Question

Suppose that a​ one-way ANOVA is being performed to compare the means of five populations and that the sample sizes are 17 comma 15 comma 16 comma 18 comma and 11. Determine the degrees of freedom for the​ F-statistic.

(a) the degree of freedom of the numerator _______________________
(b) the degree of freedom of the denominator______________________

Answer 1

a)
`df_(num)=k-1=5-1=4`

b)
`df_(den)=N-k=77-5=72`

Explanation:

Analysis of variance (ANOVA) "is used to analyze the differences among group means in a sample".

The sum of squares "is the sum of the square of variation, where variation is defined as the spread between each individual value and the grand mean"

If we assume that we have
`p` groups and on each group from
`j=1,\dots,p` we have
`n_j` individuals on each group we can define the following formulas of variation:

`SS_(total)=\sum_(j=1)^p \sum_(i=1)^(n_j) (x_(ij)-\bar x)^2`

`SS_(between)=SS_(model)=\sum_(j=1)^p n_j (\bar x_(j)-\bar x)^2`

`SS_(within)=SS_(error)=\sum_(j=1)^p \sum_(i=1)^(n_j) (x_(ij)-\bar x_j)^2`

And we have this property

`SST=SS_(between)+SS_(within)`

The degrees for the numerator are
`df_(num)=k-1=5-1=4`, where k represent the number of groups on this case k =5.

The degrees for the denominator are
`df_(den)=N-k=77-5=72`, where N represent the total number of people N=17+15+16+18+11=77.

And the total degrees of freedom are given by:
`df_(tot)=df_(num)+df_(den)=N-1=4+72=76`