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: e.g., 2 4 1 3 5
Finally, randomize the order of the columns: e.g., 4 3 5 1 2
Linear Model
 Linear Model: Yij =  + i + j +k + ij
–
–
–
–
–
=
βi =
j =
k =
ij =
mean effect
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
s Y1 Y2  2 MSE 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 could use the following
formula:
Yˆ ij(k) =


t Ri +Cj +Tk 2- G
 t1- t2- 

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
Better approach – 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
MSR   t  1 MSE
RE 
t * MSE
 To compare with an RBD using rows as blocks
MSC   t  1 MSE
RE 
t * MSE
 To compare with a CRD
MSR  MSC   t  1 MSE
RE 
(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
SS
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**
12
36.80
3.07
Error
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
MSR   t  1 MSE
RE 
t * MSE
(21.85+(5-1)*3.07)/(5*3.07)=2.22
(2.22-1)*100 = 122% gain in
efficiency by adding rows
 To compare with an RBD using rows as blocks
MSC   t  1 MSE
RE 
t * MSE
(4.14+(5-1)*3.07)/(5*3.07)=1.07
(1.07-1)*100 = 7% gain in
efficiency by adding columns
 To compare with a CRD
(21.85+4.14+(5-1)*3.07)/(6*3.07)=2.08
MSR  MSC   t  1 MSE (2.08-1)*100 = 108% gain
RE 
(t  1) * MSE
CRD would require 2.08*5 or 11 reps