Transcript analysis of variance and experimental design

Completely
Randomized Design
Completely Randomized Design
1. Experimental Units (Subjects) Are Assigned
Randomly to Treatments
• Subjects are Assumed Homogeneous
2. One Factor or Independent Variable
• 2 or More Treatment Levels or Classifications
3. Analyzed by One-Way ANOVA
Randomized Design Example
Factor levels
(Treatments)
Experimental
units
Factor (Training Method)
Level 3
Level 2
Level 1
  
Dependent
variable
21 hrs.
17 hrs.
31 hrs.
27 hrs.
25 hrs.
28 hrs.
(Response)
29 hrs.
20 hrs.
22 hrs.
The Linier Model
X =μ + t +ε
ij
i = 1,2,…, t
i
ij
j = 1,2,…, r
Xij = the observation in ith treatment and the jth
replication
m = overall mean
t = the effect of the ith treatment
i
eij = random error
One-Way ANOVA F-Test
1. Tests the Equality of 2 or More (t) Population Means
2. Variables
• One Nominal Scaled Independent Variable
 2 or More (t) Treatment Levels or Classifications
• One Interval or Ratio Scaled Dependent Variable
3. Used to Analyze Completely Randomized
Experimental Designs
Assumptions
1. Randomness & Independence of Errors
• Independent Random Samples are Drawn for each
condition
2. Normality
• Populations (for each condition) are Normally Distributed
3. Homogeneity of Variance
• Populations (for each condition) have Equal Variances
Hypotheses
H0: m1 = m2 = m3 = ... = mt

• All Population Means are Equal
• No Treatment Effect
Ha: Not All mi Are Equal

• At Least 1 Pop. Mean is Different
• Treatment Effect
NOT m1  m2  ...  mt
Hypotheses


H0: m1 = m2 = m3 = ... = mt
• All Population Means are
Equal
• No Treatment Effect
Ha: Not All mi Are Equal
• At Least 1 Pop. Mean is
Different
• Treatment Effect
NOT m1  m2  ...  mt
f(X)
m1 = m2 = m3
X
f(X)
m1 = m 2 m 3
X
One-Way ANOVA
Basic Idea
1. Compares 2 Types of Variation to Test
Equality of Means
2. Comparison Basis Is Ratio of Variances
3. If Treatment Variation Is Significantly Greater
Than Random Variation then Means Are Not
Equal
4. Variation Measures Are Obtained by
‘Partitioning’ Total Variation
One-Way ANOVA
Partitions Total Variation
One-Way ANOVA
Partitions Total Variation
Total variation
One-Way ANOVA
Partitions Total Variation
Total variation
Variation due to
treatment
One-Way ANOVA
Partitions Total Variation
Total variation
Variation due to
treatment
Variation due to
random sampling
One-Way ANOVA
Partitions Total Variation
Total variation
Variation due to
treatment




Sum of Squares Among
Sum of Squares Between
Sum of Squares Treatment
Among Groups Variation
Variation due to
random sampling
One-Way ANOVA
Partitions Total Variation
Total variation
Variation due to
treatment




Sum of Squares Among
Sum of Squares Between
Sum of Squares
Treatment (SST)
Among Groups Variation
Variation due to
random sampling



Sum of Squares Within
Sum of Squares Error
(SSE)
Within Groups Variation
Total Variation
SS Total   X 11  X   X 21  X     X ij  X 
2
2
2
Response, X
X
Group 1
Group 2
Group 3
Treatment Variation






2
2
2
SST  n X  X  n X  X    n X  X
1 1
2 2
t t
Response, X
X3
X
X1
Group 1
X2
Group 2
Group 3
Random (Error) Variation
SSE  X11  X1   X 21  X1     X tj  X t 
2
2
2
Response, X
X3
X1
Group 1
Group 2
X2
Group 3
SStotal=SSE+SST
SS    X ij  X .. 
n1 ni
i 1 j 1
n1 ni
  
X ij  X i.    X i.  X .. 
2
i 1 j 1
ni
n1 ni
n
1
2
2
   X ij  X i.     X i.  X .. 
i 1 j 1
i 1 j 1


