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ANOVA & sib analysis. basics of ANOVA - revision application to sib analysis intraclass correlation coefficient.

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Presentation on theme: "ANOVA & sib analysis. basics of ANOVA - revision application to sib analysis intraclass correlation coefficient."— Presentation transcript:

1 ANOVA & sib analysis

2 basics of ANOVA - revision application to sib analysis intraclass correlation coefficient

3 - analysis of variance (ANOVA) is a way of comparing the ratio of systematic variance to unsystematic variance in a study

4 - analysis of variance (ANOVA) is a way of comparing the ratio of systematic variance to unsystematic variance in a study ANOVA as regression

5 - analysis of variance (ANOVA) is a way of comparing the ratio of systematic variance to unsystematic variance in a study ANOVA as regression - research question: does exposure to content of Falconer & Mackay (1996) increase knowledge of quantitative genetics?

6 - analysis of variance (ANOVA) is a way of comparing the ratio of systematic variance to unsystematic variance in a study ANOVA as regression - research question: does exposure to content of Falconer & Mackay (1996) increase knowledge of quantitative genetics? Person Condition Nothing (N) Lectures (L)Lectures + book (LB) person person 2179 person 3168 person person 5157 μ N = 1μ L = 5μ LB = 9μ = 5

7 - analysis of variance (ANOVA) is a way of comparing the ratio of systematic variance to unsystematic variance in a study ANOVA as regression - research question: does exposure to content of Falconer & Mackay (1996) increase knowledge of quantitative genetics? Person Condition Nothing (N) Lectures (L)Lectures + book (LB) person person 2179 person 3168 person person 5157 μ N = 1μ L = 5μ LB = 9μ = 5 person score

8 - analysis of variance (ANOVA) is a way of comparing the ratio of systematic variance to unsystematic variance in a study ANOVA as regression - research question: does exposure to content of Falconer & Mackay (1996) increase knowledge of quantitative genetics? outcome ij = model + error ij Person Condition Nothing (N) Lectures (L)Lectures + book (LB) person person 2179 person 3168 person person 5157 μ N = 1μ L = 5μ LB = 9μ = 5 person score

9 Dummy coding:

10 Condition Dummy variable Dummy1 (lec) Dummy2 (lecbook) Nothing (N)00 Lectures (L)10 Lectures + book (LB)01 Dummy coding:

11 outcome ij = model + error ij Condition Dummy variable Dummy1 (lec) Dummy2 (lecbook) Nothing (N)00 Lectures (L)10 Lectures + book (LB)01 Dummy coding:

12 outcome ij = model + error ij knowledge ij = b 0 + b 1 *dummy1 j + b 2 *dummy2 j + ε ij i = 1, … N, N = number of people per condition = 5 j = 1, … M, M = number of conditions = 3 Condition Dummy variable Dummy1 (lec) Dummy2 (lecbook) Nothing (N)00 Lectures (L)10 Lectures + book (LB)01 Dummy coding:

13 outcome ij = model + error ij knowledge ij = b 0 + b 1 *dummy1 j + b 2 *dummy2 j + ε ij i = 1, … N, N = number of people per condition = 5 j = 1, … M, M = number of conditions = 3 knowledge i1 = b 0 + b 1 *dummy1 1 + b 2 *dummy2 1 + ε i1 Condition Dummy variable Dummy1 (lec) Dummy2 (lecbook) Nothing (N)00 Lectures (L)10 Lectures + book (LB)01 Dummy coding:

14 outcome ij = model + error ij knowledge ij = b 0 + b 1 *dummy1 j + b 2 *dummy2 j + ε ij i = 1, … N, N = number of people per condition = 5 j = 1, … M, M = number of conditions = 3 knowledge i1 = b 0 + b 1 *dummy1 1 + b 2 *dummy2 1 + ε i1 = b 0 + b 1 *0 + b 2 *0 + ε i1 Condition Dummy variable Dummy1 (lec) Dummy2 (lecbook) Nothing (N)00 Lectures (L)10 Lectures + book (LB)01 Dummy coding:

