In: Statistics and Probability
Let us suppose that we have n treatments which is to be applied in m samples. let us consider a one way model with n treatments and m samples as
Then total sum of square is represented by SST given by
The total sum of square is divided into two parts namely between and within sum of squares which is given by
In above case the first one represent the within sum of square and second one represents the between sum of squares given by
There is total number of m*n observation so degree of freedom for total sum of squares =(m*n-1).
Also there are m samples so (m-1) degree of freedom for between sum of square,
Degree of freedom for within sum of squares = Total degree of freedom - Degree of freedom for between sum of squares. =(m*n-1)-(m-1)=m(n-1).
Now, we need mean sum of squares for testing of one way anova given by
Also we know that F value in case of one way anova is the ratio of Between mean sum of squares and Within mean sum of squares.
Therefore the anova table for all the sum of squares with degree of freedom is given by
Now we have tabulated F which is obtained on the basis of above table. Also F(m-1,m(n-1)) is there which is obtained by making use of F distribution for given degree of freedom. Let us say it a distributional value.
Now If the tabulated value is less than the distributional value then we accept the null hypothesis with no difference in mean otherwise we reject the null hypothesis.