1 / 19

The Purpose of an Experiment

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 (m 1 ) 1 cm/ha applied late (m 2 ) 2 cm/ha applied early (m 3 ) 2 cm/ha applied late (m 4 ).

duman
Download Presentation

The Purpose of an Experiment

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. 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)

  2. 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)

  3. 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

  4. L(L) = L+ ta V(L) 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: Skj = 0 • To estimate the contrast: • 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

  5. 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 jth mean in the linear contrast

  6. 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?

  7. 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

  8. 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

  9. 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?

  10. Table of Contrasts: • Contrast C PB PS NPB NPSSum • 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

  11. 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

  12. Under these conditions • V(L) = [(Skj2)/r] * MSE for equal replication • Caution – old notes, old homework, and old exams • Calculations are based on totals • SSL = MSL = • V(L) = (rSkj2) * MSE for equal replication Contrast of Means

  13. Drawing the contrasts - Not always easy • Divide and conquer 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

  14. Revisiting the PxK Experiment Treatment P0K0 P1K0 P0K1 P1K1 Means (3 reps) 12 16 14 17 SST=44.25 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 44.25 SS(Li)= r*Li2 / Sj kij2

  15. 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

  16. 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 357.1342.9323.9341.3

  17. 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 ANOVA • Treatments are highly significant so we can divide into contrasts

  18. Control C • Hand weeding W1 • Preemergent spray W2 • Postemergent spray W3 • Hand weeding + P PW1 • Preemergent spray + P PW2 • Postemergent spray + P PW3 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 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

  19. 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 Related