Presentation on theme: "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)"— Presentation transcript:
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 = t t H 0 : L = 0 H a : L 0 This would be a legitimate contrast if and only if: j = 0 To estimate the contrast: L(L) = L+ t V(L) The variance of a contrast V(L) = ( k j 2 )/r * MSE for equal replication V(L) = (k 1 2 /r 1 + k 2 2 /r 2 +... k t 2 /r t )*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 Where j = the j th mean in the linear contrast
For Example: A 2 factor factorial: P(2 levels), K(2 levels) –no P, no K (P 0 K 0 ) –20 kg/ha P, no K (P 1 K 0 ) –no P, 20 kg/ha K(P 0 K 1 ) –20 kg/ha P, 20 kg/ha K (P 1 K 1 ) 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 ContrastP 0 K 0 P 1 K 0 P 0 K 1 P 1 K 1 Sum P vs K0+1-100 Alone vs Together0-1-1+20 None vs Some-3+1+1+10 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: ContrastP 0 K 0 P 1 K 0 P 0 K 1 P 1 K 1 Sum P (Main Effect)-1+1-1+10 K (Main Effect)-1-1+1+10 PK (Interaction)+1-1-1+10 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: ContrastCPBPSNPBNPSSum None vs some-4+1+1+1+10 Band vs broadcast 0+1-1+1-10 N vs no N 0-1-1+1+10 N vs (B vs S) 0-1+1+1-10 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 – 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) = [( j 2 )/r] * MSE for equal replication Caution – old notes, old homework, and old exams – Calculations are based on totals – SSL = MSL = – V(L) = (r j 2 ) * MSE for equal replication – Don’t use these formulas when calculations are based on means
Drawing the contrasts - Not always easy Between group comparison NumberSet 1Set 2Single df contrast 1g1,g2,g3g4,g52(G1+G2+G3)-3(G4+G5) 2g1g2,g32G1-(G2+G3) 3g2g3G2-G3 4g4g5G4-G5 Divide and conquer Note that the coefficients could also be fractions –For example
Revisiting the PxK Experiment ContrastP 0 K 0 P 1 K 0 P 0 K 1 P 1 K 1 LSS(L) P vs K 0+1-1 026.00 Alone vs Together 0-1-1+248.00 None vs Some-3+1+1+11130.25 TreatmentP 0 K 0 P 1 K 0 P 0 K 1 P 1 K 1 Means (3 reps)12161417 SST=44.25 44.25 SS(L i )= r*L i 2 / j k ij 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. –ControlC –Hand weedingW1 –Preemergent sprayW2 –Postemergent sprayW3 –Hand weeding + phosphorusPW1 –Preemergent spray + phosphorusPW2 –Postemergent spray + phosphorusPW3
ANOVA SourcedfSSMSF Total20121,670.29 Block23,903.721,951.8611.86 Treatment6115,792.2919,298.72117.30** Error121,974.28164.52 Treatments are highly significant so we can divide into contrasts
Orthogonal Contrasts CW1W2W3PW1PW2PW3 Contrast196.7351.7311303439.7399388LSS(L)F 1-61111111012.373201.34444.94** 2 0-1-1-111126134060.50207.03** 3 02-1-12-1-1181.78250.7050.15** 4 00-110-11-19270.751.64ns 5 0-211+2-1-132.250.01ns 6 001-10-11-36.750.04ns 1 =Some vs none 2 =P vs no P 3 =Hand vs Chemical 4 =pre vs post emergence 5 =Interaction 2 x 3 6 =Interaction 2 x 4 ControlC Hand weedingW1 Preemergent sprayW2 Postemergent sprayW3 Hand weeding + PPW1 Preemergent spray + PPW2 Postemergent spray + PPW3
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.