Latin squares design in R

The Latin square design is used where the researcher desires to control the variation in an experiment that is related to rows and columns in the field.
Remember that:
* Treatments are assigned at random within rows and columns, with each treatment once per row and once per column.
* There are equal numbers of rows, columns, and treatments.
* Useful where the experimenter desires to control variation in two different directions

The formula used for this kind of three-way ANOVA are:

Source of variation	Degrees of freedoma	Sums of squares (SSQ)	Mean square (MS)	F
Rows (R)	r-1	SSQR	SSQR/(r-1)	MSR/MSE
Columns (C)	r-1	SSQC	SSQC/(r-1)	MSC/MSE
Treatments (Tr)	r-1	SSQTr	SSQTr/(r-1)	MSTr/MSE
Error (E)	(r-1)(r-2)	SSQE	SSQE/((r-1)(r-2))
Total (Tot)	r2-1	SSQTot
awhere r = number of (treatments=rows=columns).

Suppose you want to analyse the productivity of 5 kind on fertilizer, 5 kind of tillage, and 5 kind of seed. The data are organized in a latin square design, as follow:

             treatA  treatB  treatC  treatD  treatE
fertilizer1  "A42"   "C47"   "B55"   "D51"   "E44"         
fertilizer2  "E45"   "B54"   "C52"   "A44"   "D50"         
fertilizer3  "C41"   "A46"   "D57"   "E47"   "B48"         
fertilizer4  "B56"   "D52"   "E49"   "C50"   "A43"         
fertilizer5  "D47"   "E49"   "A45"   "B54"   "C46"  

The three factors are: fertilizer (fertilizer1:5), tillage (treatA:E), seed (A:E). The numbers are the productivity in cwt / year.

Now create a dataframe in R with these data:

fertil <- c(rep("fertil1",1), rep("fertil2",1), rep("fertil3",1), rep("fertil4",1), rep("fertil5",1))
treat <- c(rep("treatA",5), rep("treatB",5), rep("treatC",5), rep("treatD",5), rep("treatE",5))
seed <- c("A","E","C","B","D", "C","B","A","D","E", "B","C","D","E","A", "D","A","E","C","B", "E","D","B","A","C")
freq <- c(42,45,41,56,47, 47,54,46,52,49, 55,52,57,49,45, 51,44,47,50,54, 44,50,48,43,46)
 
mydata <- data.frame(treat, fertil, seed, freq)

mydata

    treat  fertil seed freq
1  treatA fertil1    A   42
2  treatA fertil2    E   45
3  treatA fertil3    C   41
4  treatA fertil4    B   56
5  treatA fertil5    D   47
6  treatB fertil1    C   47
7  treatB fertil2    B   54
8  treatB fertil3    A   46
9  treatB fertil4    D   52
10 treatB fertil5    E   49
11 treatC fertil1    B   55
12 treatC fertil2    C   52
13 treatC fertil3    D   57
14 treatC fertil4    E   49
15 treatC fertil5    A   45
16 treatD fertil1    D   51
17 treatD fertil2    A   44
18 treatD fertil3    E   47
19 treatD fertil4    C   50
20 treatD fertil5    B   54
21 treatE fertil1    E   44
22 treatE fertil2    D   50
23 treatE fertil3    B   48
24 treatE fertil4    A   43
25 treatE fertil5    C   46

We can re-create the original table, using the matrix function:

matrix(mydata$seed, 5,5)

     [,1] [,2] [,3] [,4] [,5]
[1,] "A"  "C"  "B"  "D"  "E" 
[2,] "E"  "B"  "C"  "A"  "D" 
[3,] "C"  "A"  "D"  "E"  "B" 
[4,] "B"  "D"  "E"  "C"  "A" 
[5,] "D"  "E"  "A"  "B"  "C" 

matrix(mydata$freq, 5,5)

     [,1] [,2] [,3] [,4] [,5]
[1,]   42   47   55   51   44
[2,]   45   54   52   44   50
[3,]   41   46   57   47   48
[4,]   56   52   49   50   43
[5,]   47   49   45   54   46

Before proceeding with the analysis of variance of this Latin square design, you should perform a Boxplot, aimed to have an idea of what we expect:

par(mfrow=c(2,2))
plot(freq ~ fertil+treat+seed, mydata)

Note that the differences considering the fertilizer is low; it is medium considering the tillage, and is very high considering the seed.
Now confirm these graphics observations, with the ANOVA table:

myfit <- lm(freq ~ fertil+treat+seed, mydata)
anova(myfit)

Analysis of Variance Table

Response: freq
          Df  Sum Sq Mean Sq F value   Pr(>F)    
fertil     4  17.760   4.440  0.7967 0.549839    
treat      4 109.360  27.340  4.9055 0.014105 *  
seed       4 286.160  71.540 12.8361 0.000271 ***
Residuals 12  66.880   5.573                     
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 

Well, the boxplot was useful. Look at the significance of the F-test.
- The difference between group considering the fertilizer is not significant (p-value > 0.1);
- The difference between group considering the tillage is quite significant (p-value < 0.05);
- The difference between group considering the seed is very significant (p-value < 0.001);