Transcript Contrasts - Oregon State

The Purpose of an Experiment
 We usually want to answer certain questions
posed by the objectives of the experiment
 Irrigation experiment (a 2 x 2 factorial)
– 1 cm/ha applied early(m1)
– 1 cm/ha applied late (m2)
– 2 cm/ha applied early(m3)
– 2 cm/ha applied late (m4)
Questions we might ask:
 Is there a yield difference between 2 cm/ha and
1 ha-cm of water applied?
(m1 + m2)/2
vs
(m3 + m4)/2
 Does it make a difference whether the water is
applied early or late?
(m1 + m3)/2
vs
(m2 + m4)/2
 Does the difference between 1 and 2 cm/ha
depend on whether the water is applied early or
late?
(m1 - m2)
vs
(m3 - m4)
To test the hypothesis
 We would test the hypothesis that the grouped
means are equal or said another way - that the
difference between the two groups is zero.
(m1 + m2)/2
=
(m3 + m4)/2
or
[(m1 + m2)/2] - [(m3 + m4)/2] = 0
Contrasts of Means
 Think of the contrast as a linear function of the means
L = k1m1 + k2m2 + . . . ktmt
 H0: L = 0
Ha: L0
 This would be a legitimate contrast if and only if:
 To estimate the contrast:
Skj = 0
L  k1 Y1 + k 2 Y 2 + ...k t Y t
 The variance of a contrast
V(L) = (Skj2)/r * MSE for equal replication
V(L) = (k12/r1 + k22/r2 + . . . kt2/rt )*MSE for unequal replication
 Interval estimate of a contrast
L(L) = L+ ta
V(L)
Orthogonal Contrasts
 With t treatments, it is possible to form t-1 contrasts that
are statistically independent of each other (i.e., one
contrast conveys no information about the other)
 In order to be statistically independent, they must be
orthogonal
 Contrasts are orthogonal if and only if the sum of the
products of the coefficients equals 0
κ
1j
κ 2j  0
j
Where j = the jth mean in the linear contrast
For Example:
 A 2 factor factorial: P(2 levels), K(2 levels)
–
–
–
–
no P, no K
20 kg/ha P, no K
no P, 20 kg/ha K
20 kg/ha P, 20 kg/ha K
(P0K0)
(P1K0)
(P0K1)
(P1K1)
 Questions we might ask
– Any difference between P and K when used alone?
– Any difference when the two fertilizers are used alone
versus together?
– Any difference when there is no fertilizer versus when
there is some?
Table of Contrasts
Contrast
P0K0
P1K0
P0K1
P1K1
Sum
P vs K
0
+1
-1
0
0
Alone vs Together
0
-1
-1
+2
0
None vs Some
-3
+1
+1
+1
0
Test for Orthogonality
(0x0) + (1x-1) + (-1x-1) + (0x2) = 0 - 1 + 1 + 0 =
0
(0x-3) + (1x1) + (-1x1) + (0x1) = 0 + 1 - 1 + 0 =
0
(0x -3) + (-1 x 1) + (-1 x 1) + (2 x 1) = 0 - 1 - 1 + 2 =
0
Or another set:
Contrast
P0K0
P1K0
P0K1
P1K1
Sum
P (Main Effect)
-1
+1
-1
+1
0
K (Main Effect)
-1
-1
+1
+1
0
PK (Interaction)
+1
-1
-1
+1
0
(-1 x -1) + (1 x -1) + (-1 x 1) + (1 x 1) = 1 - 1 - 1 + 1 =
0
(-1 x 1) + (1 x -1) + (-1 x -1) + (1 x 1) = -1 - 1 + 1 + 1 =
0
(-1 x 1) + (-1 x -1) + (1 x -1) + (1 x 1) = -1 + 1 - 1 + 1 =
0
Test for Orthogonality
Here is another example:
 A fertilizer experiment with 5 treatments:
–
–
–
–
–
C
PB
PS
NPB
NPS
=
=
=
=
=
Control, no fertilizer
Banded phosphate
Broadcast (surface) phosphate
Nitrogen with banded phosphate
Nitrogen with broadcast phosphate
 We might want to answer the following:
–
–
–
–
Is there a difference between fertilizer and no fertilizer?
Does the method of P application make a difference?
Does added N make a difference?
Does the effect of N depend on the method of P application?
