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More on Latin Square Design
Example: Latin Square
The effect of five different ingredients (A, B, C, D, E) on reaction time of a chemical process is being studied. Each batch of new material is only large enough to permit five runs to be made. Furthermore, each run requires approximately 1.5 hours, so only five runs can be made in one day. The experimenter decides to run the experiment as a Latin square so that day and batch effects can be systematically controlled. She obtains the data that follow.
Analyze the data from this experiment (use = 0.05) and draw conclusions.
Batch
1
2
3
4
5
1
A=8
C=11
B=4
D=6
E=4
2
B=7
E=2
A=9
C=8
D=2
Day
3
D=1
A=7
C=10
E=6
B=3
4
C=7
D=3
E=1
B=6
A=8
5
E=3
B=8
D=5
A=10
C=8
Example: ANOVA
Source DF
Catalyst 4
Batch
4
Day
4
Error
12
Total
24
SS
141.440
15.440
12.240
37.520
206.640
MS
35.360
3.860
3.060
3.127
F
11.31
1.23
0.98
P
0.000
0.348
0.455
The result identifies the ingredients as having a significant effect on reaction time.
Replicating The Latin Square Design
Suppose one Latin Square is not enough. That is we need a larger sample size to satisfy the precision requirements of the experiment. We can replicate the Latin Square. That is, run an additional (few) Latin Squares.
There are 3 cases to consider illustrated by a 3x3 chemical recipe experiment. Suppose we want a total of 27 runs.
Original Latin Square Design
Operator
1
2
3
1
Batch
A
B
C
2
B
C
A
3
C
A
B
Case 1a.
Suppose we can get all 27 runs with 3 batches and 3 operators.
(i.e. no physical restrictions that would disallow it)
operators
 1
2
3
1  AAA BBB CCC
Batch 2  BBB CCC AAA
3  CCC AAA BBB
Or can be drawn differently
Treatments
A
B
C

Batch 1 XXX


Oper1 Batch 2 
XXX

Batch 3 

XXX

Batch 1 
XXX

Oper2 Batch 2 

XXX
Batch 3 XXX



Batch 1 

XXX
Oper3 Batch 2 XXX


Batch 3 
XXX


Case 1a.
In this case the df would breakdown like the following, where n is the number of replicated Latin squares
SS df Rows p1 Columns p1 Treatments p1
Error
np23p+2
Total
np21
Case 1b.
If we run a multiple of p replicates we could do something like:
Which is the same as running these 3 Latin Squares:
Treatments
A
B
C

