Solution. Machines become fixed, with \(\sum \tau _i = 0\), and operators (nested, random) give the middle column of Theorem 12.1.2: \[ E(MS_{\mathrm {Machines}}) = \sigma ^2 + 2\sigma ^2_\beta + \frac {6\sum \tau _i^2}3 . \] The tests are numerically unchanged: machines against operators (\(F_0 = 3.42\), \(P = 0.073\)) and operators against error (\(F_0 = 4.17\)). But the machine hypothesis is now \(\tau _1 = \dots = \tau _4 = 0\), about these four machines, and instead of \(\sigma ^2_\tau \) one estimates the effects \(\hat \tau _i = \bar y_{i\cdot \cdot } - \bar y_{\cdot \cdot \cdot }\): \[ \hat \tau _1 = 68.67 - 65.83 = 2.83 , \quad \hat \tau _2 = 17.33 , \quad \hat \tau _3 = -3.33 , \quad \hat \tau _4 = -16.83 . \]

Solution. The design is nested, with samples within batches and determinations within samples. The model is \[ y_{ijk} = \mu + \tau _i + \beta _{j(i)} + \varepsilon _{(ij)k} , \] with batches \(\tau _i \sim N(0, \sigma ^2_\tau )\), samples \(\beta _{j(i)} \sim N(0, \sigma ^2_\beta )\) and determination errors \(\varepsilon \sim N(0, \sigma ^2)\); \(a = 5\), \(b = 2\), \(n = 2\), and the degrees of freedom are \(4\), \(5\) and \(10\). The expected mean squares are \(\sigma ^2 + 2\sigma ^2_\beta + 4\sigma ^2_\tau \), \(\sigma ^2 + 2\sigma ^2_\beta \) and \(\sigma ^2\).

Samples: \(F_0 = 6.2/1.4 = 4.43 > F_{0.05, 5, 10} = 3.33\) (\(P = 0.022\)), significant. Batches: \(F_0 = 24.0/6.2 = 3.87 < F_{0.05, 4, 5} = 5.19\) (\(P = 0.085\)), not significant. The estimates are \[ \hat \sigma ^2 = 1.4 , \qquad \hat \sigma ^2_\beta = \frac {6.2 - 1.4}2 = 2.4 , \qquad \hat \sigma ^2_\tau = \frac {24.0 - 6.2}4 = 4.45 . \] As with the machines, the largest component is not significant, because only five batches were sampled.

Solution. V2 against V1 under I3 is a comparison of subplot levels at one whole-plot level: the difference is \(7.475 - 6.375 = 1.10\) with standard error \(0.134\) on \(18\) degrees of freedom, so \(t = 8.2\), far beyond \(t_{0.025, 18} = 2.10\).

I3 against I1 for V2 compares whole-plot levels at one subplot level: the difference is \(7.475 - 5.45 = 2.03\) with standard error \(0.366\). The degrees of freedom are \[ \hat \nu = \frac {\bigl (2(0.0358) + 0.7305\bigr )^2}{\bigl (2(0.0358)\bigr )^2/18 + 0.7305^2/6} = 7.2 , \] and \(t = 5.5 > t_{0.025, 7.2} = 2.35\). Both differences are significant, but the second, three times the size of its standard error rather than eight, is judged against the much larger whole-plot variation.

Solution. With location random and inoculation and spacing fixed, the restricted rule gives \begin {align*} E(MS_L) &= \sigma ^2 + 8\sigma ^2_L , & E(MS_S) &= \sigma ^2 + 2\sigma ^2_{LS} + 6Q_S , \\ E(MS_\tau ) &= \sigma ^2 + 4\sigma ^2_{L\tau } + 12Q_\tau , & E(MS_{LS}) &= \sigma ^2 + 2\sigma ^2_{LS} , \\ E(MS_{L\tau }) &= \sigma ^2 + 4\sigma ^2_{L\tau } , & E(MS_{\tau S}) &= \sigma ^2 + 3Q_{\tau S} . \end {align*}