15 outcome ij = model + error ij knowledge ij = b 0 + b 1 *dummy1 j + b 2 *dummy2 j + ε ij i = 1, … N, N = number of people per condition = 5 j = 1, … M, M = number of conditions = 3 knowledge i1 = b 0 + b 1 *dummy1 1 + b 2 *dummy2 1 + ε i1 = b 0 + b 1 *0 + b 2 *0 + ε i1 = b 0 + ε i1 Condition Dummy variable Dummy1 (lec) Dummy2 (lecbook) Nothing (N)00 Lectures (L)10 Lectures + book (LB)01 Dummy coding:

16 outcome ij = model + error ij knowledge ij = b 0 + b 1 *dummy1 j + b 2 *dummy2 j + ε ij i = 1, … N, N = number of people per condition = 5 j = 1, … M, M = number of conditions = 3 knowledge i1 = b 0 + b 1 *dummy1 1 + b 2 *dummy2 1 + ε i1 = b 0 + b 1 *0 + b 2 *0 + ε i1 = b 0 + ε i1 knowledge i2 = b 0 + b 1 *dummy1 2 + b 2 *dummy2 2 + ε i2 Condition Dummy variable Dummy1 (lec) Dummy2 (lecbook) Nothing (N)00 Lectures (L)10 Lectures + book (LB)01 Dummy coding:

17 outcome ij = model + error ij knowledge ij = b 0 + b 1 *dummy1 j + b 2 *dummy2 j + ε ij i = 1, … N, N = number of people per condition = 5 j = 1, … M, M = number of conditions = 3 knowledge i1 = b 0 + b 1 *dummy1 1 + b 2 *dummy2 1 + ε i1 = b 0 + b 1 *0 + b 2 *0 + ε i1 = b 0 + ε i1 knowledge i2 = b 0 + b 1 *dummy1 2 + b 2 *dummy2 2 + ε i2 = b 0 + b 1 *1 + b 2 *0 + ε i2 Condition Dummy variable Dummy1 (lec) Dummy2 (lecbook) Nothing (N)00 Lectures (L)10 Lectures + book (LB)01 Dummy coding:

18 outcome ij = model + error ij knowledge ij = b 0 + b 1 *dummy1 j + b 2 *dummy2 j + ε ij i = 1, … N, N = number of people per condition = 5 j = 1, … M, M = number of conditions = 3 knowledge i1 = b 0 + b 1 *dummy1 1 + b 2 *dummy2 1 + ε i1 = b 0 + b 1 *0 + b 2 *0 + ε i1 = b 0 + ε i1 knowledge i2 = b 0 + b 1 *dummy1 2 + b 2 *dummy2 2 + ε i2 = b 0 + b 1 *1 + b 2 *0 + ε i2 = b 0 + b 1 + ε i2 Condition Dummy variable Dummy1 (lec) Dummy2 (lecbook) Nothing (N)00 Lectures (L)10 Lectures + book (LB)01 Dummy coding:

19 outcome ij = model + error ij knowledge ij = b 0 + b 1 *dummy1 j + b 2 *dummy2 j + ε ij i = 1, … N, N = number of people per condition = 5 j = 1, … M, M = number of conditions = 3 knowledge i1 = b 0 + b 1 *dummy1 1 + b 2 *dummy2 1 + ε i1 = b 0 + b 1 *0 + b 2 *0 + ε i1 = b 0 + ε i1 knowledge i2 = b 0 + b 1 *dummy1 2 + b 2 *dummy2 2 + ε i2 = b 0 + b 1 *1 + b 2 *0 + ε i2 = b 0 + b 1 + ε i2 knowledge i3 = b 0 + b 1 *dummy1 3 + b 2 *dummy2 3 + ε i3 Condition Dummy variable Dummy1 (lec) Dummy2 (lecbook) Nothing (N)00 Lectures (L)10 Lectures + book (LB)01 Dummy coding:

20 outcome ij = model + error ij knowledge ij = b 0 + b 1 *dummy1 j + b 2 *dummy2 j + ε ij i = 1, … N, N = number of people per condition = 5 j = 1, … M, M = number of conditions = 3 knowledge i1 = b 0 + b 1 *dummy1 1 + b 2 *dummy2 1 + ε i1 = b 0 + b 1 *0 + b 2 *0 + ε i1 = b 0 + ε i1 knowledge i2 = b 0 + b 1 *dummy1 2 + b 2 *dummy2 2 + ε i2 = b 0 + b 1 *1 + b 2 *0 + ε i2 = b 0 + b 1 + ε i2 knowledge i3 = b 0 + b 1 *dummy1 3 + b 2 *dummy2 3 + ε i3 = b 0 + b 1 *0 + b 2 *1 + ε i3 = b 0 + b 2 + ε i3 Condition Dummy variable Dummy1 (lec) Dummy2 (lecbook) Nothing (N)00 Lectures (L)10 Lectures + book (LB)01 Dummy coding:

