1. 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

2. Advantages and Disadvantages
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
Another variation – use a smaller Latin Square (3x3 or 4x4), but repeat it. Analysis can then be combined across the two replicates.

3. 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.
Usually done this way for economic reasons (limited number of cows).
One potential problem is carryover from one treatment to the next on the same cow.

4. Are there any potential problems with this design?
The “Latin Square” Cow
A simple type of “crossover design”
Early
Mid
Late
This is a simple type of ‘crossover’ design, commonly used in human and animal subjects.
Are there any potential problems with this design?

5. 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 B C D A
‘Row’ 1 2 3 4
‘Column’
With four treatments, there would only be 6 df remaining for the error term
With five treatments, dfe = =12
Practically speaking, use only when you have more than four but fewer than ten treatments
a minimum of 12 df for error

6. Randomization First row in alphabetical order  A B C D E
Subsequent rows - shift letters one position
A B C D E C D E A B A B D C E
B C D E A A B C D E D E B A C
C D E A B D E A B C B C E D A
D E A B C B C D E A E A C B D
E A B C D E A B C D C D A E B
Randomize the order of the rows: ie
Finally, randomize the order of the columns: ie

7. Linear Model Linear Model: Yij =  + i + j +k + ij
 = mean effect
βi = ith block effect
j = jth column effect
k = kth treatment effect
ij = random error
Each treatment occurs once in each block and once in each column
r = c = t
N = t2
This is not a three-way factorial

8. 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

9. The ANOVA Source df SS MS F Total t2-1 SSTot =
Row t-1 SSR= SSR/(t-1) MSR/MSE
Column t-1 SSC= SSC/(t-1) MSC/MSE
Treatment t-1 SST = SST/(t-1) MST/MSE
Error (t-1)(t-2) SSE = SSE/(t-1)(t-2)
SSTot-SSR- SSC-SST

10. Means and Standard Errors
Standard Error of a treatment mean
Confidence interval estimate
Standard Error of a difference
Confidence interval estimate
t to test difference between
two means

11. Oh NO!!! not Missing Plots ^
If only one plot is missing, you can use the following formula: ^
Yij(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

12. 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
To compare with a CRD
RE = MSR + MSC + (t-1)MSE
(t+1)MSE

13. 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?

14. Data collection A B D C E 33.8 33.7 30.4 32.7 24.4 D E B A C
D E B A C
C D A E B
E A C B D
B C E D A

15. ANOVA of Dry Matter Yield
Source df SS MS F
Total
Rows **
Columns
Treatments **
Error
Rows = water
Columns = shade

16. Report of Statistical Analysis
I source A B C D None SE
Mean Yield
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

17. 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
To compare with a CRD
RE = MSR + MSC + (t-1)MSE
(t+1)MSE
(21.85+(5-1)3.07)/(5x3.07)=2.22
(2.22-1)*100 = 122% gain in efficiency by adding rows
Rows = water
Columns = shade
(4.14+(5-1)3.07)/(5x3.07)=1.07
(1.07-1)*100 = 7% gain in efficiency by adding columns
( (5-1)3.07)/(6x3.07)=2.08
(2.08-1)*100 = 108% gain
CRD would require 2.08*5 or 11 reps