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# Anova Formula

Anova is a statistical test which analyzes variance. It is helpful in making comparison of two or more means which enables a researcher to draw various results and predictions about two or more sets of data. Anova test includes one-way anova, two-way anova or multiple anova depending upon the type and arrangement of the data. One-way anova has the following test statistics:

Where,

$F$ = Anova Coefficient

$MST$ = Mean sum of squares due to treatment

$MSE$ = Mean sum of squares due to error.

Formula for MST is given below:

Where,

$SST$ = Sum of squares due to treatment

$p$ = Total number of populations

$n$ = Total number of samples in a population.

Formula for MSE is given below:

Where,

$SSE$ = Sum of squares due to error

$S$ = Standard deviation of the samples

$N$ = Total number of observations.

 Related Calculators Anova Calculator anova effect size calculator Acceleration Formula Calculator Area of a Circle Formula Calculator

## Anova Formula Problems

Few problems based on Anova formula are given below:

### Solved Examples

Question 1: Following data is given about cricket teams of three countries:

 Countries Number of Players Average Runs Standard Deviations India 11 60 15 New Zealand 11 50 10 South Africa 11 70 12

Find Anova coefficient?

Solution:

Construct the following table:

 Cricket Teams n x S S2 India 11 60 15 225 New Zealand 11 50 10 100 South Africa 11 70 12 144

n = 11
p = 3
N = 33
$\bar{x}$ = $\frac{60+50+70}{3}$ = 60
$SST=\sum n(x-\bar{x})^{2}$
$SST=11(60-60)^{2}+11(50-60)^{2}+11(70-60)^{2}$
= 2200
$MST$ = $\frac{SST}{p-1}$
$MST$ = $\frac{2200}{3-1}$
= 1100
$SSE=\sum (n-1)S^{2}$
SSE = 10*225 + 10*100 + 10*144
= 4690
$MSE$ = $\frac{SSE}{N-p}$
$MSE$ = $\frac{4690}{33-3}$
MSE = 156.33
$F$ = $\frac{MST}{MSE}$
$F$ = $\frac{1100}{156.33}$
= 7.036

Question 2: The following data is given:

 Plant Name Number of plants Average Flowers Standard Deviation Rose 5 12 2 Marigold 5 16 1 Lily 5 20 4

Calculate the Anova coefficient.

Solution:

Construct the following table:

 Plant name n x S S2 Rose 5 12 2 4 Marigold 5 16 1 1 Lily 5 20 4 16

p = 3
n = 5
N = 15
$\bar{x}$ = 16
$SST$ = $\sum n(x-\bar{x})^{2}$
$SST$ = $5(12-16)^{2}+5(16-16)^{2}+11(20-16)^{2}$
= 160
$MST$ = $\frac{SST}{p-1}$
$MST$ = $\frac{160}{3-1}$
= 80
$SSE$ = $\sum (n-1)S^{2}$
SSE = 4*4 + 4*1 + 4*16
= 84
$MSE$ = $\frac{SSE}{N-p}$
$MSE$ = $\frac{84}{15-3}$
MSE = 7
$F$ = $\frac{MST}{MSE}$
$F$ = $\frac{80}{7}$
= 11.429

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