Question

A large department store examined a sample of 18 credit card sales and recorded the amounts charged for each of three types of credit cards: MasterCard, Visa, and Discover. Six MasterCard sales, seven Visa, and five Discover sales were recorded. The store used an ANOVA to test if the mean sales for each credit card were equal. What are the degrees of freedom for the F statistic?

a. 3 in the numerator, 18 in the denominator

b. 18 in the numerator, 3 in the denominator

c. 6 in the numerator, 15 in the denominator

d. 2 in the numerator, 15 in the denominator

Answer 1

d. 2 in the numerator, 15 in the denominator

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
`3` groups and on each group from
`j=1,\dots,n_j` 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`

We want to test this hypothesis:

`\mu_(Master card)=\mu_(Visa)=\mu_(Discover)`

And we have this property

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

The degrees of freedom for the numerator on this case is given by
`df_(num)=df_(within)=k-1=3-1=2` where k =3 represent the number of groups.

The degrees of freedom for the denominator on this case is given by
`df_(den)=df_(between)=N-K=18-3=15`.

And the total degrees of freedom would be
`df=N-1=18 -1 =17`

On this case the correct answer would be 2 for the numerator and 15 for the denominator.

d. 2 in the numerator, 15 in the denominator