# ANOVA Test Procedure

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 .