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
yij     i   ij
i = 1,2,…, t
j = 1,2,…, r
yij = the observation in ith treatment and the jth
replication
 = overall mean
 = the effect of the ith treatment
i
ij = random error
One-Way ANOVA FTest
One-Way ANOVA F-Test
1. Tests the Equality of 2 or More (p) Population
Means
2. Variables
• One Nominal Scaled Independent Variable
 2 or More (p) Treatment Levels or Classifications
• One Interval or Ratio Scaled Dependent Variable
3. Used to Analyze Completely Randomized
Experimental Designs
One-Way ANOVA F-Test
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
One-Way ANOVA F-Test
Hypotheses
H0: 1 = 2 = 3 = ... = t

• All Population Means are Equal
• No Treatment Effect
Ha: Not All i Are Equal

• At Least 1 Pop. Mean is Different
• Treatment Effect
 NOT 1  2  ...  t
One-Way ANOVA F-Test
Hypotheses


H0: 1 = 2 = 3 = ... = t
• All Population Means are
Equal
• No Treatment Effect
Ha: Not All i Are Equal
• At Least 1 Pop. Mean is
Different
• Treatment Effect
 NOT 1  2  ...  t
f(X)
1 = 2 = 3
X
f(X)
1 =  2  3
X
Why Variances?




Observe one sample from each treatment
group
Their means may be slightly different
How different is enough to conclude
population means are different?
Depends on variability within each population
• Higher variance in population  higher variance in
means
• Statistical tests are conducted by comparing
variability between means to variability within
each sample
Two Possible
Experiment Outcomes
Same treatment variation
Different random variation
Pop 1 Pop 2 Pop 3

A
Pop 4
Pop 5
Pop 6
Reject equality of means!
Pop 1 Pop 2 Pop 3
Pop 4
Pop 5
Can’t reject equality of means!
Pop 6
Two More Possible
Experiment Outcomes
Same treatment variation
Different random variation
Pop 1 Pop 2 Pop 3

A
Different treatment variation
Same random variation
Pop 1 Pop 2 Pop 3
B
Reject
Reject
Pop 4
Pop 5
Pop 6
Pop 4
Pop 5
Can’t reject equality of means!
Pop 6
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
SS=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, SS=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 H0
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
Wax
Wax
Type 2
Type 3
27
30
29
28
31
33
28
31
30
30
29
28
30
32
31
29.0
2.5
30.4
3.3
30.0
2.5
Example: Home Products, Inc.

Hypotheses
H0: 1 = 2 = 3
Ha: Not all the means are equal
where:
1 = mean number of washes for Type 1 wax
2 = mean number of washes for Type 2 wax
3 = 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
Squares
5.2
33.2
38.4
Degrees of
Freedom
2
12
14
Mean
Squares
2.60
2.77
F
.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 (consequences?)
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 (see
Chapter 14) 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
 = overall mean
i = the effect of the ith treatment
No interaction
rj = the effect of the jth block
between blocks
ij = random error
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.. )   (y ij  yi.  y.j  y.. ) 2
2
i 1
2
j1
i 1 j1
SST  SSTreatments  SS Blocks  SS E
Extension of the ANOVA to the RCBD
The degrees of freedom for the sums of squares in
SST  SSTreatments  SS Blocks  SS E
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 = SST + SSB + SSE

The total degrees of freedom, nT - 1, are partitioned
such that k - 1 degrees of freedom go to treatments,
b - 1 go to blocks, and (k - 1)(b - 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
SSTottr - 1
Mean
Squares
SST/t-1
F
MST/MSE
SSE/(t-1)(r-1)
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
31
30
30
30
29
29
29
29
28
33
31
29
26
25
26
29.8
28.8
28.4
Blocks
Means
30.333
29.333
28.667
31.000
25.667
Example: Eastern Oil Co.

Mean Square Due to Treatments
The overall sample mean is 29. Thus,
SST= 5[(29.8 - 29)2+ (28.8 - 29)2+ (28.4 - 29)2]= 5.2

MST = 5.2/(3 - 1) = 2.6
Mean Square Due to Blocks
SSB = 3[(30.333 - 29)2 + . . . + (25.667 - 29)2] = 51.33

MSB = 51.33/(5 - 1) = 12.8
Mean Square Due to Error
SSE = 62 - 5.2 - 51.33 = 5.47
MSE = 5.47/[(3 - 1)(5 - 1)] = .68
Example: Eastern Oil Co.

Rejection Rule
Using test statistic:
Using p-value:
Reject H0 if F > 4.46
Reject H0 if p-value < .05
Assuming = .05, F.05 = 4.46 (2 d.f. numerator and 8 d.f.
denominator)
Example: Eastern Oil Co.