Table of Contrasts:
Contrast
C
PB
PS NPB NPS Sum
None vs some
-4
+1
+1
+1
+1
0
Band vs broadcast 0
+1
-1
+1
-1
0
N vs no N
0
-1
-1
+1
+1
0
N vs (B vs S)
0
-1
+1
+1
-1
0
Test for Orthogonality
(-4x0) + (1x1) + (+1x-1) + (1x1) + (1x-1) = 0+1-1+1-1=
(-4x0) + (1x-1) + (1x-1) + (1x1) + (1x1) = 0-1-1+1+1=
(-4x0) + (1x-1) + (1x1) + (1x1) + (1x-1) = 0-1+1+1-1=
(0x0) + (1x-1) + (-1x-1) + (1x1) + (-1x1) = 0-1+1+1-1=
(0x0) + (1x-1) + (-1x1) + (1x1) + (-1x-1) = 0-1-1+1+1=
(0x0) + (-1x-1) + (-1x1) + (1x1) + (1x-1) = 0+1-1+1-1=
0
0
0
0
0
0
Contrast of Means
 We can partition the treatment SS into SS for
contrasts with the following conditions:
– Treatments are equally replicated
– t = number of treatments
– r = number of replications per treatment
– Y j = mean of yields on the j-th treatment
(
– SST  r  j Y j - Y
)
2
 Contrast SS will add up to SST (if you have a complete
set of t-1 orthogonal contrasts)
 It is not necessary to have a complete set of contrasts
provided that those in your subset are orthogonal to
each other
Contrast of Means
Under these conditions
– SSL  MSL 
r (  k jYj )
 kj
2
2
r * L2

2
 kj
– V(L) = [(Skj2)/r] * MSE for equal replication
Caution – old notes, old homework, and old exams
– Calculations are based on totals
2
(  κ jTj )  L2
– SSL = MSL =
r  κ 2j
r  κ 2j
– V(L) = (rSkj2) * MSE for equal replication
– Don’t use these formulas when calculations are based on means
Drawing the contrasts - Not always easy
 Divide and conquer
Between group comparison
Number
1
Set 1
Set 2
g1,g2,g3 g4,g5
Single df contrast
2(G1+G2+G3)-3(G4+G5)
2
g1
g2,g3
2G1-(G2+G3)
3
g2
g3
G2-G3
4
g4
g5
G4-G5
 Note that the coefficients could also be fractions
– For example
1
1
( G1 + G2 + G3 ) - ( G4 + G5 )
3
2
Revisiting the PxK Experiment
Treatment
Means (3 reps)
P0K0
P1K0
P0K1
P1K1
12
16
14
17
Contrast
P0K0 P1K0 P0K1 P1K1
SST=44.25
L
SS(L)
P vs K
0
+1
-1
0
2
6.00
Alone vs Together
0
-1
-1
+2
4
8.00
-3
+1
+1
+1
11
30.25
None vs Some
SS(Li)= r*Li2 / Sj kij2
44.25
Another example
 A weed scientist wanted to study the effect of a new, all-
purpose herbicide to control grassy weeds in lentils. He
decided to try the herbicide both as a preemergent and
as a postemergent spray. He also wanted to test the
effect of phosphorus.
–
–
–
–
–
–
–
Control
Hand weeding
Preemergent spray
Postemergent spray
Hand weeding + phosphorus
Preemergent spray + phosphorus
Postemergent spray + phosphorus
C
W1
W2
W3
PW1
PW2
PW3
Treatment Means
Treatment
I
II
III
Mean
C
218
180
192
196.7
W1
357
353
345
351.7
W2
325
311
297
311.0
W3
321
297
291
303.0
PW1
462
458
399
439.7
PW2
407
409
381
399.0
PW3
410
392
362
388.0
Mean
357.1
342.9
323.9
341.3
ANOVA
Source
df
Total
20
121,670.29
Block
2
3,903.72
1,951.86
11.86
Treatment
6
115,792.29
19,298.72
117.30**
12
1,974.28
164.52
Error
SS
MS
F
 Treatments are highly significant so we can
divide into contrasts
Control
Hand weeding
Preemergent spray
Postemergent spray
Hand weeding + P
Preemergent spray + P
Postemergent spray + P
Orthogonal Contrasts
C
W1
W2
Contrast 196.7 351.7 311
W3
C
W1
W2
W3
PW1
PW2
PW3
PW1 PW2 PW3
303 439.7 399
388
L
SS(L)
F
1
-6
1
1
1
1
1
1
2
0
-1
-1
-1
1
1
1
261
3
0
2
-1
-1
2
-1
-1
181.7
4
0
0
-1
1
0
-1
1
-19
270.75
1.64ns
5
0
-2
1
1
+2
-1
-1
3
2.25
0.01ns
6
0
0
1
-1
0
-1
1
-3
6.75
0.04ns
1 = Some vs none
2 = P vs no P
3 = Hand vs Chemical
1012.3 73201.34 444.94**
34060.50 207.03**
8250.70
4 = pre vs post emergence
5 = Interaction 2 x 3
6 = Interaction 2 x 4
50.15**
So What Does This Mean?
 The treated plots outyielded the untreated check plots
 Fertilized plots outyielded unfertilized plots by an
average of about 87 kg/plot
 Hand weeding resulted in higher yield than did
herbicide regardless of when applied by about 45
kg/plot
 No difference in effect between preemergence and
postemergence application of the herbicide
 The differences in weed control treatments did not
depend on the presence or absence of phosphorus
fertilizer
More on Analysis of Experiments with Complex
Treatment Structure
 Augmented Factorials
– Balanced factorial treatments plus controls
 Incomplete Factorials
– Some treatment combinations left out (not of interest)
 Simultaneous Confidence Intervals
– Multivariate T tests for multiple contrasts
– Better control of Type I error (familywise error or FWE)
– Orthogonality is not essential
 References
– Marini, 2003. HortSci. 38:117-120.
– Piepho, Williams, & Fleck, 2006. HortSci. 41:117-120.
– Schaarschmidt & Vaas, 2009. HortSci. 44:188-195.