 2   X ij  X i.  X i.  X .. 
n1 ni
i 1 j 1
But
  X ij  X i.  X i.  X .. 
n1 ni
i 1 j 1
   X i.  X ..   X ij  X i. 
n1
ni
i 1
j 1
n1
   X i.  X .. n X i.  n X i. 
i 1
0
Thus, SStotal=SSE+SST
SS    X ij  X i.      X i.  X .. 
n1 ni
2
n1 ni
2
i 1 j 1
i 1 j 1
   X ij  X i.    ni  X i.  X .. 
n1 ni
i 1 j 1
 SSE  SST
2
n1
i 1
2
One-Way ANOVA F-Test
Test Statistic
1. Test Statistic
• F = MST / MSE


MST Is Mean Square for Treatment
MSE Is Mean Square for Error
2. Degrees of Freedom
 1 = t -1
 2 = tr - t
 t = # Populations, Groups, or Levels
 tr = Total Sample Size
One-Way ANOVA
Summary Table
Source of Degrees Sum of
Variation
of
Squares
Freedom
Mean
Square
(Variance)
F
Treatment
t-1
SST
MST =
SST/(t - 1)
MST
MSE
Error
tr - t
SSE
MSE =
SSE/(tr - t)
Total
tr - 1
SS(Total) =
SST+SSE
ANOVA Table for a
Completely Randomized Design
Source of
Variation
Treatments
Sum of
Squares
SST
Degrees of
Freedom
Mean
Squares
t-1
SST/t-1
SSE/tr-t
Error
SSE
tr - t
Total
SSTot
tr - 1
F
MST/MSE
The F distribution

Two parameters
• increasing either one decreases F-alpha (except for
v2<3)
• I.e., the distribution gets smashed to the left

0
F ( v1 , v2 )

F
One-Way ANOVA F-Test Critical
Value
If means are equal, F =
MST / MSE  1. Only
reject large F!
Reject H0

Do Not
Reject H 0
0
F ( t 1, tr
F
-t)
Always One-Tail!
© 1984-1994 T/Maker Co.
Example: Home Products, Inc.

Completely Randomized Design
Home Products, Inc. is considering marketing a long-lasting
car wax. Three different waxes (Type 1, Type 2, and Type 3)
have been developed.
In order to test the durability of these waxes, 5 new cars
were waxed with Type 1, 5 with Type 2, and 5 with Type 3.
Each car was then repeatedly run through an automatic
carwash until the wax coating showed signs of deterioration.
The number of times each car went through the carwash is
shown on the next slide.
Home Products, Inc. must decide which wax to market. Are
the three waxes equally effective?
Example: Home Products, Inc.
Wax
Observation
1
2
3
4
5
Sample Mean
Sample Variance
Type 1
27
30
29
28
31
29.0
2.5
Wax
Type 2
Wax
Type 3
33
28
31
30
30
29
28
30
32
31
30.4
3.3
30.0
2.5
Example: Home Products, Inc.

Hypotheses
H0: m1 = m2 = m3
Ha: Not all the means are equal
where:
m1 = mean number of washes for Type 1 wax
m2 = mean number of washes for Type 2 wax
m3 = mean number of washes for Type 3 wax
Example: Home Products, Inc.

Mean Square Between Treatments
Since the sample sizes are all equal:
_
_
_
= (x + x + x )/3 = (29 + 30.4 + 30)/3 = 29.8
μ=
1
2
3
SSTR= 5(29–29.8)2+ 5(30.4–29.8)2+ 5(30–29.8)2= 5.2

MSTR = 5.2/(3 - 1) = 2.6
Mean Square Error
SSE = 4(2.5) + 4(3.3) + 4(2.5) = 33.2
MSE = 33.2/(15 - 3) = 2.77
Example: Home Products, Inc.

Rejection Rule
Using test statistic:
Using p-value:
Reject H0 if F > 3.89
Reject H0 if p-value < .05
where F.05 = 3.89 is based on an F distribution with 2
numerator degrees of freedom and 12 denominator degrees
of freedom
Example: Home Products, Inc.