Test Statistic
F = MST/MSE = 2.6/.68 = 3.82
Conclusion
Since 3.82 < 4.46, we cannot reject H0. There is not
sufficient evidence to conclude that the miles per gallon
ratings differ for the three gasoline blends.
Using Excel’s Anova:
Two-Factor Without Replication Tool



Step 1
Step 2
Step 3
Select the Tools pull-down menu
Choose the Data Analysis option
Choose Anova: Two Factor Without
Replication from the list of Analysis Tools
… continued
Using Excel’s Anova:
Two-Factor Without Replication Tool

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

A
B
C
D
E
F
Value
Worksheet
(top
portion)
1 Automobile Blend X Blend Y Blend Z
2
3
4
5
6
1
2
3
4
5
31
30
29
33
26
30
29
29
31
25
30
29
28
29
26
G
Using Excel’s Anova:
Two-Factor Without Replication Tool

A
B
C
D
E
F
Value
Worksheet
(middle
portion)
8 Anova: Two-Factor Without Replication
9
10
11
12
13
14
15
16
17
18
19
SUMMARY
Count
1
2
3
4
5
Blend X
Blend Y
Blend Z
Sum
3
3
3
3
3
91
88
86
93
77
Average
30.3333
29.3333
28.6667
31
25.6667
5
5
5
149
144
142
29.8
28.8
28.4
Variance
0.33333
0.33333
0.33333
4
0.33333
6.7
5.2
2.3
G
Using Excel’s Anova:
Two-Factor Without Replication Tool

A
B
C
D
E
F
Value
Worksheet
(bottom
portion)
8 ANOVA
9
10
11
12
13
14
Variation
Rows
Columns
Error
Total
SS
51.3333
5.2
5.46667
62
df
G
MS
F
P-value
F crit
4 12.8333 18.7805 0.0004 3.83785
2
2.6 3.80488 0.06899 4.45897
8 0.68333
14
Using Excel’s Anova:
Two-Factor Without Replication Tool

Conclusion Using the p-Value
• The value worksheet shows that the p-value is
.06899
• The rejection rule is “Reject H0 if p-value <
.05”
• Thus, we cannot reject H0 because the p-value
= .06899 >  = .05
• There is not 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
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
If you can block on two (perpendicular) sources of variation (rows x
columns) you can reduce experimental error when compared to the
A B C D
RCBD
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:
• 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
ABC
BCA
CAB
Selected Latin
4x4
ABCD ABCD
BADC
BCDA
CDBA
CDAB
DCAB
DABC
5x5
ABCDE
BAECD
CDAEB
DEBAC
ECDBA
Squares
6x6
ABCDEF
BFDCAE
CDEFBA
DAFECB
FEBADC
ABCD
BDAC
CADB
DCBA
ABCD
BADC
CDAB
DCBA
In a Latin square You have 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
Example
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:
1
2
Drivers
3
4
5
1
5.83
D
4.80
P
7.43
F
6.60
R
11.24
C
2
6.22
P
7.56
D
11.29
C
9.54
F
6.34
R
Week
3
7.67
F
10.34
C
7.01
R
11.11
D
11.30
P
4
9.43
C
5.82
R
10.48
D
10.84
P
12.58
F
5
6.57
R
9.86
F
9.27
P
15.05
C
16.04
D
The Linier Model
yij k      k  ri   j   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
 = overall mean
k = the effect of the ith treatment
No interaction
ri = the effect of the ith row
between rows,
columns and
th
j = the effect of the j column
treatments
ij(k) = random error
 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 Table for a Latin Square
Experiment
Sourc
e
Treat
Rows
S.S.
d.f.
SST
SSRo
t-1
t-1
w
Cols
SSCol
t-1
Error
Total
SSE
SST
(t-1)(t-2)
t2 - 1
M.S.
F
MST MST /MSE
MSRow
MSRow
/MSE
MSCol
MSCol
/MSE
MSE
pvalue
The Anova Table for Example
Source
S.S.
d.f.
M.S.
F
p-value
Week
51.17887
4
12.79472
16.06
0.0001
Driver
69.44663
4
17.36166
21.79
0.0000
Car
70.90402
4
17.72601
22.24
0.0000
Error
9.56315
12
0.79693
Total
201.09267
24
Example
In this Experiment the 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 and
 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 and
 F represents the high protein-beef diet.
The weight gain after a fixed period is measured for
each of the test animals and is tabulated below:
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 for Example
Source
S.S.
d.f.
M.S.
F
p-value
Inwt
1767.0836
5
353.41673
111.1
0.0000
App
2195.4331
5
439.08662
138.03
0.0000
Diet
4183.9132
5
836.78263
263.06
0.0000
Error
63.61999
20
3.181
8210.0499
35
Total
Diet SS partioned into main effects for
Source and Level of Protein
Source
S.S.
d.f.
M.S.
F
p-value
Inwt
1767.0836
5
353.41673
111.1
0.0000
App
2195.4331
5
439.08662
138.03
0.0000
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
B
Cb
Df
Ec
Bb
Cf
Dc
E
Fd
Cc
D
Ed
F
G
Dd
E
F
Gb
Af
E
Fb
Gf
Ac
B
Ff
Gc
A
Bd
C
G
Ad
B
C
Db
F
G
Ab
Bc
Cd
D
Ef
Gd
A
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 GreacoLatin 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(), 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      k  l  ri   j   ij 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
 = overall mean
k = 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
ij(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.