Information Technology Reference
In-Depth Information
Note that the denominator refers to the sum of squared coefficients,
which we will designate with a lower case
c
. Substituting our values into
the previous equation, we have the following:
86)
2
7(
−
613
.
SS
A
comp
1
=
(4)
2
1)
2
1)
2
1)
2
1)
2
+
−
+
−
+
−
+
−
(
(
(
(
2, 637, 768
.
60
=
20
=
131,888.43
.
(7.15)
The remaining procedures for computing degrees of freedom, mean
squares, and
F
are straightforward. There is only 1
df
in this comparison
(even though there are five treatment groups) because we are comparing
the zero months of study group with the combined two to eight months
of study groups. Hence,
df
A
comp
1
=
1 ,
SS
A
comp
1
131, 888
.
43
MS
A
comp
1
=
=
=
131,888.43
.
1
Thus,
MS
A
comp
MS
S
/
A
=
.
131, 888
43
F
A
comp
1
=
=
99.48
.
(7.16)
1, 325
.
73
This
F
value is evaluated with 1 and
df
S
/
A
(which is 30 in the present
example). Since our observed
F
of 99.48 is greater than the critical value of
F
, at the .05 level (critical value
4.17), we conclude that this statistically
significant
F
indicates that students in the combined study conditions
scored higher on the SAT then did the no-study students.
=
7.27.2 HYPOTHESIS 2
The second hypothesis contrasts students who studied for two months
with those who prepared for four months. The coefficients for this com-
parison are as follows:
01
−
100
.
Thus,
ˆ
ψ
2
=
(
c
1
)(
Y
1
)
+
(
c
2
)(
Y
2
)
+
(
c
3
)(
Y
3
)
+
(
c
4
)(
Y
4
)
+
(
c
5
)(
Y
5
)
=
(0)(412
.
86)
+
(1)(474
.
29)
+
(
−
1)(552
.
86)
+
(0)(614
.
29)
+
(0)(623
.
86)
= −
78.57
(7.17)