Test Statistic
F = MST/MSE = 2.6/2.77 = .939

Conclusion
Since F = .939 < F.05 = 3.89, we cannot reject H0.
There is insufficient evidence to conclude that the
mean number of washes for the three wax types are
not all the same.

Example: Home Products, Inc.
ANOVA Table
Source of
Variation
Treatments
Error
Total
Sum of Degrees of
Squares Freedom Squares
5.2
33.2
38.4
2
12
14
Mean
F
2.60
2.77
.9398
Using Excel’s ANOVA:
Single Factor Tool

Value Worksheet (top portion)
A
1
2
3
4
5
6
7
Observation
1
2
3
4
5
B
Wax
Type 1
27
30
29
28
31
C
Wax
Type 2
33
28
31
30
30
D
Wax
Type 3
29
28
30
32
31
E
Using Excel’s ANOVA:
Single Factor Tool

8
9
10
11
12
13
14
15
16
17
18
19
20
21
A
B
C
D
E
F
Value
Worksheet
(bottom
portion)
Anova: Single Factor
SUMMARY
Groups
Wax Type 1
Wax Type 2
Wax Type 3
ANOVA
Source of Variation
Between Groups
Within Groups
Total
Count
5
5
5
SS
5.2
33.2
38.4
G
Sum Average Variance
145
29
2.5
152
30.4
3.3
150
30
2.5
df
MS
F
P-value F crit
2
2.6 0.939759 0.41768 3.88529
12 2.76667
14
Using Excel’s ANOVA:
Single Factor Tool

Conclusion Using the p-Value
• The value worksheet shows a p-value of .418
• The rejection rule is “Reject H0 if p-value < .05”
• Because .418 > .05, we cannot reject H0. There is
insufficient evidence to conclude that the mean
number of washes for the three wax types are not all
the same.
RCBD
(Randomized Complete Block Design)
Randomized Complete Block
Design
 An experimental design in which there is one independent
variable, and a second variable known as a blocking
variable, that is used to control for confounding or
concomitant variables.
 It is used when the experimental unit or material are
heterogeneous
 There is a way to block the experimental units or materials
to keep the variability among within a block as small as
possible and to maximize differences among block
 The block (group) should consists units or materials which
are as uniform as possible
Randomized Complete Block Design
 Confounding or concomitant variable are not being
controlled by the analyst but can have an effect on the
outcome of the treatment being studied
 Blocking variable is a variable that the analyst wants
to control but is not the treatment variable of interest.
 Repeated measures design is a randomized block
design in which each block level is an individual item or
person, and that person or item is measured across all
treatments.
The Blocking Principle
 Blocking is a technique for dealing with nuisance factors
 A nuisance factor is a factor that probably has some
effect on the response, but it is of no interest to the
experimenter…however, the variability it transmits to the
response needs to be minimized
 Typical nuisance factors include batches of raw material,
operators, pieces of test equipment, time (shifts, days,
etc.), different experimental units
 Many industrial experiments involve blocking (or should)
 Failure to block is a common flaw in designing an
experiment
The Blocking Principle
 If the nuisance variable is known and controllable, we
use blocking
 If the nuisance factor is known and uncontrollable,
sometimes we can use the analysis of covariance to
statistically remove the effect of the nuisance factor from
the analysis
 If the nuisance factor is unknown and uncontrollable
(a “lurking” variable), we hope that randomization
balances out its impact across the experiment
 Sometimes several sources of variability are combined
in a block, so the block becomes an aggregate variable
Partitioning the Total Sum of Squares
in the Randomized Block Design
SStotal
(total sum of squares)
SSE
(error sum of squares)
SST
(treatment
sum of squares)
SSB
(sum of squares
blocks)
SSE’
(sum of squares
error)
A Randomized Block Design
Single Independent Variable
.
.
Blocking
Variable
MST
MSE
Individual
observations
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
The Linier Model
y  μ  τ ρ ε
ij
i
j ij
i = 1,2,…, t
j = 1,2,…,r
yij = the observation in ith treatment in the jth block
m = overall mean
ti = the effect of the ith treatment
rj = the effect of the jth block
eij = random error
No interaction
between blocks
and treatments
Extension of the ANOVA to the RCBD
ANOVA partitioning of total variability:
t
r
t
r

