ANOVA

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