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

– 1 cm/ha applied late

– 2 cm/ha applied early

– 2 cm/ha applied late

Questions we might ask: Is there a yield difference between 2 cm/ha and

1 ha-cm of water applied?vs

Does it make a difference whether the water is applied early or late?vs

Does the difference between 1 and 2 cm/ha depend on whether the water is applied early or late? vs

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.

=

or

=

Contrasts of Means Think of the contrast as a linear function of the means

L = tt

H0: L = 0 Ha: L0

This would be a legitimate contrast if and only if: j = 0

To estimate the contrast: 1 21 2 ... ttL k Y k Y k Y

L(L) = L+ t V(L)

The variance of a contrastV(L) = (kj

2)/r * MSE for equal replication

V(L) = (k12/r1 + k2

2/r2 + . . . kt2/rt )*MSE for unequal replication

Interval estimate of a contrast

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

 

0κκ 2jj

1j

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 (P0K0)– 20 kg/ha P, no K (P1K0)– no P, 20 kg/ha K (P0K1)– 20 kg/ha P, 20 kg/ha K (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

Test for Orthogonality

(-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

Here is another example: A fertilizer experiment with 5 treatments:

– C = Control, no fertilizer– PB = Banded phosphate– PS = Broadcast (surface) phosphate– NPB = Nitrogen with banded phosphate– NPS = 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= 0

(-4x0) + (1x-1) + (1x-1) + (1x1) + (1x1) = 0-1-1+1+1= 0

(-4x0) + (1x-1) + (1x1) + (1x1) + (1x-1) = 0-1+1+1-1= 0

(0x0) + (1x-1) + (-1x-1) + (1x1) + (-1x1) = 0-1+1+1-1= 0

(0x0) + (1x-1) + (-1x1) + (1x1) + (-1x-1) = 0-1-1+1+1= 0

(0x0) + (-1x-1) + (-1x1) + (1x1) + (1x-1) = 0+1-1+1-1= 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– = mean of yields on the j-th treatment

–

jY

2

jjSST r Y Y

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

–

– V(L) = [(j2)/r] * MSE for equal replication

22

j j

2 2j j

r Y r *LSSL MSL

Caution – old notes, old homework, and old exams

– Calculations are based on totals

– SSL = MSL =

– V(L) = (rj2) * MSE for equal replication

L

22

j j

2 2j j

κ T

r κ r κ

Drawing the contrasts - Not always easy

Between group comparison

Number Set 1 Set 2 Single df contrast

1 g1,g2,g3 g4,g5 2(G1+G2+G3)-3(G4+G5)

2 g1 g2,g3 2G1-(G2+G3)

3 g2 g3 G2-G3

4 g4 g5 G4-G5

Divide and conquer

Revisiting the PxK Experiment

Contrast P0K0 P1K0 P0K1 P1K1 L SS(L)

P vs K 0 +1 -1 0 2 6.00

Alone vs Together 0 -1 -1 +2 4 8.00

None vs Some -3 +1 +1 +1 11 30.25

Treatment P0K0 P1K0 P0K1 P1K1

Means (3 reps) 12 16 14 17SST=44.25

44.25SS(Li)= r*Li2 / j kij

2

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 C– Hand weeding W1– Preemergent spray W2– Postemergent spray W3– Hand weeding + phosphorus PW1– Preemergent spray + phosphorus PW2– Postemergent spray + phosphorus 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 SS MS F

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**

Error 12 1,974.28 164.52

Treatments are highly significant so we can divide into contrasts

Orthogonal Contrasts

C W1 W2 W3 PW1 PW2 PW3

Contrast 196.7 351.7 311 303 439.7 399 388 L SS(L) F

1 -6 1 1 1 1 1 1 1012.3 73201.34 444.94**

2 0 -1 -1 -1 1 1 1 261 34060.50 207.03**

3 0 2 -1 -1 2 -1 -1 181.7 8250.70 50.15**

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 none2 = P vs no P3 = Hand vs Chemical

4 = pre vs post emergence5 = Interaction 2 x 36 = Interaction 2 x 4

Control CHand weeding W1Preemergent spray W2Postemergent spray W3Hand weeding + P PW1Preemergent spray + P PW2Postemergent spray + P PW3

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