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 ofvariation | Degrees offreedom ^{a} | Sums ofsquares (SSQ) | Meansquare (MS) | F |

Rows (R) | r-1 | SSQ_{R} | SSQ_{R}/(r-1) | MS_{R}/MS_{E} |

Columns (C) | r-1 | SSQ_{C} | SSQ_{C}/(r-1) | MS_{C}/MS_{E} |

Treatments (Tr) | r-1 | SSQ_{Tr} | SSQ_{Tr}/(r-1) | MS_{Tr}/MS_{E} |

Error (E) | (r-1)(r-2) | SSQ_{E} | SSQ_{E}/((r-1)(r-2)) | |

Total (Tot) | r^{2}-1 | SSQ_{Tot} | ||

^{a}where 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);