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Fixed and Random Effects. Theory of Analysis of Variance Source of variationdfEMS Between treatmentsn-1  e 2 + k  t 2 Within treatmentsnk-n e2e2 Totalnk-1.

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Presentation on theme: "Fixed and Random Effects. Theory of Analysis of Variance Source of variationdfEMS Between treatmentsn-1  e 2 + k  t 2 Within treatmentsnk-n e2e2 Totalnk-1."— Presentation transcript:

1 Fixed and Random Effects

2 Theory of Analysis of Variance Source of variationdfEMS Between treatmentsn-1  e 2 + k  t 2 Within treatmentsnk-n e2e2 Totalnk-1 [  e 2 + k  t 2 ]/  e 2 = 1, if k  t 2 = 0

3 Setting Expected Mean Squares  The expected mean square for a source of variation (say X) contains.  the error term.  a term in  2 x.  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 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) |X A - X B |/sed[x] >= t p/2

12 Least Significant Difference |X A - X B |/sed[x] >= t p/2 LSD = t p/2 x sed[x] t 0.025 = 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 X E + 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 se[x] =  (  2 /n)  (94,773/4) = 153.9 W = 4.64 x 153.9 = 714.1 W = q(p,f) x se[x]

19 Tukey’s Multiple Range Test

20 Tukey’s Multiple Comparisons

21 Duncan’s Multiple Range Test

22

23

24

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  c i = 0  [c 1i x c 2i ] = 0  c i = 0 -1 -1 +1 +1 --  c i = 0  c i = 0 -1 +1 -1 +1 --  c i = 0  c i = 0 +1 -1 -1 +1 --  c i = 0

32 Calculating Orthogonal Contrasts d.f. (single contrast) = 1 S.Sq(contrast) = M.Sq = [  c i x Y i ] 2 /n  c i 2 ]

33 Orthogonal Contrasts - Example

34 S.Sq = [  c i x Y i ]/[n  c i 2 ] S.Sq(1) [(-1)64.1+(-1)76.6+(1)40.1+(1)47.8] 2 / n  c i 2 = 52.8 2 /(3 x 4) = 232.32

35 S.Sq(2) [(-1) x 64.1+(+1) x 76.6] 2 /(3x2) 26.04 S.Sq(3) [(-1) x 40.1+(+1) x 47.8] 2 /(3x2) 9.88

36 Orthogonal Contrasts

37  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 #2 Tukey’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  c i 2 130 2 /(4 x 40) = 140.8 S.Sq(2)= [(-1)130+(+1)124] 2 /n  c i 2 6 2 /(4 x 2) = 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  c i 2 130 2 /(4 x 40) = 140.8 S.Sq(2)= [(-1)130+(-1)124+(-1)141+(4)186+(-1)119] 2 /n  c i 2 230 2 /(4 x 20) = 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  c 1i = 0 (-3) + (-3) + (+2) + (+2) + (+2) = 0 =  c 2i = 0 (-1) + (-1) + (-1) + (+4) + (-1) = 0 =  [c 1i x c 2i ] = 0 (-3)(-1)+(-3)(-1)+2(-1)+2(4)+2(-1) =10 =  Orthogonality

49 More Appropriate Contrasts

50 Analysis of Variance

51 Conclusions  Almost all the variation between cultivars is accounted for by the difference between cv ‘D’ and the others.  The remaining 4 cultivars are not significantly different.  Orthogonal contrast result is exactly the same are the result from Tukey’s contrasts.

52 Conclusions  Important to make the “correct” orthogonal contrasts.  Important to make contrasts which have “biological sense”.  Orthogonal contrasts should be decided prior to analyses and not dependant on the data.

53 Orthogonal Contrasts  Four Brassica species (B. napus, B. rapa, B. juncea, and S. alba).  Ten cultivars ‘nested’ within each species.  Three insecticide treatments (Thiodan, Furidan, no insecticide).  Three replicate split-plot design.

54 Analysis of Variance

55 Species and Treatment Means

56 Orthogonal Contrasts

57

58 Analysis of Variance

59 Species x Treatment Interaction

60 Species x Contrast (1)

61 Species x Contrast (2)

62 More Orthogonal Contrasts … Trend Analyses

63 Aim of Analyses of Variance  Detect significant differences between treatment means.  Determine trends that may exist as a result of varying specific factor levels.

64 Example #4  Ten yellow mustard (S. alba) cultivars.  Five different nitrogen application rates (50, 75, 100, 125, and 150)

65 Analysis of Variance

66 Orthogonal Contrasts

67 Example #4

68

69

70 Analysis of Variance

71 Trend Analyses  The F-value associates with a trend contrast is significant.  All higher order trend contrasts are not significant.

72 Example #4

73 Linear

74 Quadratic

75 Cubic

76 Quartic

77 Example #5  Two carrot cultivars (‘Orange Gold’ and ‘Bugs Delight’.  Four seeding rates (1.5, 2.0, 2.5 and 3.0 lb/acre).  Three replicates.

78 Example #5

79 Analysis of Variance

80

81 Orange Gold Bug’s Delight

82 End of Analyses of Variance Section


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