Batch 1  X  0  + 
Oper 1
Batch 2  +  X  0 
Batch 3  0  +  X 

Batch 1  +  X  0 
Oper 2
Batch 2  0  +  X 
Batch 3  X  0  + 

Batch 1  0  +  X 
Oper 3
Batch 2  X  0  + 
Batch 3  +  X  0 
Batch
operators
 1 2 3
1  A B C
2  B C A
3  C A B
X
Batch
1  B
2  C
3  A
C
A
B
A
B
C
0
Batch
1  C
2  A
3  B
A
B
C
B
C
A
+
This design has a nice balance to it. The analysis would be the same. It seems this is better than Case 1a
Case 2.
Suppose that each batch contains only enough raw material for 3 recipes. We would do this:
Operators
 1 2 3
1  A B C
Batch 2  B C A
3  C A B
 1 2 3
4  A B C
Batch 5  B C A
6  C A B
 1 2 3
7  A B C
Batch 8  B C A
9  C A B
Since we have 9 batches, SSbatches (or SS Total rows ) will have (91) df. In general we will have:
SS
df
Total rows np1 Error in the df.
Columns
p1
Double
Treatments p1 Check!
Error
(p1)(np1)
Total
np21
Often the SS(total rows) is decomposed further:
SStotalrows = SSrowswithinsquares + SSbetweensquares
Let: i index treatments (i=1,2,…,p) j index columns (j=1,2,…,p) k index rows within a square (k = 1, 2, p) l index squares (l = 1,...,n) (n is the number of replicates) p n
Then:
2
( y kl y )
SStotalrows=
k l
p
SSrowswithinsquares =
n
( y kn SSbetweensquares
=
( y kl yl ) 2
l
y ) 2
l
l
Often is SSbetweensquares called SSreplicates and
SSrowswithinsquares is called SSrows
Thus the ANOVA table looks like this:
SS
df
Rows
np1
Replicates
n1 Error in df.
Double Check
Columns
p1
Treatments
p1
Error
(p1)(np1)
Total
np21
Case 3.
In case 2, we needed p columns and np rows (or equivalently p rows and np columns). Sometimes we need np rows and np columns. For example, suppose that union rules prohibited operators from running more than three batches:
Batch
Batch
Batch
operators
 1 2 3
1  A B C
2  B C A
3  C A B operators  4 5 6
4  A B C
5  B C A
6  C A B operators  7 8 9
7  A B C
8  B C A
9  C A B
In this case the ANOVA table looks like this:
SS
df
Rows
np1
Columns
np1
Replicates
n1
Treatments
p1
Error
np22(np1)np+1
Total
np21
Repeated Measures, Crossover, and
Carryover Designs
These are popular designs for cases of testing on human subjects.
Example: You want to test three treatments (e.g. drugs, therapies, etc.)
You suspect that "subject" is an important factor to block on since each subject reacts differently and thus introduces significant error into the experiment. In the repeated measures design, we block on subjects and administer each treatment to each subject like this:
Treatment
A B C
X X X Subject 1
X X X Subject 2
X X X Subject 3
Since we take repeated measures on each subject we get the design name “repeated measures design”
What else is it called?
Since each subject receives each treatment one after the other, we might be worried that the order will affect the results, and thus try to balance the effects of order like this:
subjects
Order Position
1
2
3
1
A
B
C
2
B
C
A
3
C
A
B
4
A
B
C
5
B
C
A
6
C
A
B
.
.
This design is called a "Crossover" or “Changeover" design.
The ANOVA table for this design is typically:
SS
df
Subjects
np1
Double Check
OrderPosition p1
d.f.! Error
Treatments
p1
Error
(p1)(np1)
Total
np21
Using independent Latin Squares in
Crossover designs.
Suppose order position does not affect subjects equally. For example suppose we are testing 3 brands of running shoes. We want to see which shoe produces the faster 100yard dash times. Since each subject must run
3 100yard dashes, order position may be important due to fatigue, and thus a crossover design is warranted. Now suppose that the subjects consist of men and women. Suppose we suspect that fatigue affects men and women differently. We could run independent Latin squares for men and women order position like this:
Male subject
Female subject
 1 2 3
1  A B C
2  B C A
3  C A B order position
 1 2 3
1  A B C
2  B C A
3  C A B
In this case the ANOVA would look like this:
SS
df
Rows
n(p1)
Columns
n(p1)
Replicates
n1
Treatments
p1
Error
(p1)[n(p1)1]
Total
np21
Designs for carryover or residual effects.
Suppose we are testing drugs in a repeated measures fashion. Not only does order position count, we also expect that the previous drug will have some effect on the current one. For example we might expect residual traces of the previous drug remain in the blood stream. Here is one design for such a situation:
