Two treatments in blocks

Topic: Analysis of variance · more questions on this topic
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In a randomised block design with t=2 treatments, the treatment F test and the paired t test on the within-block differences are equivalent. Show that the F statistic is the square of that paired t statistic, and that the level-\alpha critical regions agree.

Worked answer (click to show)

Let D_i=Y_{i1}-Y_{i2} for block i=1,\ldots,b, with mean \overline D and variance s_D^2. The paired statistic is T=\overline D/(s_D/\sqrt b), on b-1 degrees of freedom.

The treatment means are \overline Y_{\cdot1} and \overline Y_{\cdot2}, with \overline Y_{\cdot1}-\overline Y_{\cdot2}=\overline D and grand mean halfway between them. So

\mathrm{SS}_{\mathrm{treatment}}=b\Bigl[\bigl(\tfrac{\overline D}2\bigr)^2+\bigl(\tfrac{\overline D}2\bigr)^2\Bigr]=\frac{b\overline D^2}{2},

on t-1=1 degree of freedom. The error residual in block i for treatment 1 is Y_{i1}-\overline Y_{i\cdot}-\overline Y_{\cdot1}+\overline Y_{\cdot\cdot}=\tfrac12(D_i-\overline D), and for treatment 2 it is the negative of that. Hence

\mathrm{SS}_{\mathrm{error}}=\sum_{i=1}^b2\cdot\tfrac14(D_i-\overline D)^2=\frac{(b-1)s_D^2}{2},

on (b-1)(2-1)=b-1 degrees of freedom, so \mathrm{MS}_{\mathrm{error}}=s_D^2/2. Then

F=\frac{b\overline D^2/2}{s_D^2/2}=\frac{\overline D^2}{s_D^2/b}=T^2.

F is F_{1,b-1} and T is t_{b-1}, and F_{1,b-1;1-\alpha}=t_{b-1;1-\alpha/2}^2, so F\ge F_{1,b-1;1-\alpha} exactly when |T|\ge t_{b-1;1-\alpha/2}. The two level-\alpha tests reject on the same data. For the tyres, T=12.83 and F=164.6=12.83^2.