Example 12.3.1. A study was made of the effect of inoculation on the root yield of sugar beet planted at different in-row spacings. At each of three locations, one of two equal plots was chosen at random to be inoculated. Each plot was divided into four subplots, and the spacings \(4\), \(6\), \(12\) and \(18\) cm were randomised to them. Inoculation is a fixed factor, and location and spacing are to be treated as random. The yields were

Design. This is a split-plot design with locations as blocks: inoculation is the whole-plot factor, randomised to the two plots at each location, and spacing the subplot factor, randomised to the four subplots of each plot.

Model. With \(i\) for location (\(3\) levels, random), \(j\) for inoculation (\(2\), fixed) and \(k\) for spacing (\(4\), random), \[ y_{ijk} = \mu + L_i + \tau _j + (L\tau )_{ij} + S_k + (LS)_{ik} + (\tau S)_{jk} + \varepsilon _{ijk} , \] where \(L_i \sim N(0, \sigma ^2_L)\), \(S_k \sim N(0, \sigma ^2_S)\), the interactions are normal with variances \(\sigma ^2_{L\tau }\), \(\sigma ^2_{LS}\), \(\sigma ^2_{\tau S}\) (those with \(\tau \) restricted to sum to zero over \(j\)), \(\sum _j\tau _j = 0\), and \(\varepsilon _{ijk} \sim N(0, \sigma ^2)\) is the residual, which also contains the negligible three-factor interaction. \((L\tau )_{ij}\) is the whole-plot error.

Expected mean squares. By the restricted rule, with \(Q_\tau = \sum _j\tau _j^2\), \begin {align*} E(MS_L) &= \sigma ^2 + 2\sigma ^2_{LS} + 8\sigma ^2_L , \\ E(MS_\tau ) &= \sigma ^2 + 4\sigma ^2_{L\tau } + 3\sigma ^2_{\tau S} + 12Q_\tau , \\ E(MS_{L\tau }) &= \sigma ^2 + 4\sigma ^2_{L\tau } , \\ E(MS_S) &= \sigma ^2 + 2\sigma ^2_{LS} + 6\sigma ^2_S , \\ E(MS_{LS}) &= \sigma ^2 + 2\sigma ^2_{LS} , \\ E(MS_{\tau S}) &= \sigma ^2 + 3\sigma ^2_{\tau S} . \end {align*}

For instance \((L\tau )\) does not enter \(E(MS_L)\), because its extra subscript \(j\) belongs to a fixed factor, but \((LS)\) does, with coefficient \(2\), the number of levels of \(j\).

Analysis. With \(C = 426^2/24 = 7561.5\) and \(\sum y^2 = 7722\), \begin {align*} SS_L &= \frac {145^2 + 138^2 + 143^2}8 - C = 3.25 , \\ SS_\tau &= \frac {189^2 + 237^2}{12} - C = 96.0 , \\ SS_S &= \frac {106^2 + 114^2 + 110^2 + 96^2}6 - C = 29.83 , \end {align*}

and with the given \(SS_{L\tau } = 4.75\), \(SS_{LS} = 4.417\), \(SS_{\tau S} = 20\), \(SS_E = 2.25\) and \(SST = 160.5\):

Locations and spacings are tested against location \(\times \) spacing. For locations, \(F_0 = 2.21 < F_{0.05, 2, 6} = 5.14\): no significant variation between locations. For spacings, \(F_0 = 13.51 > F_{0.05, 3, 6} = 4.76\) (\(P = 0.004\)): spacing matters. The inoculation \(\times \) spacing interaction is also large (\(P = 0.002\)): the totals for inoculated plots are \(50\), \(53\), \(48\), \(38\) across the spacings, and for uninoculated plots \(56\), \(61\), \(62\), \(58\), so the loss of yield from inoculation grows with the spacing, from \(6\) at \(4\) cm to \(20\) at \(18\) cm (totals over the three locations).

No mean square has the expectation \(\sigma ^2 + 4\sigma ^2_{L\tau } + 3\sigma ^2_{\tau S}\) needed to test inoculation, so a quasi-\(F\) test is required: \begin {align*} F_0 &= \frac {MS_\tau }{MS_{L\tau } + MS_{\tau S} - MS_E} \\ &= \frac {96.0}{2.375 + 6.667 - 0.375} = \frac {96.0}{8.667} = 11.08 , \end {align*}

with \(\hat \nu = 8.667^2/(2.375^2/2 + 6.667^2/3 + 0.375^2/6) = 4.25\) denominator degrees of freedom. Since \(F_{0.05, 1, 4.25} = 7.36\) (\(P = 0.027\)), inoculation has a significant effect: on average it lowered yield from \(19.75\) to \(15.75\).