 (yij  y..)  (yi.  y..)  (y.j  y..)  (yij  yi.  y.j  y..)
2
i 1 j1

2
i 1 j1
t
r
t
r
 r  (yi.  y..)  t  (y.j  y..)  (yij  yi.  y.j  y..)2
2
i 1
2
j1
i 1 j1
SST  SSTreatments  SSBlocks  SSE
Extension of the ANOVA to the RCBD
The degrees of freedom for the sums of squares in
SST  SSTreatments  SSBlocks  SSE
are as follows:
tr  1  (t  1)  (r  1)  [(t  1)(r  1)]


Ratios of sums of squares to their degrees of freedom result
in mean squares, and
The ratio of the mean square for treatments to the error
mean square is an F statistic  used to test the hypothesis of
equal treatment means
ANOVA Procedure


The ANOVA procedure for the randomized block design
requires us to partition the sum of squares total (SST) into
three groups: sum of squares due to treatments, sum of
squares due to blocks, and sum of squares due to error.
The formula for this partitioning is
SSTot = SSTreatment + SSBlock + SSE

The total degrees of freedom, tr - 1, are partitioned such that
t - 1 degrees of freedom go to treatments, r - 1 go to blocks,
and (t - 1)(r - 1) go to the error term.
ANOVA Table for a
Randomized Block Design
Source of
Variation
Sum of
Squares
Degrees of
Freedom
Treatments
SST
t–1
Blocks
SSB
r-1
Error
SSE
(t - 1)(r - 1)
Total
SSTot
tr - 1
Mean
Squares
SST/t-1
SSE/(t-1)(r-1)
F
MST/MSE
Example: Eastern Oil Co.
Randomized Block Design
Eastern Oil has developed three new blends of
gasoline and must decide which blend or blends to
produce and distribute. A study of the miles per
gallon ratings of the three blends is being conducted
to determine if the mean ratings are the same for the
three blends.
Five automobiles have been tested using each of
the three gasoline blends and the miles per gallon
ratings are shown on the next slide.
Example: Eastern Oil Co.
Automobile
(Block)
1
2
3
4
5
Treatment
Means
Type of Gasoline (Treatment)
Blend X
Blend Y
Blend Z
32
31
30
29
28
27
30
30
27
32
31
30
27
25
26
30
29
28
Blocks
Means
31
28
29
31
26
29
Example: Eastern Oil Co.

Mean Square Due to Treatments
The overall sample mean is 29. Thus,
SSTreatment = 5[(30-29)2+ (29-29)2+ (28-29)2]= 10.00
MSTreatment = 10/(3 - 1) = 5

Mean Square Due to Blocks
SSBlock = 3[(31-29)2 + . . . + (26-29)2] = 54
MSBlock= 54/(5 - 1) = 13.5

Mean Square Due to Error
CF = (435)2 /(3x5) =12615
SStotal = (32)2+ …+ (26)2 – 12615 = 62
SSE = 68 - 10 – 54 = 4
MSE = 4/[(3 - 1)(5 - 1)] = .5
Example: Eastern Oil Co.

Rejection Rule
Using test statistic:
Reject H0 if F > 4.46
Assuming α = .05, F.05 = 4.46 (2 d.f. numerator
and 8 d.f. denominator)
Example: Eastern Oil Co.


