4.4 Latin Square Design
In this design we group the experimental units simultaneously in two ways corresponding to two
mutually perpendicular direction as rows and columns.
Each row and each column is complete replication each treatment appears once in each row and each
column.
If there are \(k\) treatments, experimental area is divided into \(k\) rows and \(k\) columns results in \(k^2\) plots or
experimental units.
The treatments are then assigned at random to plots.
Example 4.8. Analyse the Latin square design below, taking \(\alpha =0.01\), where \(A\), \(B\), \(C\), \(D\) are four types of fertiliser used on a rice crop. The subscript on each letter is the yield recorded on that plot.
| Col 1 | Col 2 | Col 3 | Col 4 | |
| Row 1 | \(A\) 4 | \(B\) 3 | \(C\) 5 | \(D\) 3 |
| Row 2 | \(B\) 5 | \(A\) 4 | \(D\) 5 | \(C\) 2 |
| Row 3 | \(C\) 4 | \(D\) 2 | \(A\) 3 | \(B\) 3 |
| Row 4 | \(D\) 2 | \(C\) 1 | \(B\) 2 | \(A\) 2 |
Solution. First check the design really is a Latin square: each of \(A\), \(B\), \(C\), \(D\) must appear exactly once in every row and every column. It does. That is what allows both row and column variation to be removed from the error term.
Four sets of totals are needed. As in the randomised block design, the treatment totals must be gathered by letter, not by position – the letters move about the square, which is the whole point of it.
| 1 | 2 | 3 | 4 | Total | |
| Row totals | 15 | 16 | 12 | 7 | 50 |
| Column totals | 15 | 10 | 15 | 10 | 50 |
| Treatment | \(A\) | \(B\) | \(C\) | \(D\) | Total |
| Total | 13 | 13 | 12 | 12 | 50 |
All three sets total \(50\), which is the first check to make.
With \(k=4\), \(N=k^2=16\) and \(T=50\): \[CF=\frac {T^2}{N}=\frac {2500}{16}=156.25,\] \[SS_{\text {Total}}=\sum \sum x^2-CF=180-156.25=23.75,\] \[SS_{\text {Rows}}=\frac {15^2+16^2+12^2+7^2}{4}-CF=12.25,\] \[SS_{\text {Columns}}=\frac {15^2+10^2+15^2+10^2}{4}-CF=6.25,\] \[SS_{\text {Treatments}}=\frac {13^2+13^2+12^2+12^2}{4}-CF=0.25,\] \[SS_{\text {Error}}=23.75-12.25-6.25-0.25=5.00.\] Each total is divided by \(k=4\), the number of observations behind it.
| Variation due to | df | \(SS\) | \(MS\) | \(F\) |
| Rows | \(k-1=3\) | \(12.25\) | \(4.08\) | \(4.90\) |
| Columns | \(k-1=3\) | \(6.25\) | \(2.08\) | \(2.50\) |
| Treatments | \(k-1=3\) | \(0.25\) | \(0.08\) | \(0.10\) |
| Error | \((k-1)(k-2)=6\) | \(5.00\) | \(0.83\) | |
| Total | \(N-1=15\) | \(23.75\) |
The degrees of freedom add: \(3+3+3+6=15\). Note how few are left for error – only 6 from 16 observations, because three sets of three have been spent on rows, columns and treatments. This is the price of removing two sources of variation, and it is why Latin squares smaller than \(4\times 4\) are rarely worth running.
At \(\alpha =0.01\) the critical value is \(F_{3,6,\,0.01}=9.78\) in every case.
- Treatments: \(F=0.10\). Nowhere near. There is no evidence of any difference between the four fertilisers (\(P=0.96\)).
- Rows: \(F=4.90\), still short of \(9.78\) at the \(1\%\) level (\(P=0.047\)).
- Columns: \(F=2.50\) (\(P=0.16\)).
Remark 4.9. The treatment \(F\) of \(0.10\) is worth dwelling on. The fertiliser totals are \(13, 13, 12, 12\) – almost identical – while the row totals run from \(7\) to \(16\). Nearly all the variation in this field is down to where a plot sits, and almost none to what was put on it.
Notice also that the row effect, at \(P=0.047\), would have been significant at \(5\%\) though not at the \(1\%\) demanded here. That is the design earning its keep: had rows been ignored and a one-way analysis run, that \(12.25\) would have been left in the error term, inflating it and making the fertiliser comparison less sensitive still.
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