LATIN SQUARE DESIGN (LS)
Facts about the LS Design
-With the Latin Square design you are able to control variation in two directions.
-Treatments are arranged in rows and columns
-Each row contains every treatment.
-Each column contains every treatment.
-The most common sizes of LS are 5x5 to 8x8
Advantages of the LS Design
1. You can control variation in two directions.
2. Hopefully you increase efficiency as compared to the RCBD.
Disadvantages of the LS Design
1. The number of treatments must equal the number of replicates.
2. The experimental error is likely to increase with the size of the square.
3. Small squares have very few degrees of freedom for experimental error.
4. You can’t evaluate interactions between:
a. Rows and columns
b. Rows and treatments
c. Columns and treatments.
Effect of the Size of the Square on Error Degrees of Freedom
SOV Df 2x2 3x3 4x4 5x5 8x8
Rows r-1 1 2 3 4 7
Columns r-1 1 2 3 4 7
Treatments r-1 1 2 3 4 7
Error (r-1)(r-2) 0 2 6 12 42
Total r2 - 1 3 8 15 24 63
Where r = number of rows, columns, and treatments.
-One way to increase the Error df for small squares is to use more than one square in the
experiment (i.e. repeated squares).
1
Example
Two 4x4 Latin squares.
SOV Df
Squares sq – 1 = 1
* Row(square) sq(r-1) = 6
* Column(square) sq(r-1) = 6
Treatment r-1 = 3
Square x Treatment (sq-1)(r-1) = 3
* Error sq(r-1)(r-2) = 12
Total sqr2 – 1 = 31
*Additive across squares.
Where sq = number of squares.
Examples of Uses of the Latin Square Design
1. Field trials in which the experimental error has two fertility gradients running
perpendicular each other or has a unidirectional fertility gradient but also has residual
effects from previous trials.
Gradient 2
Gradient 1
2. Animal science feed trials.
3. Insecticide field trial where the insect migration has a predictable direction that is
perpendicular to the dominant fertility gradient of the experimental field.
4. Greenhouse trials in which the experimental pots are arranged in a straight line
perpendicular to the glass walls, such that the difference among rows of pots and
distace from the glass wall are expected to be the major sources of variability.
A D C B B C A D D A B C C B D A
2
Randomization Procedure
-Depends on the type of Latin Square you use.
3x3 Latin Square
-Start with the standard square and randomize all columns and all but the first row.
1 2 3 3 1 2
1 A B C C A B
Randomize columns
2 B C A A B C
3 C A B B C A
Standard square
Randomize all but the first row
C A B
B C A
A B C
4x4 Latin Square
-Randomly choose a standard square.
-Randomize all columns and all but the first row.
5x5 Latin Square
-Randomly choose a standard square.
-Randomize all columns and rows.
3
Analysis of a Single Latin Square
Example
Grain yield of three maize hybrids (A, B, and D) and a check (C).
Row Column 1 Column 2 Column 3 Column 4 Row (∑ R )
1 1.640 (B) 1.210 (D) 1.425 (C) 1.345 (A) 5.620
2 1.475 (C) 1.185 (A) 1.400 (D) 1.290 (B) 5.350
3 1.670 (A) 0.710 (C) 1.665 (B) 1.180 (D) 5.225
4 1.565 (D) 1.290 (B) 1.655 (A) 0.660 (C) 5.170
Column total (∑ C ) 6.350 4.395 6.145 4.475 21.365
Step 1. Calculate treatment totals.
Treatment Total
A 5.855
B 5.885
C 4.270
D 5.355
Step 2. Compute the Correction Factor (CF).
Y..2
CF =
r2
21.365 2
=
42
= 28.53
Step 3. Calculate the Total SS
TotalSS = ∑ Yij2 − CF
= (1.64 2 + 1.210 2 + 1.425 2 + ... + 0.66 2 ) − CF
= 1.4139
4
Step 4. Calculate the Row SS
RowSS =
∑ Row 2
− CF
r
(5.62 2 + 5.35 2 + 5.225 2 + 5.17 2 )
= − CF
4
= 0.0302
Step 5. Calculate the Column SS.
Col.SS =
∑ Col 2
− CF
r
(6.35 2 + 4.395 2 + 6.145 2 + 4.475 2 )
= − CF
4
= 0.8273
Step 6. Calculate the Treatment SS
TrtSS =
∑Y 2
i.
− CF
r
(5.855 2 + 5.885 2 + 4.270 2 + 5.355 2 )
= − CF
4
= 0.4268
Step 7. Calculate the Error SS
Error SS = Total SS – Row SS – Column SS – Trt SS
= 0.1296
5
Step 8. Complete the ANOVA table
SOV Df SS MS F
Row r-1 = 3 0.030
Column r-1 = 3 0.827
Trt r-1 = 3 0.427 0.142 Trt MS/Error MS = 6.60*
Error (r-1)(r-2) = 6 0.129 0.0215
Total r2-1 = 15 1.414
Step 9. Calculate the LSD.
2 ErrorMS
LSD = tα
2 r
2(.0215)
= 2.447
4
= 0.254
Linear Model
Yij (t ) = µ + β i + κ j + τ t + ε ij (t )
where: µ = the experiment mean.
β i = the row effect,
κ j = the column effect,
τ t = the treatment effect, and
ε ij (t ) = the random error.
