5.4 Relative efficiency
How much did blocking gain? The question is how large the error variance would have been if the same units had been used in a completely randomised design. Section 1 answers it: under complete randomisation, the variance that matters is the variance \(S^2\) of the units about their overall mean, and that can be estimated from the RCBD mean squares.
Proposition 5.4.1. Under the additive model with fixed block effects, \[ \hat \sigma ^2_{\mathrm {CR}} = \frac {(b - 1)MS_{\mathrm {Blocks}} + b(t - 1)MSE}{tb - 1} \] is unbiased for \(\sigma ^2 + t\sum _j\beta _j^2/(tb - 1)\), the expected variance of the \(tb\) units about their overall mean in the absence of treatment effects. The relative efficiency of the RCBD to the completely randomised design is estimated by \[ RE = \frac {\hat \sigma ^2_{\mathrm {CR}}}{MSE} . \]
Proof. By Theorem 5.2.3, \begin {align*} E\bigl [(b - 1)MS_{\mathrm {Blocks}} + b(t - 1)MSE\bigr ] &= (b - 1)\sigma ^2 + t\sum _j\beta _j^2 + b(t - 1)\sigma ^2 \\ &= (tb - 1)\sigma ^2 + t\sum _j\beta _j^2 . \end {align*}
On the other hand, without treatment effects the unit values are \(\beta _j + \varepsilon _{ij}\), and the expected value of their sample variance over the \(tb\) units is \(\sigma ^2 + \sum _i\sum _j\beta _j^2/(tb - 1) = \sigma ^2 + t\sum _j\beta _j^2/(tb - 1)\), since the \(\beta _j\) sum to zero. □
\(RE\) is the number of replicates a completely randomised design would need for each replicate of the RCBD to achieve the same precision. Because the RCBD has fewer error degrees of freedom, \(f_1 = (t - 1)(b - 1)\) against \(f_2 = t(b - 1)\), Fisher suggested multiplying \(RE\) by \[ \frac {(f_1 + 1)(f_2 + 3)}{(f_2 + 1)(f_1 + 3)} , \] a factor slightly below \(1\) that accounts for the extra uncertainty in a variance estimated on fewer degrees of freedom.
Example 5.4.2. For the fertiliser data, \[ \hat \sigma ^2_{\mathrm {CR}} = \frac {4(20.675) + 15(0.808)}{19} = 4.99 , \qquad RE = \frac {4.99}{0.808} = 6.17 . \] With \(f_1 = 12\) and \(f_2 = 16\) the correction factor is \((13)(19)/\bigl ((17)(15)\bigr ) = 0.969\), giving \(5.98\). A completely randomised experiment would have needed about six times as many plots, around \(120\) instead of \(20\), for the same precision.
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