Geoscience Reference
In-Depth Information
among means based on the same number (
n
) of observations, the sum of squares has one degree of
freedom and is computed as
ˆ
2
2
nQ
a
SS
=
∑
1
df
i
This sum of squares divided by the mean square for error provides an
F
test of the comparison.
Thus, in testing A and B vs. C, D, and E we have
ˆ
Q
=
3134
(.)
+
3144
(
.)
−
2116
(.)
−
2114
(
.)
−
2118
(.)
=
13 .
and
2
(.)
() () ()
5138
952 2
. 0
30
S
df
=
++−+−+−
=
=
31 74
.
3
3
2
2
2
33
2
2
2
1
Then, dividing by the error mean square gives the
F
value for testing the contrast:
31 74
148
.
F
=
=
21 446
.
ith1and 20 degrees of freedom
.
This exceeds the tabular value of
F
(4.35) at the 0.05 probability level. If this is the level at which we
decided to test, we would reject the hypothesis that the mean of treatments A and B does not differ
from the mean of treatments C, D, and E.
If
Q
is expressed in terms of the treatment totals rather than their means so that
(
)
+
(
)
+
ˆ
Qa Xa X
∑∑
T
=
…
1
1
2
2
then the equation for the single degree of freedom sum of squares is
ˆ
2
Q
na
T
SS
=
(
)
∑
2
1
df
i
The results will be the same as those obtained with the means. For the test of A and B vs. C, D, and E,
ˆ
Q
T
=
367372
() () () () ()
+
−
258257
−
−
259 9
=
And,
2
69
4761
1
S
df
=
=
=
31 74
.
, as before
2
2
2
2
2
50
53
++−+−+−
3
() () ()
2
2
2
1
Working with the totals saves the labor of computing means and avoids possible rounding errors.
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