Test Statistic
F = MSTreatment/MSE = 5/.5 = 10
Conclusion
Since 10.00 > 4.46, we reject H0. There is sufficient
evidence to conclude that the miles per gallon ratings differ
for the three gasoline blends.
Using Excel’s Anova



Step 1 Select the Tools pull-down menu
Step 2 Choose the Data Analysis option
Step 3 Choose Anova: Two Factor Without
Replication from the list of Analysis Tools
Using Excel’s Anova

Step 4 When the Anova: Two Factor Without
Replication dialog box appears:
Enter A1:D6 in the Input Range box
Select Labels
Enter .05 in the Alpha box
Select Output Range
Enter A8 (your choice) in the Output Range box
Click OK
Using Excel’s Anova:
Two-Factor Without Replication Tool

Value Worksheet (top portion)
1
2
3
4
5
6
7
Automobile Blend X Blend Y Blend Z
1
32
31
30
2
29
28
27
3
30
30
27
4
32
31
30
5
27
25
26
Using Excel’s Anova:
Two-Factor Without Replication Tool
Value Worksheet
(middle portion)
1
3
93
10
 11
12
13
14
15
16
17
18
19
20
21
22
SUMMARY
Count
2
3
4
5
Blend X
Blend Y
Blend Z
Sum
3
3
3
3
84
87
93
78
Average
31
28
29
31
26
5
5
5
150
145
140
30
29
28
Using Excel’s Anova
A
8
9
10
11
12
13
14
ANOVA
Variation
Rows
Columns (Treatments)
Error
Total
B
C
SS
df
D
54
10
4
4
2
8
68
14
MS
13.5
5
0.5
E
G
Fcalc
F table
26 3.83785
10 4.45897
Using Excel’s Anova:
Two-Factor Without Replication Tool

Conclusion Using F table
• The value worksheet shows that F table for column
is 4.459
• The rejection rule is “Reject H0 if F calculated > F
Table”
• Thus, we reject H0 because F calculated > F Table
for  = .05
• There is sufficient evidence to conclude that the
miles per gallon ratings differ for the three gasoline
blends
Similarities and differences between
CRD and RCBD: Procedures




RCBD: Every level of “treatment” encountered by each
experimental unit; CRD: Just one level each
Descriptive statistics and graphical display: the same as
CRD
Model adequacy checking procedure: the same except:
specifically, NO Block x Treatment Interaction
ANOVA: Inclusion of the Block effect; dferror change from
t(r – 1) to (t – 1)(r – 1)
Latin Square Design
Definition


A Latin square is a square array of objects (letters A, B,
C, …) such that each object appears once and only
once in each row and each column.
Example - 4 x 4 Latin Square.
ABCD
BCDA
CDAB
DABC
The Latin Square Design







This design is used to simultaneously control (or eliminate) two
sources of nuisance variability (confounding variables)
It is called “Latin” because we usually specify the treatment by the
Latin letters
“Square” because it always has the same number of levels (t) for the
row and column nuisance factors
A significant assumption is that the three factors (treatments and two
nuisance factors) do not interact
More restrictive than the RCBD
Each treatment appears once and only once in each row and column
A B (rows
C D
If you can block on two (perpendicular) sources of variation
x
columns) you can reduce experimental error when compared
D theA
B C to
RCBD
C
D
A
B
D
A
B
C
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
Early
Mid
Early
Mid
Late
Late
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: 2 4 1 3 5
Finally, randomize the order of the columns: 4 3 5 1 2
Analysis



Set up a two-way table and compute
the row and column totals
Compute a table of treatment totals and
means
Set up an ANOVA table divided into
sources of variation
•
•
•
•

Rows
Columns
Treatments
Error
Significance tests
• FT tests difference among treatment
Advantages and Disadvantages

