# Anova Formula

Analysis of variance, or ANOVA, is a strong statistical technique that is used to show the difference between two or more means or components through significance tests. It also shows us a way to make multiple comparisons of several populations means. The Anova test is performed by comparing two types of variation, the variation between the sample means, as well as the variation within each of the samples. The below mentioned formula represents one-way Anova test statistics:
$\begin{array}{l}S S_{\text {total }}=\sum_{j=1}^{n}\left(\bar{X}_{j}-\bar{X}\right)^{2} \quad \\S S_{\text {within }}=\sum_{j=1}^{k} \sum_{j=1}^{l}\left(X-\bar{X}_{j}\right)^{2} \\ S S_{\text {between }}=\sum_{j=1}^{k}\left(\bar{X}_{j}-\bar{X}\right)^{2}\end{array}$
F = MST/MSE
MST = SST/ p-1
MSE = SSE/N-p

Where,

F = Anova Coefficient

MSB = Mean sum of squares between the groups

MSW = Mean sum of squares within the groups

SST = total Sum of squares

k = Total number of populations

n = The total number of samples in a population

SSW = Sum of squares within the groups

SSB = Sum of squares between the groups

S = Standard deviation of the samples

N = Total number of observations

## Solved Examples

For reference please go through the below-mentioned example:

Question 1:

The following data is given:

 Types of Animals Number of animals Average Domestic animals Standard Deviation Dogs 5 12 2 Cats 5 16 1 Hamsters 5 20 4

Calculate the Anova coefficient.

Solution:

Construct the following table:

 Animal name n x S $S^{2}$ Dogs 5 12 2 4 Cats 5 16 1 1 Hamster 5 20 4 16

p = 3
n = 5
N = 15
x̄ = 16
SST = ∑n (x−x̄)2
SST= 5(12−16)2+5(16−16)2+11(20−16)2
= 160

MST = $\frac{SST}{p−1}$

MST = $\frac{160}{3−1}$

MST = 80

SSE = ∑ (n−1)$S^{2}$

SSE = 4$\times$4 + 4$\times$1 + 4$\times$16

SSE = 84

MSE= $\frac{SSE}{N−p}$

MSE=$\frac{8415}{38415−3}$

MSE = 7

F = $\frac{MST}{MSE}$

F = $\frac{80}{7}$

F = 11.429