Two-Way Within-Subjects Design - Analysis of Variance Designs - page 296

df A × S = ( a − 1)( n − 1) = (2 − 1)(6 − 1) = (1)(5) = 5

df B × S = ( b − 1)( n − 1) = (3 − 1)(6 − 1) = (2)(5) = 10

df A × B × S = ( a − 1)( b − 1)( n − 1) = (2 − 1)(3 − 1)(6 − 1)

= (1)(2)(5) = 10

df T = ( a )( b )( n ) − 1 = (2)(3)(6) − 1 = 36 − 1 = 35 .

11.6.3 CALCULATING MEAN SQUARES AND F RATIOS
Mean squares are calculated by dividing each sum of squares by its respec-
tive degrees of freedom. Note that we do not calculate a mean square total
value. You will recall from our earlier discussion that a two-factor within-
subjects design produces three F ratios. What is unique about this analysis
is that each F ratio requires a separate error term for the denominator to
control for the different sources of variability influencing the scores of the
dependentmeasureunderstudy.Thus,ourthree F ratios are formed with
the following error terms. F A : MS A × S , F B : MS B × S , and F A × B : MS A × B × S .
Thus,

MS A
MS A × S =

F A = 1.76 (11.9)

MS B
MS B × S =

78 . 70

F B = = 52.82 (11.10)

1 . 49

MS A × B
MS A × B × S =

8 . 53

F A × B =

33 =

25.85 . (11.11)

0 .

11.6.4 EVALUATING THE F RATIOS
We test the null hypothesis for each of our three observed (calculated) F
ratios by evaluating each F value with critical values of F (see Appendix C)
at the following degrees of freedom: ( df effect , df error ), at a particular alpha
level. Thus, our F observed for Factor A was F A = 76 and is evaluated

with (1, 5) degrees of freedom. The critical value of F at these degrees
of freedom (with alpha set at .05) is 6.61. Since our F A is less than the
critical value, we do not reject the null hypothesis, and conclude that type
of vehicle does not affect the error rate.
For the main effect of Factor B ,our F B was 52.82 and is evaluated
with (2, 10) degrees of freedom. The critical value of F is 4.10 at the .05
level. Since our F B is greater than the critical value, we reject the null and
conclude that the amount of alcohol affects the error rate.
Lastly, we evaluate the interaction effect ( F A × B =

1 .

85) with (2, 10)
degrees of freedom. These happen to be the same degrees of freedom
we had with F B ;hence,werejectthenullhypothesisandconcludethat

25 .