Latin Square Analysis of Variance
Menu location: Analysis_Analysis of Variance_Latin.
This function calculates ANOVA for a special three factor design known as Latin squares.
The Latin square design applies when there are repeated exposures/treatments and two other factors. This design avoids the excessive numbers required for full three way ANOVA.
An example of a Latin square design is the response of 5 different rats (factor 1) to 5 different treatments (repeated blocks A to E) when housed in 5 different types of cage (factor 2):
Rat | ||||||
1 | 2 | 3 | 4 | 5 | ||
1 | A | E | C | D | A | |
2 | E | B | A | B | C | |
Cage | 3 | C | D | E | D | D |
4 | D | C | B | C | B | |
5 | B | A | D | A | E |
This special sort of balancing means that the systematic variation between rows, or similarity between columns, does not affect the comparison of treatments.
The Latin square is probably under used in most fields of research because text book examples tend to be restricted to agriculture, the area which spawned most original work on ANOVA. Agricultural examples often reflect geographical designs where rows and columns are literally two dimensions of a grid in a field. Rows and columns can be any two sources of variation in an experiment. In this sense a Latin square is a generalisation of a randomized block design with two different blocking systems. Armitage and Berry (1994) discuss medical applications of this method. Further details are given by Cochran and Cox (1957).
The variance ratio test statistics given by StatsDirect for this design are valid only when an additive model applies (Armitage and Berry, 1994).
Automatic column comparisons are not given here. If you want to make linear contrasts between row or column means then you can use the residual mean square of the Latin square as the variance estimate. This estimate is not reliable if the additive model does not apply.
Technical Validation
The Latin square ANOVA for three factors without interaction is calculated as follows (Armitage and Berry, 1994; Cochran and Cox, 1957):
- where X_{ijk} is the observation from the ith row of the jth column with the kth treatment, G is the grand total of all observations, R_{i} is the total for the ith row, C_{j} is the total for the jth column, T_{k} is the total for the kth treatment, SS_{total} is the total sum of squares, SS_{rows} is the sum of squares due to the rows, SS_{columns} is the sum of squares due to the columns, SS_{treatments} is the sum of squares due to the treatments and a is the number of rows, columns or treatments.
Example
From Armitage and Berry (1994, p. 236).
Test workbook (ANOVA worksheet: Observations; Rabbit; Position; Order).
Armitage quotes a paper which reported an experiment that had been designed as a Latin square. The skins of rabbits' backs were inoculated with a diffusing factor in six separate sites. Six rabbits were therefore used and the order in which the sites were inoculated was done six different ways. The outcome measured was area of blister (cm²). The overall objective was to see whether or not the order of administration affected this outcome. The experimental design and data are represented in the Latin square below.
Rabbit | |||||||
1 | 2 | 3 | 4 | 5 | 6 | ||
a | iii | v | iv | i | vi | ii | |
7.9 | 8.7 | 7.4 | 7.4 | 7.1 | 8.2 | ||
b | iv | ii | vi | v | iii | i | |
6.1 | 8.2 | 7.7 | 7.1 | 8.1 | 5.9 | ||
c | i | iii | v | vi | ii | iv | |
7.5 | 8.1 | 6 | 6.4 | 6.2 | 7.5 | ||
Position | |||||||
d | vi | i | iii | ii | iv | v | |
6.9 | 8.5 | 6.8 | 7.7 | 8.5 | 8.5 | ||
e | ii | iv | i | iii | v | vi | |
6.7 | 9.9 | 7.3 | 6.4 | 6.4 | 7.3 | ||
f | v | vi | ii | iv | i | iii | |
7.3 | 8.3 | 7.3 | 5.8 | 6.4 | 7.7 |
To analyse these data in StatsDirect you must first enter them in the workbook with a separate column for the observations, the column classifier (factor 1), the row classifier (factor 2) and the treatment/Latin/randomised classifier (factor 3):
Observation | Rabbit | Position | Order |
7.9 | 1 | 1 | 3 |
6.1 | 1 | 2 | 4 |
7.5 | 1 | 3 | 1 |
6.9 | 1 | 4 | 6 |
6.7 | 1 | 5 | 2 |
7.3 | 1 | 6 | 5 |
8.7 | 2 | 1 | 5 |
8.2 | 2 | 2 | 2 |
8.1 | 2 | 3 | 3 |
8.5 | 2 | 4 | 1 |
9.9 | 2 | 5 | 4 |
8.3 | 2 | 6 | 6 |
7.4 | 3 | 1 | 4 |
7.7 | 3 | 2 | 6 |
6 | 3 | 3 | 5 |
6.8 | 3 | 4 | 3 |
7.3 | 3 | 5 | 1 |
7.3 | 3 | 6 | 2 |
7.4 | 4 | 1 | 1 |
7.1 | 4 | 2 | 5 |
6.4 | 4 | 3 | 6 |
7.7 | 4 | 4 | 2 |
6.4 | 4 | 5 | 3 |
5.8 | 4 | 6 | 4 |
7.1 | 5 | 1 | 6 |
8.1 | 5 | 2 | 3 |
6.2 | 5 | 3 | 2 |
8.5 | 5 | 4 | 4 |
6.4 | 5 | 5 | 5 |
6.4 | 5 | 6 | 1 |
8.2 | 6 | 1 | 2 |
5.9 | 6 | 2 | 1 |
7.5 | 6 | 3 | 4 |
8.5 | 6 | 4 | 5 |
7.3 | 6 | 5 | 6 |
7.7 | 6 | 6 | 3 |
Alternatively, open the test workbook using the file open function of the file menu. Then select Latin square from the analysis of variance section of the analysis menu. First select the observations data, then the column, row and treatment classifiers respectively.
For this example:
Latin square test
Factors: Rabbit, Position, Order.
Source of Variation | Sum Squares | DF | Mean Square |
Rows | 3.833333 | 5 | 0.766667 |
Columns | 12.833333 | 5 | 2.566667 |
Treatments | 0.563333 | 5 | 0.112667 |
Residual | 13.13 | 20 | 0.6565 |
Total | 30.36 | 35 |
F (rows) = 1.167809, P = .3592
F (columns) = 3.909622, P = .0124
F (treatments) = 0.171617, P = .9701
Here we see that the order of administration does not have a statistically significant effect on blistering but that inter-rabbit variation was significant.