Solution. The first thing to get right is which totals matter. The blocks are the rows \(R_1, R_2, R_3\); the treatments are the varieties \(A\) to \(E\). Because the varieties are randomised within each block, they appear in a different order in each row – so the total of a column position is not the total of a treatment, and means nothing at all. The treatment totals have to be gathered by label.

Collected by variety:

There are \(r=3\) blocks and \(c=5\) treatments, so \(N=15\) and \(T=60\). \[CF=\frac {T^2}{N}=\frac {3600}{15}=240,\] \[SS_{\text {Total}}=\sum \sum x_{ij}^2-CF=254-240=14,\] \[SS_{\text {Blocks}}=\frac {1}{c}\sum _i B_i^2-CF=\frac {20^2+22^2+18^2}{5}-240 =\frac {1208}{5}-240=1.60,\] \[SS_{\text {Treatments}}=\frac {1}{r}\sum _j \tau _j^2-CF =\frac {14^2+13^2+13^2+11^2+9^2}{3}-240=\frac {736}{3}-240=5.33,\] \[SS_{\text {Error}}=14-1.60-5.33=7.07.\] Each total is divided by the number of observations that went into it: block totals by \(c=5\), treatment totals by \(r=3\).

The degrees of freedom add: \(2+4+8=14\). This is the check that catches the commonest mistake in a two-way table, and it is worth making every time.

At \(\alpha =0.01\):

• Treatments. \(F=1.51\) against \(F_{4,8,\,0.01}=7.01\). We fail to reject: there is no evidence of a difference between the varieties (\(P=0.29\)).
• Blocks. \(F=0.91\) against \(F_{2,8,\,0.01}=8.65\). We fail to reject: there is no evidence of a difference between blocks (\(P=0.44\)).