ^2_{\tau \beta } , \\ \Cov \bigl ((\tau \beta )_{ij}, (\tau \beta )_{i'j}\bigr ) &= 0 - \sigma ^2_{\tau \beta }/a - \sigma ^2_{\tau \beta }/a + \sigma ^2_{\tau \beta }/a = -\sigma ^2_{\tau \beta }/a . \end {align*}

The interactions in one column must sum to zero, so if one fixed level does unusually well with a given random level, the others must do worse on average: they are negatively correlated. Interactions in different columns are independent.

Solution. Both factors are now random, and Theorem 11.3.2 applies. The sums of squares and mean squares are unchanged. The interaction is still tested against error (\(F_0 = 1.49\), not significant). Both main effects are now tested against \(MS_{AB} = 0.2762\) on \(6\) degrees of freedom: \begin {align*} \text {operators: } F_0 &= 3.4393/0.2762 = 12.45 > F_{0.05, 3, 6} = 4.76 , \\ \text {filters: } F_0 &= 2.3169/0.2762 = 8.39 > F_{0.05, 2, 6} = 5.14 \quad (P = 0.018) . \end {align*}

Both are still significant, but the evidence on filters is weaker than in the restricted mixed model (\(P = 0.0002\)), because the denominator has larger expectation and fewer degrees of freedom. The estimates are now variance components for both factors: \[ \hat \sigma ^2_\tau = \frac {3.4393 - 0.2762}9 = 0.351 , \qquad \hat \sigma ^2_\beta = \frac {2.3169 - 0.2762}{12} = 0.170 , \] with \(\hat \sigma ^2_{\tau \beta } = 0.030\) and \(\hat \sigma ^2 = 0.185\). What changes is the question: instead of asking which of these four operators differ, the analysis now says that operators in general are the largest source of variability, so training or standardising the procedure is where to start.

Solution. The power is \(P\bigl (F_{t-1, 3t} > F_{0.05, t-1, 3t}/(1 + 4\times 0.5)\bigr ) = P\bigl (F_{t-1, 3t} > F_{0.05, t-1, 3t}/3\bigr )\). Evaluating it,

so \(t = 15\) levels (60 observations) are needed. For \(N = 24\) and \(\theta = 1\) the powers are \(0.73\), \(0.73\), \(0.70\) and \(0.56\). Few levels with many replicates loses error degrees of freedom little but leaves \(t - 1\) small; many levels with \(r = 2\) makes \(1 + r\theta \) small. Here \(t = 4\) to \(6\) is best for testing, but for estimating \(\sigma ^2_\tau \) the number of levels matters most, as Example 11.2.4 showed.

Solution. The denominator is \(MS_{AB} + MS_{AC} - MS_{ABC} = 12.4 + 15.8 - 4.6 = 23.6\), with degrees of freedom \(6\), \(2\) and \(6\) for the three mean squares. So \[ F_0 = \frac {95.0}{23.6} = 4.03 , \qquad \hat \nu = \frac {23.6^2}{12.4^2/6 + 15.8^2/2 + 4.6^2/6} = 3.6 . \] Since \(F_{0.05, 2, 3.6} = 7.67\) (\(P = 0.12\)), \(H_0\) is not rejected. The alternative statistic gives the same conclusion: \[ F_0' = \frac {95.0 + 4.6}{12.4 + 15.8} = 3.53 , \] on \(2.2\) and \(5.3\) degrees of freedom, with \(P = 0.10\). Testing against \(MS_{AB}\) alone would give \(F_0 = 95.0/12.4 = 7.66 > F_{0.05, 2, 6} = 5.14\), a spurious rejection: it ignores the contribution \(bn\sigma ^2_{\tau \gamma }\) of the \(AC\) interaction to \(E(MS_A)\).