Advantage:
• Allows the experimenter to control two sources of variation

Disadvantages:
• Error degree of freedom (df) is small if there are only a few
treatments
• The experiment becomes very large if the number of
treatments is large
• The statistical analysis is complicated by missing plots and
mis-assigned treatments
Latin Square Designs
3x3
Selected Latin Squares
4x4
ABC
BCA
CAB
ABCD
BADC
CDBA
DCAB
5x5
ABCDE
BAECD
CDAEB
DEBAC
ECDBA
6x6
ABCDEF
BFDCAE
CDEFBA
DAFECB
FEBADC
ABCD
BCDA
CDAB
DABC
ABCD
BDAC
CADB
DCBA
ABCD
BADC
CDAB
DCBA
In a Latin square, there are three factors:
 Treatments (t) (letters A, B, C, …)
 Rows (t)
 Columns (t)
The number of treatments = the number of rows =
the number of columns = t.
The row-column treatments are represented by cells in
a t x t array.
The treatments are assigned to row-column
combinations using a Latin-square arrangement
The Linier Model
yijk   m  t k  ri   j  e ijk 
i = 1,2,…, t
j = 1,2,…, t
k = 1,2,…, t
yij(k) = the observation in ith row and the jth column
receiving the kth treatment
m = overall mean
tk = the effect of the kth treatment
No interaction
th
ri = the effect of the i row
between rows,
columns and
j = the effect of the jth column
treatments
eij(k) = random error
Latin Square
 A Latin Square experiment is assumed to be a
three-factor experiment.
 The factors are rows, columns and treatments.
 It is assumed that there is no interaction between
rows, columns and treatments.
 The degrees of freedom for the interactions is used
to estimate error.
The Anova for a Latin Square
Source
S.S.
d.f.
M.S.
Treat
SST
t-1
MST
SSRow
SSCol
SSE
SST
t-1
t-1
Rows
Cols
Error
Total
(t-1)(t-2)
t2 - 1
MSRow
MSCol
MSE
Fcal
Ftable
MST /MSE Fα, (t-1);
(t-1)(t-2)
Example
1.
A courier company is interested in deciding between
five brands (D,P,F,C and R) of car for its next purchase
of fleet cars.
 The brands are all comparable in purchase price.
 The company wants to carry out a study that will
enable them to compare the brands with respect to
operating costs.
 For this purpose they select five drivers (Rows).
 In addition the study will be carried out over a five
week period (Columns = weeks).
 Each week a driver is assigned to a car using
randomization and a Latin Square Design.
 The average cost per mile is recorded at the end of
each week and is tabulated below:
Week
1
2
Drivers
3
4
5
1
7
D
3
P
7
F
6
R
12
C
2
4
P
8
D
9
C
11
F
8
R
3
7
F
10
C
8
R
11
D
9
P
4
9
C
6
R
12
D
12
P
11
F
5
8
R
8
F
10
P
14
C
15
D
The Anova Table
Source
S.S.
d.f.
M.S.
Week
Driver
50.0
4
12.5
80.0
4
20.0
Car
55.2
4
13.8
Error
12.8
12
1.07
Total
198
24
Fcalc
12.94
Ftable
3.26
Example
2.
we are again interested in how weight gain (Y) in rats
is affected by Source of protein (Beef, Cereal, and
Pork) and by Level of Protein (High or Low).
There are a total of t = 3 X 2 = 6 treatment
combinations of the two factors.
 Beef -High Protein
 Cereal-High Protein
 Pork-High Protein
 Beef -Low Protein
 Cereal-Low Protein
 Pork-Low Protein
In this example we will consider using a Latin Square
design
Six Initial Weight categories are identified for the test
animals in addition to Six Appetite categories.
 A test animal is then selected from each of the 6 X
6 = 36 combinations of Initial Weight and Appetite
categories.
 A Latin square is then used to assign the 6 diets to
the 36 test animals in the study.
In the latin square the letter
 A represents the high protein-cereal diet
 B represents the high protein-pork diet
 C represents the low protein-beef diet
 D represents the low protein-cereal diet
 E represents the low protein-pork diet
 F represents the high protein-beef diet.
The weight gain after a fixed period is
measured for each of the test animals
1
2
Initial
Weight
Category
3
4
5
6
1
62.1
A
86.2
B
63.9
C
68.9
D
73.8
E
101.8
F
Appetite Category
2
3
4
84.3
61.5
66.3