21 outcome ij = model + error ij knowledge ij = b 0 + b 1 *dummy1 j + b 2 *dummy2 j + ε ij knowledge i1 = b 0 + b 1 *dummy1 1 + b 2 *dummy2 1 + ε i1 = b 0 + b 1 *0 + b 2 *0 + ε i1 = b 0 + ε i1 knowledge i2 = b 0 + b 1 *dummy1 2 + b 2 *dummy2 2 + ε i2 = b 0 + b 1 *1 + b 2 *0 + ε i2 = b 0 + b 1 + ε i2 knowledge i3 = b 0 + b 1 *dummy1 3 + b 2 *dummy2 3 + ε i3 = b 0 + b 1 *0 + b 2 *1 + ε i3 = b 0 + b 2 + ε i3 Condition Dummy variable Dummy1 (lec) Dummy2 (lecbook) Nothing (N)00 Lectures (L)10 Lectures + book (LB)01 Therefore: Dummy coding:

22 outcome ij = model + error ij knowledge ij = b 0 + b 1 *dummy1 j + b 2 *dummy2 j + ε ij knowledge i1 = b 0 + b 1 *dummy1 1 + b 2 *dummy2 1 + ε i1 = b 0 + b 1 *0 + b 2 *0 + ε i1 = b 0 + ε i1 knowledge i2 = b 0 + b 1 *dummy1 2 + b 2 *dummy2 2 + ε i2 = b 0 + b 1 *1 + b 2 *0 + ε i2 = b 0 + b 1 + ε i2 knowledge i3 = b 0 + b 1 *dummy1 3 + b 2 *dummy2 3 + ε i3 = b 0 + b 1 *0 + b 2 *1 + ε i3 = b 0 + b 2 + ε i3 Condition Dummy variable Dummy1 (lec) Dummy2 (lecbook) Nothing (N)00 Lectures (L)10 Lectures + book (LB)01 Therefore: → μ condition1 = b 0 b 0 is the mean of condition 1 (N) Dummy coding:

23 outcome ij = model + error ij knowledge ij = b 0 + b 1 *dummy1 j + b 2 *dummy2 j + ε ij knowledge i1 = b 0 + b 1 *dummy1 1 + b 2 *dummy2 1 + ε i1 = b 0 + b 1 *0 + b 2 *0 + ε i1 = b 0 + ε i1 knowledge i2 = b 0 + b 1 *dummy1 2 + b 2 *dummy2 2 + ε i2 = b 0 + b 1 *1 + b 2 *0 + ε i2 = b 0 + b 1 + ε i2 knowledge i3 = b 0 + b 1 *dummy1 3 + b 2 *dummy2 3 + ε i3 = b 0 + b 1 *0 + b 2 *1 + ε i3 = b 0 + b 2 + ε i3 Condition Dummy variable Dummy1 (lec) Dummy2 (lecbook) Nothing (N)00 Lectures (L)10 Lectures + book (LB)01 Therefore: → μ condition1 = b 0 b 0 is the mean of condition 1 (N) → μ condition2 = b 0 + b 1 = μ condition1 + b 1 μ condition2 - μ condition1 = b 1 Dummy coding:

24 outcome ij = model + error ij knowledge ij = b 0 + b 1 *dummy1 j + b 2 *dummy2 j + ε ij knowledge i1 = b 0 + b 1 *dummy1 1 + b 2 *dummy2 1 + ε i1 = b 0 + b 1 *0 + b 2 *0 + ε i1 = b 0 + ε i1 knowledge i2 = b 0 + b 1 *dummy1 2 + b 2 *dummy2 2 + ε i2 = b 0 + b 1 *1 + b 2 *0 + ε i2 = b 0 + b 1 + ε i2 knowledge i3 = b 0 + b 1 *dummy1 3 + b 2 *dummy2 3 + ε i3 = b 0 + b 1 *0 + b 2 *1 + ε i3 = b 0 + b 2 + ε i3 Condition Dummy variable Dummy1 (lec) Dummy2 (lecbook) Nothing (N)00 Lectures (L)10 Lectures + book (LB)01 Therefore: → μ condition1 = b 0 b 0 is the mean of condition 1 (N) → μ condition2 = b 0 + b 1 = μ condition1 + b 1 b 1 is the difference in means of μ condition2 - μ condition1 = b 1 condition 1 (N) and condition 2 (L) Dummy coding:

