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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:

Anova Formula
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:
Formula for Anova
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:
Analysis of Variance Formula

Where,

$SSE$ = Sum of squares due to error

$S$ = Standard deviation of the samples

$N$ = Total number of observations.

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Anova Formula Problems

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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 plantsAverage FlowersStandard 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   
 S
Rose
5
1224
Marigold
5
16
1
1
Lily
5
204
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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