B
C
D
91.9
69.2
64.5
F
D
C
71.1
69.6
90.4
D
E
F
77.2
97.3
72.1
A
F
E
73.3
78.6
101.9
C
A
B
83.8
110.6
87.9
E
B
A
5
73.0
E
80.8
A
100.7
B
81.7
C
111.5
F
93.5
D
6
104.7
F
83.9
E
93.2
A
114.7
B
95.3
D
103.8
C
The Anova Table
S.S.
d.f.
M.S.
Fcalc
Ftable
Inwt
1767.0836
5
353.41673
111.1
2.71
App
2195.4331
5
439.08662
138.03
2.71
Diet
4183.9132
5
836.78263
263.06
2.71
Error
63.61999
20
3.181
8210.0499
35
Source
Total
Diet SS partioned into main effects for Source and
Level of Protein
Source
Ftable
S.S.
d.f.
M.S.
Fcalc
Inwt
1767.0836
5
353.41673
111.1
App
2195.4331
5
439.08662
138.03
Source
631.22173
2
315.61087
99.22
0.0000
Level
2611.2097
1
2611.2097
820.88
0.0000
SL
941.48172
2
470.74086
147.99
0.0000
Error
63.61999
20
3.181
8210.0499
35
Total
Graeco-Latin Square
Designs
Mutually orthogonal Squares
Definition
A Greaco-Latin square consists of two latin squares (one using
the letters A, B, C, … the other using greek letters a, b, c, …)
such that when the two latin square are supper imposed on each
other the letters of one square appear once and only once with
the letters of the other square. The two Latin squares are called
mutually orthogonal.
Example: a 7 x 7 Greaco-Latin Square
A
Bb
Cc
Dd
Ee
Ff
G
Be
Cf
D
E
Fb
Gc
Ad
Cb
Dc
Ed
Fe
Gf
A
B
Df
E
F
Gb
Ac
Bd
Ce
Ec
Fd
Ge
Af
B
C
Db
F
G
Ab
Bc
Cd
De
Ef
Gd
Ae
Bf
C
D
Eb
Fc
The Graeco-Latin Square Design
√ This design is used to simultaneously control (or
eliminate) three sources of nuisance variability
√ It is called “Graeco-Latin” because we usually specify the
third nuisance factor, represented by the Greek letters,
orthogonal to the Latin letters
√ A significant assumption is that the four factors
(treatments, nuisance factors) do not interact
√ If this assumption is violated, as with the Latin square
design, it will not produce valid results
√ Graeco-Latin squares exist for all t ≥ 3 except t = 6
Note:
At most (t –1) t x t Latin squares L1, L2, …, Lt-1 such
that any pair are mutually orthogonal.
It is possible that there exists a set of six 7 x 7
mutually orthogonal Latin squares L1, L2, L3, L4, L5,
L6 .
The Greaco-Latin Square Design
- An Example
A researcher is interested in determining the effect of
two factors:
 the percentage of Lysine in the diet and
 percentage of Protein in the diet have on Milk
Production in cows.
Previous similar experiments suggest that interaction
between the two factors is negligible.
For this reason it is decided to use a Greaco-Latin
square design to experimentally determine the two
effects of the two factors (Lysine and Protein).
Seven levels of each factor is selected
• 0.0(A), 0.1(B), 0.2(C), 0.3(D), 0.4(E), 0.5(F), and
0.6(G)% for Lysine and
• 2(), 4(b), 6(c), 8(d), 10(e), 12(f) and 14()% for
Protein.
• Seven animals (cows) are selected at random for the
experiment which is to be carried out over seven
three-month periods.
A Greaco-Latin Square is the used to assign the 7 X 7
combinations of levels of the two factors (Lysine and Protein) to a
period and a cow. The data is tabulated on below:
The Linear Model
yijkl   m  t k  l  ri   j  eijkl 
i = 1,2,…, t
j = 1,2,…, t
k = 1,2,…, t
l = 1,2,…, t
yij(kl) = the observation in ith row and the jth column receiving the
kth Latin treatment and the lth Greek treatment
m = overall mean
tk = the effect of the kth Latin treatment
l = the effect of the lth Greek treatment
ri = the effect of the ith row
j = the effect of the jth column
eij(k) = random error
No interaction between rows, columns, Latin
treatments and Greek treatments
 A Greaco-Latin Square experiment is assumed to
be a four-factor experiment.
 The factors are rows, columns, Latin treatments
and Greek treatments.
 It is assumed that there is no interaction between
rows, columns, Latin treatments and Greek
treatments.
 The degrees of freedom for the interactions is used
to estimate error.