Inoculation now has an exact test against its whole-plot error: \(F_0 = 96.0/2.375 = 40.4 > F_{0.05, 1, 2} = 18.51\) (\(P = 0.024\)). Spacing is still tested against location \(\times \) spacing (\(F_0 = 13.51\)), and inoculation \(\times \) spacing against the residual (\(F_0 = 17.78\)). Location is tested against the residual: \(F_0 = 1.625/0.375 = 4.33 < F_{0.05, 2, 6} = 5.14\). The conclusions are the same, but the inoculation test has only \(2\) denominator degrees of freedom, a reminder that three locations give little replication of the whole-plot factor.

Solution. (a) A split-plot design in blocks. Each shift is a block. Within a shift, the oven run at one temperature is a whole plot, so temperature is the whole-plot factor; the three specimens in a run are the subplots, and time is the subplot factor. Temperature is changed only once per shift.

(b) With \(i\) for shift (\(4\), random), \(j\) for temperature (\(2\), random) and \(k\) for time (\(3\), fixed), \[ y_{ijk} = \mu + S_i + T_j + (ST)_{ij} + \gamma _k + (S\gamma )_{ik} + (T\gamma )_{jk} + \varepsilon _{ijk} , \] with \(S_i\), \(T_j\), \((ST)_{ij}\), \((S\gamma )_{ik}\), \((T\gamma )_{jk}\) normal with mean zero and variances \(\sigma ^2_S, \dots , \sigma ^2_{T\gamma }\) (those involving time restricted to sum to zero over \(k\)), \(\sum _k\gamma _k = 0\), and \(\varepsilon _{ijk} \sim N(0, \sigma ^2)\), all independent. \((ST)_{ij}\) is the whole-plot error. By the restricted rule, with \(Q_\gamma = \sum _k\gamma _k^2/2\), \begin {align*} E(MS_S) &= \sigma ^2 + 3\sigma ^2_{ST} + 6\sigma ^2_S , & E(MS_T) &= \sigma ^2 + 3\sigma ^2_{ST} + 12\sigma ^2_T , \\ E(MS_{ST}) &= \sigma ^2 + 3\sigma ^2_{ST} , & E(MS_\gamma ) &= \sigma ^2 + 2\sigma ^2_{S\gamma } + 4\sigma ^2_{T\gamma } + 8Q_\gamma , \\ E(MS_{S\gamma }) &= \sigma ^2 + 2\sigma ^2_{S\gamma } , & E(MS_{T\gamma }) &= \sigma ^2 + 4\sigma ^2_{T\gamma } , \end {align*}

and \(E(MSE) = \sigma ^2\).

(c) The degrees of freedom are \(3\) (shift), \(1\) (temperature), \(3\) (shift \(\times \) temperature), \(2\) (time), \(6\) (shift \(\times \) time), \(2\) (temperature \(\times \) time) and \(6\) (error), \(23\) in all. With \(MSE = 244.417/6 = 40.74\):

• shift \(\times \) temperature: \(MS = 80.15\), \(F_0 = 1.97 < F_{0.05, 3, 6} = 4.76\);
• shift \(\times \) time: \(MS = 79.74\), \(F_0 = 1.96 < F_{0.05, 6, 6} = 4.28\);
• temperature \(\times \) time: \(MS = 397.63\), \(F_0 = 9.76 > F_{0.05, 2, 6} = 5.14\) (\(P = 0.013\)).

The effect of time depends on the temperature. The remaining \(2645.083\) is the total of the three main-effect sums of squares, which the information given does not separate. Shift and temperature are tested against shift \(\times \) temperature, and time needs a quasi-\(F\) test, \[ F_0 = \frac {MS_\gamma }{MS_{S\gamma } + MS_{T\gamma } - MSE} , \] since no single mean square has expectation \(\sigma ^2 + 2\sigma ^2_{S\gamma } + 4\sigma ^2_{T\gamma }\).