Introduction
to
- Read More: Estimation of Missing Value in LS Design
- Read More: Relative Efficiency Pros and Cons of LSD
- Read More: Estimation of Parameters in LSD
- Read More: Expectation in LSD
- Read More: Graeco-Latin Square Design
lecture - 14
What is a Latin square?
A Latin square is a table filled with p X p different alphabets or symbols in such a way that each alphabet or symbol appears once and only once in each row and exactly once in each column. Here are a few examples.
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a |
b |
|
b |
a |
|
a |
b |
c |
|
b |
c |
a |
|
c |
a |
b |
|
a |
b |
c |
d |
|
b |
c |
d |
a |
|
C |
d |
a |
b |
|
d |
a |
b |
c |
A Latin square in which the treatments in the first
row and in the first column are arranged in alphabetical or numerical order.
|
a |
b |
c |
|
b |
c |
a |
|
c |
a |
b |
From a standard Latin square of order p, p! (p - 1)! Different Latin squares can be obtained by making p! permutations of columns and (p - 1)! Permutations of rows, which leaves the first row in place.
If two Latin
squares of the same order but with different symbols are such that when they
are superimposed on each other, every ordered pair of symbols (different)
occurs exactly once in the Latin square, then they are called orthogonal.
Introduction to Latin Square Design
(LS Design)
A randomised complete block design is used to control
for extraneous sources of variation in the experiment. If there are two sources
of variation in the experimental material, there is a chance that the treatments
will be influenced by two nuisance factors. The Latin square design is suitable
for controlling extraneous variations by blocking the experimental material in
two contrast directions.
In Latin square design, the experimental material is
divided into rows and columns, each having the same number of experimental
units, which is equal to the number of treatments. The treatments are assigned
at random to the rows and columns in such a way that each treatment appears once
and only once in each row and column.
If there are P treatments, the experimental material
is divided into P rows and P columns, resulting in
Suppose we have four treatments, i.e., A, B, C and D.
The standard Latin square is given below:
|
a |
b |
c |
d |
|
b |
c |
d |
a |
|
c |
d |
a |
b |
|
d |
a |
b |
c |
Example:
Suppose a
researcher wants to check different brands of petrol consumption per kilometre. He takes 3 different brands of petrol, represented by A, B, and C.
The factors affecting the consumption of petrol are:
·
Driver
·
Brand of car
The researcher controls these two sources of
extraneous variation by LS design as:
|
Brand of Car |
Driver – 1 |
Driver – 2 |
Driver – 3 |
|
1 |
A |
B |
C |
|
2 |
B |
C |
A |
|
3 |
C |
A |
B |
· * 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 to control variation in two different directions.
Experimental
Layout out of LS Design
The Latin square design is used to control two sources
of variability in the experimental units. So, both rows and columns are blocking factors. It removes the variation from the measured response variable in
both directions.
Let us have 4 treatments, i.e., A, B, C, and D (P =
4). We required experimental units.
Step – 1: Develop a standard Latin square or select any Latin square derived from a standard Latin square.
|
|
Column
1 |
Column
2 |
Column
3 |
Column
4 |
|
Row
1 |
A |
B |
C |
D |
|
Row
2 |
B |
C |
D |
A |
|
Row
3 |
C |
D |
A |
B |
|
Row
4 |
D |
A |
B |
C |
Step 2: Row randomisation:
|
Random
Number |
0.788 |
0.943 |
0.051 |
0.811 |
|
Rank
(order) |
2 |
4 |
1 |
3 |
|
Row |
1 |
2 |
3 |
4 |
Then one should rearrange the order of the rows as
follows:
|
|
Column
1 |
Column
2 |
Column
3 |
Column
4 |
|
Row
2 |
B |
C |
D |
A |
|
Row
4 |
D |
A |
B |
C |
|
Row
1 |
A |
B |
C |
D |
|
Row
3 |
C |
D |
A |
B |
Step – 3: Column randomisation
|
Random
Number |
0.134 |
0.655 |
0.210 |
0.165 |
|
Rank
(order) |
1 |
4 |
3 |
2 |
|
Column
|
1 |
2 |
3 |
4 |
Then one should rearrange the order of the columns as
follows:
|
|
Column
1 |
Column
4 |
Column
3 |
Column
2 |
|
Row
2 |
B |
A |
D |
C |
|
Row
4 |
D |
C |
B |
A |
|
Row
1 |
A |
D |
C |
B |
|
Row
2 |
C |
B |
A |
D |
Step – 4: Treatment randomization
|
Random
Number |
0.994 |
0.176 |
0.441 |
0.227 |
|
Treatment
|
|
|
|
|
|
Rank
(order) |
A |
B |
C |
D |
Assign A as treatment 4, B as treatment 1, C as
treatment 3 and D as treatment 2:
|
|
Column
1 |
Column
4 |
Column
3 |
Column
2 |
|
Row
2 |
T1 |
T4 |
T2 |
T3 |
|
Row
4 |
T2 |
T3 |
T1 |
T4 |
|
Row
1 |
T3 |
T2 |
T3 |
T1 |
|
Row
2 |
T4 |
T1 |
T4 |
T2 |
Statistical
Model and Analysis of LS Design
The Latin square design is represented by the
following linear statistical model.
Where:
Assumptions of the Latin Square Model:
Analysis
of Latin Square Design:
Let Yijk
|
SV |
df |
SS |
MS |
F |
|
Row |
P
– 1 |
SSR |
MSR |
F1 |
|
Column |
P
– 1 |
SSC |
MSC |
F2 |
|
Treatment |
P
– 1 |
SST |
MST |
F3 |
|
Error |
(P – 1) (P – 2) |
SSE |
MSE |
|
|
Total |
P^2
– 1 |
SSTotal |
|
|
Example:
An
experiment to investigate the effects of various dietary starch levels on milk
production was conducted on four cows. The four diets, T1, T2, T3, and T4 (in order of increasing starch equivalent), were fed for three weeks to each cow,
and the total yield of milk in the third week of each period was recorded (i.e., the third week to minimise carry-over effects due to the use of treatments
administered in a previous period). That is, the trial lasted 12 weeks since
each cow received each treatment, and each treatment required three weeks. The
investigator felt strongly that time period effects might be important (i.e.,
earlier periods in the experiment might influence milk yields differently
compared to later periods). Hence, the investigator wanted to block on both cow
and period. However, each cow cannot possibly receive more than one treatment
during the same time period; that is, all possible cow-period blocking
combinations could not logically be considered.
Solution: State the null and alternative hypotheses:
Express treatment in table:
ANOVA Table:
vi. Remarks: The calculated F value falls in
the acceptance region. Thus, we do not have sufficient evidence to reject all
three null hypotheses.
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