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Fixed and Random Effects

Fixed and Random Effects. Theory of Analysis of Variance. [ e 2 + k  t 2 ]/ e 2 = 1, if k  t 2 = 0. Setting Expected Mean Squares. The expected mean square for a source of variation (say X) contains. the error term. a term in  2 x.

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Fixed and Random Effects

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  1. Fixed and Random Effects

  2. Theory of Analysis of Variance [e2 + kt2]/e2 = 1, if kt2 = 0

  3. Setting Expected Mean Squares • The expected mean square for a source of variation (say X) contains. • the error term. • a term in 2x. • a variance term for other selected interactions involving X.

  4. Coefficients for EMS Coefficient for error mean square is always 1 Coefficient of other expected mean squares is # reps times the product of factors levels that do not appear in the factor name.

  5. Expected Mean Squares • Which interactions to include in an EMS? • All the factors appear in the interaction. • All the other factors in the interaction are Random Effects.

  6. Pooling Sums of Squares

  7. Multiple Comparisons

  8. Multiple Comparisons • Multiple Range Tests: • t-tests and LSD’s; • Tukey’s and Duncan’s. • Orthogonal Contrasts.

  9. One-way Analysis of Variance

  10. Means and Rankings

  11. Multiple t-Test sed[x] = (22/n) (2 x 94,773/4) |XA - XB|/sed[x] >= tp/2

  12. Least Significant Difference |XA - XB|/sed[x] >= tp/2 LSD = tp/2 x sed[x] t0.025 = 2.518 LSD = 2.518 x 217.7 = 548.2

  13. Least Significant Difference Say one of the cultivars (E) is a control check and we want to ask: are any of the others different from the check? LSD = 2.518 x 217.7 = 548.2 XE+ LSD 1796 + 548.2 = 2342.2 to 1247.80

  14. Means and Rankings Range = 1796 + 548.2 = 2342.2 to 1247.80

  15. Multiple LSD Comparisons

  16. Lower Triangular Form

  17. LSD Multiple Comparisons

  18. Tukey’s Multiple Range Test W = q(p,f) x se[x] se[x] = (2/n) (94,773/4) = 153.9 W = 4.64 x 153.9 = 714.1

  19. Tukey’s Multiple Range Test

  20. Tukey’s Multiple Comparisons

  21. Duncan’s Multiple Range Test

  22. Duncan’s Multiple Range Test

  23. Duncan’s Multiple Range Test

  24. Duncan’s Multiple Range Test

  25. Duncan’s Multiple Comparisons

  26. Orthogonal Contrasts

  27. AOV Orthogonal Contrasts

  28. Tukey’s Multiple Range Test

  29. Consider that cultivars A and B were developed in Idaho and C and D developed in California • Do the two Idaho cultivars have the same yield potential? • Do the two California cultivars have the same yield potential? • Are Idaho cultivars higher yielding than California cultivars?

  30. Analysis of Variance

  31. Orthogonality ci = 0 [c1i xc2i] = 0 -1 -1 +1 +1 -- ci = 0 -1 +1 -1 +1 -- ci = 0 +1 -1 -1 +1 -- ci = 0

  32. Calculating Orthogonal Contrasts d.f. (single contrast) = 1 S.Sq(contrast) = M.Sq = [ci x Yi]2/nci2]

  33. Orthogonal Contrasts - Example

  34. S.Sq = [cix Yi]/[n ci2] S.Sq(1) [(-1)64.1+(-1)76.6+(1)40.1+(1)47.8]2/ n ci2 = 52.82/(3x4) = 232.32

  35. S.Sq(2) [(-1)x64.1+(+1) x 76.6]2/(3x2) 26.04 S.Sq(3) [(-1)x40.1+(+1) x 47.8]2/(3x2) 9.88

  36. Orthogonal Contrasts

  37. Orthogonal Contrasts • Five dry bean cultivars (A, B, C, D, and E). • Cultivars A and B are drought susceptible. • Cultivars C, D and E are drought resistant. • Four Replicate RCB, one location • Limited irrigation applied.

  38. Analysis of Variance

  39. Orthogonal Contrast Example #2Tukey’s Multiple Range Test

  40. Orthogonal Contrasts • Is there any difference in yield potential between drought resistant and susceptible cultivars? • Is there any difference in yield potential between the two drought susceptible cultivars? • Are there any differences in yield potential between the three drought resistant cultivars?

  41. Orthogonal Contrasts

  42. S.Sq(1)=[(-3)130+(-3)124+(2)141+(2)186+(2)119]2 /nci2 1302/(4x40) = 140.8 S.Sq(2)=[(-1)130+(+1)124]2 /nci2 62/(4x2) = 4.5 S.Sq(Rem) = S.Sq(Cult)-S.Sq(1)-S.Sq(2) 728.2-140.8-4.5 = 582.9 (with 2 d.f.)

  43. Analysis of Variance

  44. Partition Contrast(rem)

  45. Analysis of Variance

  46. Alternative Contrasts !!!!

  47. S.Sq(1)=[(-3)130+(-3)124+(2)141+(2)186+(2)119]2 /nci2 1302/(4x40) = 140.8 S.Sq(2)=[(-1)130+(-1)124+(-1)141+(4)186+(-1)119]2 /nci2 2302/(4x20) = 661.2 S.Sq(Rem) = S.Sq(Cult)-S.Sq(1)-S.Sq(2) 728.2-140.8-661.2 = -73.8 (Oops !!!) (with 2 d.f.)

  48. Orthogonality c1i = 0 (-3) + (-3) + (+2) + (+2) + (+2) = 0 =  c2i = 0 (-1) + (-1) + (-1) + (+4) + (-1) = 0 =  [c1i x c2i] = 0 (-3)(-1)+(-3)(-1)+2(-1)+2(4)+2(-1) =10 =

  49. More Appropriate Contrasts

  50. Analysis of Variance

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