ANOVA Test Procedure

ANOVA Test

ANNOVA was performed to determine whether there are deviations due to varieties of chilli and effect of solvents. Table 6.1 is represented again. The detailed calculations are given below.

chilli type
Solvents
Acetone
Hexane
Ethyelenedichloride
% of Extraction
% of Extraction
% of Extraction
Byadgi chilli
19.988
16.576
17.890
Ramnad Mundu
16.930
16.998
18.436
Sattur Chilli
18.105
17.390
17.070
Guntur Sannam- S4 Type
19.110
16.309
18.090

1) X1 = 18.533 X2 = 16.816 X3 = 17.87

2) Total : 4 + 4 + 4 = 12 = n

3) Find ∑x = 19.988 + 16.93 + 18.105 + 19.11 + 16.567 + 16.998 + 17.39 + 16.309 + 17.89 + 18.436 + 17.07 + 18.09
= 212.883

4) Find X = ∑x / n = 17.74

5) ∑x2 = {19.9882 + 16.932 + 18.1052 + 19.112 + 16.5672 + 16.9982
+ 17.392 + 16.3092 + 17.892 + 18.4362 + 17.072 + 18.092}
= S x2
= 3789.483

6) ( ∑x )2 / n = 3776.598

7) S x2 – ( ∑x )2 / n = 12.885

8) [ ( 19.988 + 16.93 + 18.105+ 19.11)2 / 4 + ( 16.567 + 16.998 + 17.39 + 16.309 )2 / 4 + ( 17.89 + 18.436 + 17.07 + 18.09)2 / 4 ] – ( ∑x )2 / n
= [1373.925 + 1131.11 + 1277.56 ] – 3776.598
= 3782.595 – 3776.598
= 5.999

9) Find : 7) – 8) = 12.885 – 5.999
= 6.886

10) Degree of freedom : a) n-1 = 12-1 =11
b) 3-1 = 2
Find : 11-2 = 9

11) Find Variance:
Between groups = 8) / 2 = y1 = 2.99
Within groups = 9) / 11 = y2 = 0.626

Sum of Squares
Degree of Freedom
Mean Square
Between Groups
5.998
2
2.999
Within Groups
6.886
9
0.626

Therefore , F = 2.999/0.626 = 4.79

We have to check F table at 2 degree of freedom and at 9 degree of freedom at 5% significant level

Comparing F test values from the analysis with the standard table ( Table .6.3) 2 & 9 degrees of freedom F 2/9 at 95% confidence = 4.3

The analysis showed F annova = 4.79

Since Fannova > Ftable , this shows there is effect of sources of chilli and solvents .