Geoscience Reference
In-Depth Information
5
∑
(
2
Blocktotals
No.ofplots p
)
er block
2
2
2
−=
+++
−= −
60
59
…
49
4.
Block
S
df
=
CT
CT
3675 5
.
3645
=
30 .
4
4
5.
Error
SS
=
Total
SS
−
Clone
SS
−
Block
S
df
=
45.5
12
df
19
df
3
df
4
Note that in obtaining the error
SS
by subtraction, we get a partial check on ourselves by subtracting
clone and block
df
's from the total
df
to see if we come out with the correct number of error
df
. If
these don't check, we have probably used the wrong sums of squares in the subtraction.
Mean squares are again calculated by dividing the sums of squares by the associated number of
degrees of freedom. Tabulating the results of these computations
Source of Variation
Degrees of Freedom
Sums of Squares
Mean Squares
Blocks
4
30.5
7.625
Clones
3
45.0
15.000
Error
12
45.5
3.792
Total
19
121.0
F
for clones is obtained by dividing the clone means square by the error mean square. In this case
F
= 15.000/3792 = 3.956. As this is larger than the tabular
F
of 3.49 (obtained from a distribution
of
F
table) (
F
0.05
with 3 and 12 degrees of freedom) we conclude that the difference between clones
is significant at the 0.05 level. The significance appears to be due largely to the low value of C as
compared to A, B, and D.
Comparisons among clone means can be made by the methods previously described. For exam-
ple, to test the prespecified (i.e., before examining the data) hypothesis that there is no difference
between the mean of clone C and the combined average A, B, and D we would have:
(
)
2
53
CABD
−−−
2
510
12
(
−
)
S
df
for(ABDvs. C)
++ =
+− +−
=
=
41 667
.
2
2
2
2
3
+(
1
() ()
1
1
1
Then,
41 667
3 792
.
F
=
=
10 988
.
.
Tabula r
F
at the 0.01 level with 1 and 12 degrees of freedom is 9.33. As calculated
F
is greater than
this, we conclude that the difference between C and the average of A, B, and D is significant at the
0.01 level.
The sum of squares for this single-degree-of-freedom comparison (41.667) is almost as large
as that for clones (45.0) with three degrees of freedom. This result suggests that most of the clonal
variation is attributable to the low value of C, and that comparisons between the other three means
are not likely to be significant.
DID YOU KNOW?
With only two treatments, the analysis of variance of a randomized block design is equiva-
lent to the
t
test of paired replicates. The value of
F
will be equal to the value of
t
2
and the
inferences derived from the tests will be the same. The choice of tests is a matter of personal
preference.
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