Order Position
 1 2 3 4
1  A B D C
Subject 2  B C A D
3  C D B A
4  D A C B
This is a Crossover/Latinsquare design, but it is a special one: each treatment (drug) follows each of the others exactly once. E.g. A follows B, C, and D exactly once each) Thus this design balances out the carryover effect as well!
Another way to do this is to run two Latin Squares in such a way that the order is reversed in the second.
Order Position
 1 2 3
1  B A C
Subject 2  A C B
3  C B A
 1 2 3
4  C A B
Subject 5  B C A
6  A B C
This is called a "DoubleCrossover Design“ Note that this will work for an odd number of treatments whereas the previous example would not.
GraecoLatin Squares
Latin Square allows us to balance out two blocking factors.
GraecoLatin square designs allow for three!
In depicting GraecoLatin squares, it is customary to use
Latin (A,B,C,D,..) and Greek (,,,,..) letters
Example:
I have 1 treatment variable: 4 chemical recipes (A,B,C,D)
I have 3 blocking variables :
4 Batches (1,2,3,4)
4 Operators (1,2,3,4)
4 Mixing tanks (,,,)
Based on the example, imagine the following design:
( a GraecoLatin Design)
Batch
Operator
 1
2
3
4
1  A B C D
2  B A D C
3  C D A B
4  D C B A
Note that each treatment appears exactly once: in each row, in each column, and with each greek (lower case) letter.
The ANOVA for a GraecoLatin Square design looks like:
SS
df
Rows
(p1)
Columns
(p1)
Greek letters
(p1)
Treatments
(p1)
Error
(p3)(p1)
Total p21 GraecoLatin Squares have been proven to exist for all integer p 3 except p=6. Finding a GraecoLatin Square
Design is not easy to do. It is a good idea to look them up in a book. Also GraecoLatin squares are assumed to be additive (no interaction between blocking variables).
Example: Graeco Square
The yield of a chemical process was measured using five batches of raw material, five acid concentrations, five standing times, (A, B, C, D, E) and five catalyst concentrations ( , , , , ). The GraecoLatin square that follows was used. Analyze the data from this experiment (use = 0.05) and draw conclusions.
Batch
1
2
3
4
5
1
A=26
B=18
C=20
D=15
E=10
2
B=16
C=21
D=12
E=15
A=24
Acid
3
C=19
D=18
E=16
A=22
B=17
Concentration
4
D=16
E=11
A=25
B=14
C=17
5
E=13
A=21
B=13
C=17
D=14
Example: Graeco Square
Calculate means:
Grand Mean=17.2
Batch Level Mean
1
2
3
4
5
18
17.8
17.2
16.6
16.4
Acid Concentration Level Mean
1
2
3
4
5
17.8
17.6
18.4
16.6
15.6
16.4
17.8
Catalyst Level Mean
18.6
17.0
18.2
Standing Time Level Mean (Treatment Level Mean)
A
B
C
D
E
23.6
15.6
18.8
15.0
13.0
Example: Graeco Square
Source DF
Time
4
Catalyst 4
Batch
4
Acid
4
Error
8
Total
24
SS
342.800
12.000
10.000
24.400
46.800
436.000
MS
F
85.700 14.65
3.000 0.51
2.500 0.43
6.100 1.04
5.850
P
0.001
0.729
0.785
0.443
The result identifies standing time as having a significant effect on yield.
HyperSquare Designs
This idea of GraecoLatin Square can be extended to
HyperSquare designs.
In the following example we have:
4 treatments levels (A,B,C,D)
4 levels of Blocking factor 1 (rows 1,2,3,4)
4 levels of Blocking factor 2 (colmns1,2,3,4)
4 levels of Blocking factor 3 (,,,)
4 levels of Blocking factor 4 (x,y,z,w)
 1
2
3
4
1  Ax By Cz Dw
2  Bz Aw Dx Cy
3  Cw Dz Ay Bx
4  Dy Cx Bw Az
Although the assumption of ‘additivity’ is assumed for GraecoLatin Squares and HyperSquares, it should be pointed out that as the number of blocking factors increases, the Hyper Square
Design may be unstable. As more blocking factors are added, the more numerous the potential interactions, and the less reasonable the assumption of no interaction.
As an aside, Hyper Squares are useful for scheduling things like Bridge Tournaments and gourmet clubs.
Imagine a bridge tournament with sixteen players
"named" 1,2,3,4,a,b,c,d,A,B,C,D,x,y,z,w.
The tournament will be player in 4 rounds. There will be four games per round. Every player will face each opponent exactly once in the tournament.
1  Aax1 Bby2 Ccz3 Ddw4
Round 2  Bdz1 Acw2 Dbx3 Cay4
3  Cbw1 Daz2 Ady3 Bcx4
4  Dcy1 Cdx2 Baw3 Abz4
Cool!