Question

Suppose a package delivery company purchased 17 trucks at the same time. Six trucks were purchased from manufacturer A, five from manufacturer B, and six from manufacturer C. The cost of maintaining each truck was recorded. The company used ANOVA to test if the mean maintenance cost of the trucks from each manufacturer were equal. To apply the F test, how many degrees of freedom must be in the denominator?
17

4

3

14

Answer 1

Answer:

$df_{den}=df_{between}=N-K=17-3=14$.

Step-by-step 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 (A,B,C) and on each group we have sample of size (6,5,6) respectively , 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 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 $df_{den}=df_{between}=N-K=17-3=14$.

And the total degrees of freedom would be $df=N-1=17 -1 =16$

On this case the correct answer would be 2 for the numerator and 14 for the denominator.  

Best answer

14