1. CONTRAST ORTHOGONAL

2. Multiple comparisons Results from multiple comparisons
In general, if A = B, B = C, then A = C…
Remember that the situation here is probabilistic
Failure to reject H0 does not mean that they are equal
They are not sufficiently different for us to assert that they are different
Although we don’t have evidence that A and B differ or that B and C differ, we do have evidence that A and C differ A B C
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3. Linear contrasts
Pairwise comparisons of means are a special case of linear contrasts
Comparison of one group with another group with general weights
Linear contrasts take the form of a linear combination of the means
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4. Linear contrasts Restriction
Suppose we had three means and wanted to compare the first and the second only. This could be done by having a1= 1, a2= –1, and a3= 0
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5. Linear contrasts
Linear contrasts allow us to express the sum of squared differences between the means of sets of treatments
(∑aj Yj)2
or SS contrast =
n ∑ aj2
Formula assumes equal sample sizes
sama
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6. Example Suppose we had We can easily compute
Let’s say we wanted to compare the average of treatments 1 and 2 to treatment 3. Thus
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7. Example Now suppose we wanted to compare groups 1 and 2:
Note that the sum of the two SScontrast is
In this case, we can say that the contrasts completely partition SStreat
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8. F test for contrast
Note that the absolute value of the contrast weights does not matter
Means: Y1 Y2 Y3 Y4 Y5
aj:
The significance of a contrast can be tested with an F test
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9. Contrasts The square root of the F for a simple contrast
(a1 = 1, a2 = –1) is the same value obtained in a t test
t tests are special cases of linear contrasts
Note however that if you run several contrasts, the familywise error will be much larger than 
Bonferroni correction can be used
Run fewer contrasts! Only as many as needed
If they were really a priori, probably ok without correction
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10. Orthogonal contrasts ∑ bj = 0
Some contrasts are independent of one another, while others “share” information
Independent contrasts are called orthogonal
and
or ∑ajbj = a1b1+ a2b2 + a3b3 + …… + atbt = 0
contrasts given by aj and bj are orthogonal
∑ bj = 0
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11. Orthogonal contrasts
It is useful to have orthogonal contrasts because they exactly partition SStreat
SS(L1) + SS(L2) + …… + SS(Lt-1) = SS treat
However , it is not strictly necessary that all contrasts be orthogonal
But remember that in this case the contrasts do not convey independent information
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12. Degrees of freedom For a contrast, we have
What are the numerator degrees of freedom?
Consider df for SScontrast
A contrast always compares two quantities
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13. Degrees of freedom Thus SScontrast has df = 1
Another way to think of the df is that the F for a contrast can in fact be written in the usual way
To determine dfR, we must determine the number of independent parameters in the restricted model which is associated with the contrast
Consider the following null hypothesis:
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14. Degrees of freedom The corresponding restricted model is
where 1/31+1/32 + 1/33 – 4 = 0
Model has 4 parameters but only 3 are independent
In the general case for a groups, we would have a – 1 independent parameters
Thus
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15. When unequal replication
Avarage
A (3)
B (2)
C (2)
D (3)
A vs BCD a b c d
B vs CD f g h
C vs D i J
Remember orthogonal:
and
or ∑ajbj = a1b1+ a2b2 + a3b3 + …… + atbt = 0
contrasts given by aj and bj are orthogonal
∑ bj = 0,
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16. When unequal replication
Avarage
A (3)
B (2)
C (2)
D (3)
A vs BCD a b c d
B vs CD f g h
C vs D i J
i) 0a + 0b + 2i + 3j = 0 then i = -3 and j= 2
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17. When unequal replication
Avarage
A (3)
B (2)
C (2)
D (3)
A vs BCD a b c d
B vs CD f g h
C vs D i J -3 2
i) 2i + 3j = 0 then i = -3 and j= 2
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18. When unequal replication
Avarage
A (3)
B (2)
C (2)
D (3)
A vs BCD a b c d
B vs CD f g h
C vs D i J -3 2
0+ 2f + 2g + 3h = 0
0+ 2f + 2h + 3h = 0
2f = -5h
h=g = 2
f = -5
g(-3) + 3h(2) = 0
6g = 6h  g=h
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19. When unequal replication
Avarage
A (3)
B (2)
C (2)
D (3)
A vs BCD a b c d
B vs CD f g h -5 2
C vs D i J -3
0+ 2f + 2g + 3h = 0
0+ 2f + 2h + 3h = 0
2f = -5h
h=g = 2
f = -5
g(-3) + 3h(2) = 0
6g = 6h  g=h
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20. When unequal replication
Avarage
A (3)
B (2)
C (2)
D (3)
A vs BCD a b c d -7 3
B vs CD f g h -5 2
C vs D i J -3
0+ 2f + 2g + 3h = 0
0+ 2f + 2h + 3h = 0
2f = -5h
h=g = 2
f = -5
g(-3) + 3h(2) = 0
6g = 6h  g=h
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21. Soal latihan
A=15
B=10
A=12
D=11
C=7
A=14
C=8
B=7
B=5
C=9
D=10
D=8
Jika A = pupuk SP36, B = pupuk rock phospate, C = pupuk guano, D = pupuk TSP
Tunjukkan
Apakah pupuk buatan lebih baik daripada pupuk alam
Apakah pupuk guano berbeda nyata dengan rock phospate
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22. Soal Latihan Buatlah analisis variannya (α = 5%)
Pupuk
Ulangan
Simpangan baku
Rerata
Tanpa 6 3 5
Pupuk akar 4 7
Pupuk daun Gandasil B 13
Pupuk daun Vitabloom 11
Buatlah analisis variannya (α = 5%)
Apakah pemupukan bermanfaat?
Apakah pemupukan lewat akar lebih bermanfaat dari pemupukan lewat daun?
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