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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 = .
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