Solution. Replication: each treatment is applied independently to several experimental units; it provides an estimate of the experimental error, without which treatment differences cannot be judged. Randomisation is the use of a chance mechanism to allocate treatments to units and to order the runs; it makes the errors behave as independent random variables and protects against bias from uncontrolled factors. Blocking is the grouping of similar units into blocks within which the treatments are compared; it removes the effect of a known nuisance factor from the comparisons and so increases their precision.

Solution. Irrigation is applied to whole fields, so the field is the experimental unit for irrigation, and each irrigation level has only two replicates, however many plots are measured. Fertiliser is applied to plots, so the plot is the experimental unit for fertiliser. The two factors have different experimental units and so different experimental errors; this is a split-plot experiment, analysed in Section 12. Comparing irrigation levels against the plot-to-plot variation would be pseudo-replication.

Solution. Here \(\bar u = 8\) and \(S^2 = (9 + 1 + 4 + 16)/3 = 10\), so the theorem predicts mean \(0\) and variance \(2S^2/n = 10\). The \(\binom 42 = 6\) allocations give

Each value is equally likely. The mean is \(0\) and the variance is \(2(4 + 25 + 1)/6 = 10\), as predicted.

Solution. The completely randomised design gives \(\Var (\bar y_A - \bar y_B) = 2(\sigma _\beta ^2 + \sigma ^2)/10 = 2(4)/10 = 0.8\). The paired design with \(n\) cars gives \(\Var (\bar d) = 2\sigma ^2/n = 2/n\), which is at most \(0.8\) when \(n \ge 2.5\). Three cars suffice, against twenty. In practice more would be used, to estimate \(\sigma ^2\) with reasonable precision and to check the assumptions, but the comparison shows how much blocking can save.

Solution. The effect of \(A\) is \(30 - 20 = 10\) at \(B^-\) and \(12 - 40 = -28\) at \(B^+\), so \(A = \frac 12(10 - 28) = -9\) and \(AB = \frac 12(-28 - 10) = -19\). The effect of \(B\) is \(40 - 20 = 20\) at \(A^-\) and \(12 - 30 = -18\) at \(A^+\), so \(B = \frac 12(20 - 18) = 1\). The main effects are small and the interaction is large: each factor’s effect reverses with the level of the other.

The OFAT experiment finds \(30 > 20\) and moves to \(A^+\); at \(A^+\) it finds \(12 < 30\) and keeps \(B^-\). It recommends \((A^+, B^-)\) with response \(30\), and misses \((A^-, B^+)\) with \(40\), a combination it never runs.

Solution. OFAT gives variance \(2\sigma ^2/m\), so \(m = 4\) runs at each of the \(k + 1 = 4\) conditions: \(16\) runs. A balanced factorial gives \(4\sigma ^2/N\), so \(N = 8\): one replicate of the \(2^3 = 8\) combinations. The factorial uses half the runs, in agreement with Proposition 1.5.4, and it estimates the three two-factor interactions as well.