Latin Square Design
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Transcript Latin Square Design
Latin Square Design
If you can block on two (perpendicular) sources of
variation (rows x columns) you can reduce experimental
error when compared to the RBD
More restrictive than the RBD
The total number of plots is the square of the number of
treatments
Each treatment appears once and only once in each row
and column
A B C D
B C D
A
C D A B
D A B C
Advantages and Disadvantages
Advantage:
– Allows the experimenter to control two sources of
variation
Disadvantages:
– The experiment becomes very large if the number of
treatments is large
– The statistical analysis is complicated by missing
plots and misassigned treatments
– Error df is small if there are only a few treatments
• This limitation can be overcome by repeating a small
Latin Square and then combining the experiments – a 3x3 Latin Square repeated 4 times
– a 4x4 Latin Square repeated 2 times
Useful in Animal Nutrition Studies
Suppose you had four feeds you wanted to test on dairy
cows. The feeds would be tested over time during the
lactation period
This experiment would require 4 animals (think of these
as the rows)
There would be 4 feeding periods at even intervals
during the lactation period beginning early in lactation
(these would be the columns)
The treatments would be the four feeds. Each animal
receives each treatment one time only.
The “Latin Square” Cow
A simple type of “crossover design”
Early
Mid
Early
Mid
Late
Late
Are there any potential problems with this design?
Uses in Field Experiments
When two sources of variation must be controlled
– Slope and fertility
– Furrow irrigation and shading
– If you must plant your plots perpendicular to a linear gradient
‘Row’
1
2
3
4
A B C D B C D A C D A B D A B C
1
2
3
4
‘Column’
Practically speaking, use only when you have more than
four but fewer than ten treatments
– a minimum of 12 df for error
Randomization
First row in alphabetical order
ABCDE
Subsequent rows - shift letters one position
4 3 51 2
ABCDE 2
CDEAB
ABDCE
BCDEA 4
ABCDE
DEBAC
CDEAB 1
DEABC
BCEDA
DEABC 3
BCDEA
EACBD
EABCD 5
EABCD
CDAEB
Randomize the order of the rows: ie 2 4 1 3 5
Finally, randomize the order of the columns: ie 4 3 5 1 2
Linear Model
Linear Model: Yij = + i + j +k + ij
– =
mean effect
–
–
–
–
βi =
j =
k =
ij =
ith block effect
jth column effect
kth treatment effect
random error
Each treatment occurs once in each block and once in
each column
– r=c=t
– N = t2
Analysis
Set up a two-way table and compute the row and
column means and deviations
Compute a table of treatment means and deviations
Set up an ANOVA table divided into sources of
variation
–
–
–
–
Rows
Columns
Treatments
Error
Significance tests
– FT tests difference among treatment means
– FR and FC test if row and column groupings are
effective
The ANOVA
Source
df
SS
Total
t2-1
SSTot =
MS
F
2
SSR/(t-1)
MSR/MSE
2
SSC/(t-1)
MSC/MSE
SST/(t-1)
MST/MSE
i j Yij Y
Row
t-1
t Y Y
t Y Y
SSR=
t i Yi Y
Column
t-1
SSC=
j
Treatment t-1
Error
j
SST =
k
2
2
k
(t-1)(t-2) SSE =
SSE/(t-1)(t-2)
SSTot-SSR- SSC-SST
Means and Standard Errors
sY MSE r
Standard Error of a treatment mean
Confidence interval estimate
Y t MSE r
Standard Error of a difference
sY1 Y2 2MSE r
Confidence interval estimate
t to test difference between
two means
Y Y t
1
2
2MSE / r
Y1 Y 2
t
2MSE / r
Oh NO!!! not Missing Plots
If only one plot is missing, you can use the following
formula:
^
Y
ij(k)
= t(Ri + Cj + Tk)-2G
[(t-1)(t-2)]
Where:
• Ri = sum of remaining observations in the ith row
• Cj = sum of remaining observations in the jth column
• Tk = sum of remaining observations in the kth treatment
• G = grand total of the available observations
• t = number of treatments
Total and error df must be reduced by 1
Alternatively – use software such as SAS Procedures GLM,
MIXED, or GLIMMIX that adjust for missing values
Relative Efficiency
To compare with an RBD using columns as blocks
RE = MSR + (t-1)MSE
tMSE
To compare with an RBD using rows as blocks
RE = MSC + (t-1)MSE
tMSE
To compare with a CRD
RE = MSR + MSC + (t-1)MSE
(t+1)MSE
Numerical Examples
To determine the effect of four different sources of seed
inoculum, A, B, C, and D, and a control, E, on the dry matter yield
of irrigated alfalfa. The plots were furrow irrigated and there was
a line of trees that might form a shading gradient.
A
B
D
C
E
D
E
B
A
C
C
D
A
E
B
E
A
C
B
D
B
C
E
D
A
Are the blocks for shading represented by the row or columns?
Is the gradient due to irrigation accounted for by rows or columns?
Data collection
A
33.8
B
33.7
D
30.4
C
32.7
E
24.4
D
37.0
E
28.8
B
33.5
A
34.6
C
33.4
C
35.8
D
35.6
A
36.9
E
26.7
B
35.1
E
33.2
A
37.1
C
37.4
B
38.1
D
34.1
B
34.8
C
39.1
E
32.7
D
37.4
A
36.4
ANOVA of Dry Matter Yield
Source
df
Total
24
296.66
Rows
4
87.40
21.85
Columns
4
16.56
4.14
1.35
Treatments 4
155.89
38.97
12.71**
36.80
3.07
Error
12
SS
MS
F
7.13**
Report of Statistical Analysis
I source
Mean Yield
A
B
C
D
None
SE
35.8
35.0
35.7
34.9
29.2
0.78
a
a
a
a
b
LSD=2.4
Differences among treatment means were highly
significant
No difference among inocula. However, inoculation,
regardless of source produced more dry matter than did
no inoculation
Blocking by irrigation effect was useful in reducing
experimental error
Distance from shade did not appear to have a significant
effect
Relative Efficiency
To compare with an RBD using columns as blocks
RE = MSR + (t-1)MSE
tMSE
(21.85+(5-1)3.07)/(5x3.07)=2.22
(2.22-1)*100 = 122% gain in
efficiency by adding rows
To compare with an RBD using rows as blocks
RE = MSC + (t-1)MSE
tMSE
(4.14+(5-1)3.07)/(5x3.07)=1.07
(1.07-1)*100 = 7% gain in
efficiency by adding columns
To compare with a CRD
RE = MSR + MSC + (t-1)MSE
(t+1)MSE
(21.85+4.14+(5-1)3.07)/(6x3.07)=2.08
(2.08-1)*100 = 108% gain
CRD would require 2.08*5 or 11 reps