Analysis of Variance-ANOVA

Analysis of Variance-ANOVA

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  1. Consider the following (small integers, indeed for illustration while saving space) random samples from three different populations.

With the null hypothesis:
H0: µ1 = µ2 = µ3, 
and the alternative:
Ha: at least two of the means are not equal.

= 0.05, the critical value from F-table isaAt the significance level  
0.05, 2, 12 = 3.89.

 

Sum

Mean

Sample P1

2

3

1

3

1

10

2

Sample P2

3

4

3

5

0

15

3

Sample P3

5

5

5

3

2

20

4

 


Demonstrate that, SST=SSB+SSW.
That is, the sum of squares total (SST) equals sum of squares between (SSB) the groups plus sum of squares within (SSW) the groups.

And check whether Null Hypothesis is true.

Computation of sample SST: With the grand mean = 3, first, start with taking the difference between each observation and the grand mean, and then square it for each data point.

Sum

Sample P1

1

0

4

0

4

9

Sample P2

0

1

0

4

9

14

Sample P3

4

4

4

0

1

13

 

Therefore SST = 36 with d.f = (n-1) = 15-1 = 14

Computation of sample SSB:

Second, let all the data in each sample have the same value as the mean in that sample. This removes any variation WITHIN. Compute SS differences from the grand mean.

Sum

Sample P1

1

1

1

1

1

5

Sample P2

0

0

0

0

0

0

Sample P3

1

1

1

1

1

5

 

Therefore SSB = 10, with d.f = (m-1)= 3-1 = 2 for m=3 groups.

Computation of sample SSW:

Third, compute the SS difference within each sample using their own sample means. This provides SS deviation WITHIN all samples.

Sum

Sample P1

0

1

1

1

1

4

Sample P2

0

1

0

4

9

14

Sample P3

1

1

1

1

4

8

 

SSW = 26 with d.f = 3(5-1) = 12. That is, 3 groups times (5 observations in each -1)

Results are: SST = SSB + SSW, and d.fSST = d.fSSB + d.fSSW, as expected.

Now, construct the ANOVA table for this numerical example by plugging the results of your computation in the ANOVA Table. Note that, the Mean Squares are the Sum of squares divided by their Degrees of Freedom. F-statistics is the ratio of the two Mean Squares.

The ANOVA Table

Sources of Variation

Sum of Squares

Degrees of Freedom

Mean Squares

F-Statistic

Between Samples

10

2

5

2.30

Within Samples

26

12

2.17

Total

36

14

Conclusion: There is not enough evidence to reject the null hypothesis H0.

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