ANNOVA test was performed to determine whether there are deviations due to varieties of chilli and effect of solvents.
ANOVA Test Procedure
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 .