1
Two-Way ANOVA
Example 1
• The effects of two protein sources (groundnut &
soyabean) were evaluated in 20 broiler chicks from 2
breeds (Ross & Cobb).
• Establish whether breed or dietary protein source had an
effect on performance (feed conversion efficiency)
2
Soya Groundnut
Ross
Cobb
3.5
4.2
4.7
4.0
3.5
2.5
2.9
3.1
3.9
2.7
1.7
2.0
2.8
2.2
2.5
2.0
2.2
3.1
2.7
2.1
• State null hypotheses
There is no difference in the performances
of the two breeds of broiler chicks
The two protein sources had no effect on
performance
3
Soya Groundnut
Ross
Cobb
3.5
4.2
4.7
4.0
3.5
2.5
2.9
3.1
3.9
2.7
1.7
2.0
2.8
2.2
2.5
2.0
2.2
3.1
2.7
2.1
35.0 23.3
31.1
27.2
Calculate breed and protein totals
• Calculate total sums of squares (SSTotal) and degrees of
freedom
x =
x
2 =
N =
58.3  grand total of all data
182.97
20
 182.97-169.94 = 13.03
DF Total = N-1  20 – 1 = 19
4
ANOVA table
Calculate protein sums of squares (SSProtein)
= 6.85
5
ANOVA Table
Calculate breed sums of squares (SSBreed)
= 0.77
6
ANOVA Table
• SSResidual = SSTotal – (SSBreed + SSProtein)
= 13.03 – (0.77 + 6.85)
= 5.41
Calculate residual sums of squares (SSResidual)
7
ANOVA Table
Calculate the mean squares (MS)
8
ANOVA Table
Calculate F-ratios
9
ANOVA Table
Compare calculated F-values to critical F-value
• Degrees of freedom
– Protein = 1 & Residual = 17
• Critical F-values
– P < 0.05 = 4.45
– P < 0.01 = 8.40
• Calculated F-value for protein was 21.41
• Conclusion: There were significant differences
in the performances of broilers offered the two
proteins sources (P<0.01)
10
Did the breed affect the findings?
• Degrees of freedom
– Breed = 1 & Residual = 17
• Critical F-values
– P < 0.05 = 4.45
– P < 0.01 = 8.40
• Our calculated F value (2.41) is lower than the
critical values
• Conclusion: There were no differences in the
performance of the two breeds
Two-way ANOVA
with interaction
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Interaction
• Is there a difference in the response of one factor at
different levels of the other?
• For example: A study investigated the effects of dietary fat
level in on milk fat content of three breeds of dairy cows
– Main effects  fat level & breed
– Interaction  do the breeds respond in the same way to the
different levels of fat in the diet
– Look at three scenarios on the next three slides to illustrate
interaction between breed and diet fat content
Dietary fat (g/kg)
Milk fat (%)
Breed 1
Breed 2
Extreme interaction
Breed 3
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Dietary fat (g/kg)
Milk fat (%)
Breed 1
Breed 2
Moderate interaction
Breed 3
Dietary fat (g/kg)
Milk fat (%)
Breed 1
Breed 2
No interaction
Breed 3
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Example 2
Evaluate whether supplementing fat at two levels at either
3 or 5% in the diet affects milk fat content in three breeds
(Jersey, Guernsey and Holstein) of dairy cows. Also
establish whether there is interaction between dietary fat
content and breed of cow.
Fat level
Breed
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Null hypotheses
• There is no effect of breed on milk fat
content
• There is no effect of diet fat level on milk fat
• There is no interaction between breed and
dietary fat
Fat
Breed
24.1 32.3 38.8
41.9
53.3
11.1 13.3 17.5
13.0 19.0 21.3
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Calculate Sums of Squares
 399.78 – 377.63 = 22.153
SSInteraction = SSTreatment – (SSFat level+SSBreed)
SSInteraction = 19.88 – (13.56+5.42) = 0.903
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ANOVA Table
SSResidual = SSTotal – (SSFat level+SSBreed+SSInteraction)
SSResidual = 22.15- (13.56+5.42+0.90) = 2.270
DFInteraction = DFFat Level × DFBreed
DFInteraction = 1× 2 = 2
DFResidual = DFTotal – (DFFat level+ DFBreed+DFInteraction)
DFResidual = 23 – (1+ 2 + 2) = 18
17
ANOVA Table
Calculate MS
18
ANOVA Table
Fat Level
• Degrees of freedom
– Fat  1
– Residual  18
• Critical F-values
– P < 0.05 = 4.41
– P < 0.01 = 8.29
• Calculated F-value for fat level  42.94
• Conclusion: The two fat levels caused significant
differences in milk fat content (P<0.01)
19
Breed
• Degrees of freedom
– Breed  2
– Residual  18
• Critical F-values
– P < 0.05 = 3.55
– P < 0.01 = 6.01
• Calculated F-value for breed  53.79
• Conclusion: The three breeds had significant
differences in milk fat content (P<0.05)
Interaction
• Degrees of freedom
– Interaction  2
– Residual  18
• Critical F-values
– P < 0.05 = 3.55
– P < 0.01 = 6.01
• Calculated F-value for interaction  3.58
• Conclusion: There was significant interaction
between breed & level of fat in the diet (P<0.05)