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9.2 Linear Contrasts
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EXAMPLE 9.3
Various agents are used to control weeds in crops. Of particular concern is the
overusage of chemical agents. Although effective in controlling weeds, these
agents may also drain into the underground water system and cause health problems. Thus, several new biological weed agents have been proposed to eliminate
the contamination problem present in chemical agents. Researchers conducted a
study of biological agents to assess their effectiveness in comparison to the chemical weed agents. The study consisted of a control (no agent), two biological agents
(Bio1 and Bio2), and two chemical agents (Chm1 and Chm2). Thirty 1-acre plots
of land were planted with hay. Six plots were randomly assigned to receive one of
the five treatments. The hay was harvested and the total yield in tons per acre was
recorded. The data are given in Table 9.1.
TABLE 9.1
Summary statistics for
Example 9.3
Agent
1
2
3
4
5
Type
None
Bio1
Bio2
Chm1
Chm2
yi.
si
ni
1.175
.1204
6
1.293
.1269
6
1.328
.1196
6
1.415
.1249
6
1.500
.1265
6
Determine four orthogonal contrasts and demonstrate that the total of the four
sums of squares associated with the four contrasts equals the between samples
(Treatment) sum of squares.
Solution An analysis of variance was conducted on these data yielding the results
summarized in the AOV table given in Table 9.2.
TABLE 9.2
AOV table for Example 9.3
Source
df
SS
MS
F
p-value
Treatment
Error
Totals
4
25
29
.3648
.3825
.7472
.0912
.0153
5.96
.0016
From the AOV table, we have that SSTrt .3648. We will now construct four
orthogonal contrasts in the five treatment means and demonstrate that SSTrt can be
partitioned into four terms, each representing a 1-degree of freedom sum of
square associated with a particular contrast. Table 9.3 contains the coefficients and
sum of squares for each of the four contrasts.
TABLE 9.3
Treatment
Sum of squares computations
for weed control experiment
Contrast
1(Cntrl) 2(Bio1) 3(Bio2) 4(Chm1) 5(Chm2)
a1
a2
a3
a4
a5
Control vs. Agents
Biological vs. Chemical
Bio1 vs. Bio2
Chm1 vs. Chm2
4
0
0
0
1
1
1
0
1
1
1
0
1
1
0
1
1
1
0
1
yi.
1.175
1.293
1.328
1.415
1.500
5
a i1
20
4
2
2
a2i
lˆ
SSCi
.836
.294
.035
.085
.2097
.1297
.0037
.0217
.3648
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Chapter 9 Multiple Comparisons
To illustrate the calculations involved in Table 9.3, we will compute the sum
of squares associated with the first contrast, control versus agents. First, note that
the contrast represents a comparison of the yield for the control treatment versus
the average yield of the four active agents. We initially would have written this
contrast as
l m1 (m2 m3 m4 m5)
4
1
4 m
(1) m1 2
1
1
4 m 4 m
3
4
1
4 m
5
However, we can multiply each coefficient by 4 and change the coefficients from
a1 1
a2 1
4
a3 1
4
a4 1
4
a5 1
4
to
a1 4
a2 1
a3 1
a4 1
a5 1
Next, we calculate
5
2
2
2
2
2
2
a ai (4) (1) (1) (1) (1) 20
i 1
and
lˆ (4)(1.175) (1)(1.293) (1)(1.328) (1)(1.415) (1)(1.500)
.836
Finally, we can obtain the sum of squares associated with the contrast from
SSC1 (lˆ)2
n(lˆ)2
6(.836)2
.2097
5
2
5
2
a i 1(ai ni)
a i 1ai
20
The remaining three sums of squares are calculated in a similar fashion. From
Table 9.3, we thus obtain
SSC1 SSC2 SSC3 SSC4 .2097 .1297 .0037 .0217 .3648 SSTrt
EXAMPLE 9.4
Refer to Example 9.3. Verify that the four contrasts in Table 9.3 are mutually
orthogonal.
Identify the four contrasts in Table 9.3 by lˆ1 is Control vs. Agents, lˆ2 is
Biological vs. Chemical, lˆ3 is Bio1 vs. Bio2, and lˆ4 is Chm1 vs. Chm2. Note that the
sample sizes are equal, so we need to verify that a 5i 1 ai bi 0 for the six pairs of
contrasts. (See Table 9.4.)
Solution