ONE WAY ANALYSIS OF VARIANCE Dept. of AGB Veterinary College, SHIMOGA
Analysis of variance technique involves partitioning the total variability present in a set of observations into different components produced by different sources, some of which are known and others are unknown.
STAT 512 2018-19 VCS AGB RJ
Three feed were fed to 5 cows each and the milk yield in Kg of each cow is as follows.
A 12 8 9 7 4
B 7 9 6 2 1 STAT 512 2018-19 VCS AGB RJ
C 11 13 10 8 13
• The variations are due to known effectfeeds, unknown effect- Like age, sex etc. • The ANOVA is used to isolate the variation due to known and unknown causes.
STAT 512 2018-19 VCS AGB RJ
Steps in the calculation of ANOVA
(X)2
1. Calculation of correction factor =
N
2. Calculation of total sum of squares
(∑ Yij2- C.F) that is( Y112+ Y122………… Y332)- C.F 3.Calculation Treatment sum of squares
2 ΣT1 n1
ΣT2 +
n2
2
ΣT3 +
2 - C.F
n3
STAT 512 2018-19 VCS AGB RJ
4.Calculation Error sum of squares (T.S.S- Tr S.S)
5. Treatment Mean square
Treatment S S Treatment df 6. Error Mean square
Error S S Error df STAT 512 2018-19 VCS AGB RJ
A
B
C
12
7
11
8
9
13
9
6
10
7
2
8
4
1
13
(X)2 1. Calculation of correction factor = N 2. Calculation of total sum of squares 3.Calculation Treatment sum of squares
4.Calculation Error sum of squares (T.S.S- Tr S.S) 5. Treatment Mean square 6. Error Mean square STAT 512 2018-19 VCS AGB RJ
7. Calculation of Fob
ANOVA table SOURCE OF VARIATION
Degrees of Sum of Freedom Squares
Treatment (t-1)
Tr.SS
Error
t(r-1)
E.S.S
Total
(r)(t)-1 T.S.S
Mean sum of Fob squares
Tr.SS/ Tr.d.f E.S.S/ Error d.f
STAT 512 2018-19 VCS AGB RJ
Fob =
Treatment S S Error M S
If Fob value is greater than Ftab Value then the null hypothesis is rejected that is the treatment difference is significant.
STAT 512 2018-19 VCS AGB RJ
Example: There were three sires each mated to six dams and one progeny of each dam was recorded for their first litter size which is given below.
47
Sires 2 5 8 7 7 6 8 41
50
X=138
6
6
6
N=18
1 10 7 9 9 7 5
Total ni=No.of observation/sire
3 9 11 8
Total
9 8 5
STAT 512 2018-19 VCS AGB RJ
Steps in the calculation of ANOVA 1. Calculation of correction factor =
1 10 7 9 9 7 5 47
2 5 8 7 7 6 8 41
3 9 11 8 9 8 5 50
(X)2 N
CF = (138)2 18 Correction factor = 1058 ∑X= 138
STAT 512 2018-19 VCS AGB RJ
2. Calculation of total sum of squares
(∑ Yij2- C.F) that is( Y112+ Y122………… Y332)- C.F 1 10 7 9 9 7 5 47
2 5 8 7 7 6 8 41
3 9 11 8 9 8 5 50
∑ Yij2= 102+72+92+…….82+52
TSS= 1108-1058 =50
STAT 512 2018-19 VCS AGB RJ
3.Calculation Treatment sum of squares
T1 + T2 n n 2
2
1
2 47
6
+
2
2 41
6
+
+ T3
2
n
3
2 50
6
-
C.F
= 1065-1058 = 7 - 1058 BSS = 7
STAT 512 2018-19 VCS AGB RJ
ANOVA d.f
Sum of Mean sum of squares squares
Fob
Between Treatment
S-1 3-1=2
Bss =7
3.5/2.87 =1.21
Error
N-S Wss=43 Mse=Wss/N-S 18-3=15 =43/15=2.87
Total
18-1=17 50
Source of variation
MSs=Bss/S-1 =7/2=3.5
STAT 512 2018-19 VCS AGB RJ