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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 p-1 Columns p-1 Treatments p-1
Error
np2-3p+2
-------------------------------Total
np2-1

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 (9-1) df. In general we will have:
SS
df
--------------------------------------Total rows np-1 Error in the df.
Columns
p-1
Double
Treatments p-1 Check!
Error
(p-1)(np-1)
--------------------------------------Total
np2-1
Often the SS(total rows) is decomposed further:

SStotal-rows = SSrows-within-squares + SSbetween-squares

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 )
SStotal-rows=

 k l

p

SSrows-within-squares =

n


( y kn SSbetween-squares

=

( y kl  yl ) 2

l

 y ) 2
l

l

Often is SSbetween-squares called SSreplicates and
SSrows-within-squares is called SSrows

Thus the ANOVA table looks like this:
SS
df
--------------------------------------Rows
np-1
Replicates
n-1 Error in df.
Double Check
Columns
p-1
Treatments
p-1
Error
(p-1)(np-1)
--------------------------------------Total
np2-1

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
np-1
Columns
np-1
Replicates
n-1
Treatments
p-1
Error
np2-2(np-1)-n-p+1
--------------------------------------Total
np2-1

Repeated Measures, Cross-over, and
Carry-over 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 "Cross-over" or “Change-over" design.

The ANOVA table for this design is typically:
SS
df
--------------------------------------Subjects
np-1
Double Check
Order-Position p-1
d.f.! Error
Treatments
p-1
Error
(p-1)(np-1)
--------------------------------------Total
np2-1

Using independent Latin Squares in
Cross-over 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 100-yard dash times. Since each subject must run
3 100-yard dashes, order position may be important due to fatigue, and thus a cross-over 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(p-1)
Columns
n(p-1)
Replicates
n-1
Treatments
p-1
Error
(p-1)[n(p-1)-1]
--------------------------------------Total
np2-1

Designs for carry-over 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 Cross-over/Latin-square 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 carry-over 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 "Double-Crossover Design“ Note that this will work for an odd number of treatments whereas the previous example would not.

Graeco-Latin Squares
Latin Square allows us to balance out two blocking factors.
Graeco-Latin square designs allow for three!
In depicting Graeco-Latin 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 Graeco-Latin 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 Graeco-Latin Square design looks like:
SS
df
---------------------------------------Rows
(p-1)
Columns
(p-1)
Greek letters
(p-1)
Treatments
(p-1)
Error
(p-3)(p-1)
---------------------------------------Total p2-1 Graeco-Latin Squares have been proven to exist for all integer p  3 except p=6. Finding a Graeco-Latin Square
Design is not easy to do. It is a good idea to look them up in a book. Also Graeco-Latin 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 Graeco-Latin 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.

Hyper-Square Designs
This idea of Graeco-Latin Square can be extended to
Hyper-Square 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 | Ax By Cz Dw
2 | Bz Aw Dx Cy
3 | Cw Dz Ay Bx
4 | Dy Cx Bw Az

Although the assumption of ‘additivity’ is assumed for Graeco-Latin Squares and Hyper-Squares, 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!