ANOVA is one-way analysis of variance that is used to determine any differences between the means of two or more groups. ANOVA calculator givesĀ statistical test that calculates SST, MST, SSE, MSE and Anova Coefficient. Plug in the number of samples(n), mean(x) of observation and standard deviation(S) to get those values.

**Step 1 :**From the problem, note the values and create the table which contains $n$, $x$, $S$ and $S^{2}$.

**Step 2 :**Find $\bar{x}$ and Sum of square due to treatment(SST) by using the formula,

SST = $\sum n(x-\bar{x})^{2}$

Where, $\bar{x}$ are the mean of all the observation.

**Step 3 :**Find the mean sum of square due to tretment (MST), Sum of square due to error(SSE) and Mean sum of square due to error(MSE) by using the formula,

MST = $\frac{SST}{p-1}$

SSE = $\sum(n-1)S^2$

MSE = $\frac{SSE}{N-p}$

Where,

p = total number of populations ,

n = number of samples,

N = total number of observations,

**Step 4 :**At last find the Anova coefficient(F) by using the formula,

F = $\frac{MST}{MSE}$

Based on ANOVA, some of the problems are given below.### Solved Examples

**Question 1:**Match was played between three groups and scores were noted. Following data is given about football teams of three countries:

Countries |
Number of Players |
Scores | Standard Deviations |

India | 11 |
6 |
20 |

New Zealand |
11 |
5 |
18 |

South Africa |
11 |
7 |
22 |

Find Anova coefficient.

**Solution:**

Given data :

Countries |
Number of Players |
Scores | Standard Deviations |

India | 11 |
6 |
20 |

New Zealand |
11 |
5 |
18 |

South Africa |
11 |
7 |
22 |

n = 11

p = 3

N = 33

$\bar{x}$ = $\frac{6+5+7}{3}$ = 6

Sum of square due to treatment, $SST=\sum n(x-\bar{x})^{2}$

$SST=11(6-6)^{2}+11(5-6)^{2}+11(7-6)^{2}$

= 22

Sum of square due to tretment, $MST$ = $\frac{SST}{p-1}$

$MST$ = $\frac{22}{3-1}$

= 11

Sum of square due to error, $SSE=\sum (n-1)S^{2}$

SSE = 10 $\times$ 400 + 10 $\times$ 324 + 10 $\times$ 484

= 12080

Mean sum of square due to error, $MSE$ = $\frac{SSE}{N-p}$

$MSE$ = $\frac{12080}{33-3}$

MSE = 402.667

Anova coefficient, $F$ = $\frac{MST}{MSE}$

$F$ = $\frac{11}{402.667}$

= 0.027

**Question 2:**Work was assigned to 3 group of employees in a railway factory. Following are the data that contains different potential.

Group |
Number of employees |
Potential |
Standard Deviations |

1 |
6 |
10 |
15 |

2 |
6 |
14 |
11 |

3 |
6 |
11 |
20 |

Find Anova coefficient.

**Solution:**

Given data :

Group |
Number of employees |
Potential |
Standard Deviations |

1 |
6 |
10 |
15 |

2 |
6 |
14 |
11 |

3 |
6 |
11 |
20 |

n = 6

p = 3

N = 18

$\bar{x}$ = $\frac{10+14+11}{3}$ = 11.66

Sum of square due to treatment, $SST=\sum n(x-\bar{x})^{2}$

$SST=6(10-11.66)^{2}+6(14-11.66)^{2}+6(11-11.66)^{2}$

= 52

Sum of square due to tretment, $MST$ = $\frac{SST}{p-1}$

$MST$ = $\frac{52}{3-1}$

= 26

Sum of square due to error, $SSE=\sum (n-1)S^{2}$

SSE = 5 $\times$ 225 + 5 $\times$ 121 + 5 $\times$ 400

= 3730

Mean sum of square due to error, $MSE$ = $\frac{SSE}{N-p}$

$MSE$ = $\frac{3730}{18-3}$

MSE = 248.667

Anova coefficient, $F$ = $\frac{MST}{MSE}$

$F$ = $\frac{26}{248.667}$

= 0.105