Information Technology Reference
In-Depth Information
the null hypothesis is true, the researchers will reject the null hypothesis
and conclude that the means of the groups are significantly different.
That is, they will conclude that the independent variable accounts for
significantly more than 0 percent of the variance of the dependent variable.
The Critical Values table for the
F
distribution is in Appendix C. In the
present case, we have 4
df
between groups and 30
df
within groups. The
table reveals a critical value of
F
at the .05 level to be 2.69. Since our
observed (or computed)
F
of 43.47 exceeds the value for
F
corresponding
to our preset alpha level of .05, we can conclude that our obtained
F
value
is statistically significant at the .05 alpha level, indicating a statistically
significant difference between the five groups.
SPSS and SAS have access to these critical values and make these
assessments automatically. For example, when we present the SPSS output
for this analysis, we will see a column labeled
Sig
with a value .000. This
value is one that SPSS has truncated down to three decimal places. We
interpret this outcome to mean that this
F
value equals or exceeds the
tabled critical value of
F
at the .001 alpha level and would therefore
ordinarily occur less than once in every thousand occasions if the null
hypothesis is true. Please note that a
p
value can never actually be .000! A
value of zero would indicate that such an
F
value would never occur and
that there is a zero chance of the researcher committing a Type I error in
such an instance! Please remember that there is always some chance that
the researcher's results are incorrect.
When reporting the
F
ratio, we need to add a probability element to
the
F
value and its degrees of freedom. In the example analysis, using an
alpha level of .05, we would write
F
(4, 30)
=
43
.
47,
p
<.
05
.
6.5.7 ETA SQUARED
We know from the
F
ratio being statistically significant that the indepen-
dent variable can explain some of the variance of the dependent variable.
To determine what this percentage is, that is, to determine the strength of
the effect of the independent variable, we must calculate the value of eta
squared as discussed in Section 4.5.3 of Chapter 4. As you may recall, the
formulaisasfollows:
sum of squares between groups
sum of squares total
Eta squared
=
.
(6.8)
For the example data set:
SS
A
SS
T
=
230,496.57
270,268.571
=
.
853
.
This eta squared value indicates that roughly 85 percent of the total vari-
ance of the dependent variable (SAT Verbal and Quantitative composite
score) can be attributed to the independent variable (preparation time)
