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these two groups only. Hence, our
a
8. We can now proceed
computing our sum of squares, degrees of freedom, mean squares, and
F
for this pairwise comparison.
=
2and
n
=
10.10.2 COMPUTATIONS FOR A WITHIN-GROUP PAIRWISE
COMPARISON
A
2
n
T
2
(
a
)(
n
)
=
(73)
2
(41)
2
(114)
2
(2)(8)
SS
A
comp
=
+
−
−
=
64.00
8
S
2
a
T
2
(
a
)(
n
)
=
SS
s
−
(17)
2
(14)
2
(15)
2
(12)
2
(13)
2
(16)
2
(16)
2
(11)
2
+
+
+
+
+
+
+
=
2
(114)
2
(2)(8)
−
=
15.75
A
2
n
S
2
a
T
2
(
a
)(
n
)
Y
aj
−
−
SS
A
comp
×
S
=
+
=
912
−
876
.
25
−
828
+
812
.
25
=
20.00
T
2
(
a
)(
n
)
=
Y
aj
−
SS
T
=
912
−
812
.
25
=
99.75
df
a
=
a
−
1
=
2
−
1
=
1 ,
df
s
=
n
−
1
=
8
−
1
=
7
df
A
comp
×
S
=
(
a
−
1)(
n
−
1)
=
(1)(7)
=
7 ,
df
T
=
(
a
)(
n
)
−
1
=
−
=
16
1
15
SS
A
comp
df
A
comp
=
64
00
1
.
MS
A
comp
=
=
64.00
.
SS
S
df
S
=
15
75
MS
S
=
=
2.25
7
SS
A
comp
×
S
df
A
comp
×
S
=
20
.
00
MS
A
comp
×−
S
=
=
2.86
7
MS
A
comp
MS
A
comp
×
S
=
64
.
00
F
A
comp
=
86
=
22.38
.
2
.
This observed
F
is evaluated at (
df
A
comp
,
df
A
comp
×
S
)
=
F
(1, 7). The critical
value at
05 is 5.59. Our observed
F
of 22.38 exceeds the critical value.
We therefore reject the null and conclude there is a statistically significant
difference between the two means. Specifically, symptom intensity was
significantly reduced between the first and last tests.
α
=
.