# 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 .