Information Technology Reference
In-Depth Information
10.9.4 EVALUATING THE
F
RATIO AND TREATMENT
EFFECT MAGNITUDE
We test the null hypothesis for our observed (calculated)
F
ratio by eval-
uating (or comparing) the
F
value with critical values of
F
(see Appendix
C) at the following degrees of freedom: (
df
A
,
df
A
×
S
), at a particular alpha
level. Thus, for our present numerical example we have
F
(4, 28)
df
and the
critical value is 2.95 at the .05 alpha level. Because our observed
F
of 18.65
exceeds the critical value of
F
, we reject the null hypothesis and conclude
that a statistically significant treatment effect is present. More concretely,
symptom intensity appears to be a function of time of test.
To determine how much variance we are accounting for by our treat-
ment manipulation (time of test), we will compute eta squared and partial
eta squared.
The formulas for these two measures of treatment effect magnitude,
based on the present numerical example, follow:
SS
A
123
.
85
2
η
=
SS
T
=
97
=
0.616 or 62%
(10.9)
200
.
SS
A
SS
A
+
123
.
85
2
partial
η
=
SS
A
×
S
=
04
=
0.728 or 73%
.
(10.10)
170
.
Because the partial eta squared statistic is typically generated by computer
programs like SPSS and SAS, it gets reported more often than other
measures of treatment magnitude (e.g., omega squared, partial omega
squared, and eta squared). From this assessment, we can conclude that
our statistically significant
F
value accounts for about 73 percent of the
total variance. These results are conveniently depicted in Table 10.3.
10.10 PERFORMING USER-DEFINED (PLANNED)
COMPARISONS BY HAND
10.10.1 DESIGNATING THE COMPARISONS OF INTEREST
Our statistically significant
F
in our previous numerical example informs
us that there is a real difference between or among the five treatment
Table 10.3.
Summary table for one-way within-subjects design
Within-subjects effects
Source
SS
df
MS
F
η
2
18.65
∗
Symptom intensity (
A
)
123.85
4
30.96
0
.
73
Subject (
S
)
30.57
7
4.37
A
×
S
46.55
28
1.66
Total
200.97
39
Note:
Partial eta squared
=
SS
A
/
SS
T
-
SS
S
=
123.85/170.40
=
0.727.
∗
p
<
.05.