25 outcome ij = model + error ij knowledge ij = b 0 + b 1 *dummy1 j + b 2 *dummy2 j + ε ij knowledge i1 = b 0 + b 1 *dummy1 1 + b 2 *dummy2 1 + ε i1 = b 0 + b 1 *0 + b 2 *0 + ε i1 = b 0 + ε i1 knowledge i2 = b 0 + b 1 *dummy1 2 + b 2 *dummy2 2 + ε i2 = b 0 + b 1 *1 + b 2 *0 + ε i2 = b 0 + b 1 + ε i2 knowledge i3 = b 0 + b 1 *dummy1 3 + b 2 *dummy2 3 + ε i3 = b 0 + b 1 *0 + b 2 *1 + ε i3 = b 0 + b 2 + ε i3 Condition Dummy variable Dummy1 (lec) Dummy2 (lecbook) Nothing (N)00 Lectures (L)10 Lectures + book (LB)01 Therefore: → μ condition1 = b 0 b 0 is the mean of condition 1 (N) → μ condition2 = b 0 + b 1 = μ condition1 + b 1 b 1 is the difference in means of μ condition2 - μ condition1 = b 1 condition 1 (N) and condition 2 (L) → μ condition3 = b 0 + b 2 = μ condition1 + b 2 μ condition3 - μ condition1 = b 2 Dummy coding:

26 outcome ij = model + error ij knowledge ij = b 0 + b 1 *dummy1 j + b 2 *dummy2 j + ε ij knowledge i1 = b 0 + b 1 *dummy1 1 + b 2 *dummy2 1 + ε i1 = b 0 + b 1 *0 + b 2 *0 + ε i1 = b 0 + ε i1 knowledge i2 = b 0 + b 1 *dummy1 2 + b 2 *dummy2 2 + ε i2 = b 0 + b 1 *1 + b 2 *0 + ε i2 = b 0 + b 1 + ε i2 knowledge i3 = b 0 + b 1 *dummy1 3 + b 2 *dummy2 3 + ε i3 = b 0 + b 1 *0 + b 2 *1 + ε i3 = b 0 + b 2 + ε i3 Condition Dummy variable Dummy1 (lec) Dummy2 (lecbook) Nothing (N)00 Lectures (L)10 Lectures + book (LB)01 Therefore: → μ condition1 = b 0 b 0 is the mean of condition 1 (N) → μ condition2 = b 0 + b 1 = μ condition1 + b 1 b 1 is the difference in means of μ condition2 - μ condition1 = b 1 condition 1 (N) and condition 2 (L) → μ condition3 = b 0 + b 2 = μ condition1 + b 2 b 2 is the difference in means of μ condition3 - μ condition1 = b 2 condition 1 (N) and condition 3 (LB) Dummy coding:

27 Person Condition Nothing (N) Lectures (L)Lectures + book (LB) person person 2179 person 3168 person person 5157 μ N = 1μ L = 5μ LB = 9μ = 5 μNμN μLμL μ LB μ

28 Person Condition Nothing (N) Lectures (L)Lectures + book (LB) person person 2179 person 3168 person person 5157 μ N = 1μ L = 5μ LB = 9μ = 5 μNμN μLμL μ LB μ b0b0 b1b1 b2b2

29 Person Condition Nothing (N) Lectures (L)Lectures + book (LB) person person 2179 person 3168 person person 5157 μ N = 1μ L = 5μ LB = 9μ = 5 μNμN μLμL μ LB μ

30 Person Condition Nothing (N) Lectures (L)Lectures + book (LB) person person 2179 person 3168 person person 5157 μ N = 1μ L = 5μ LB = 9μ = 5 μNμN μLμL μ LB μ Sums of squares

31 Person Condition Nothing (N) Lectures (L)Lectures + book (LB) person person 2179 person 3168 person person 5157 μ N = 1μ L = 5μ LB = 9μ = 5 μNμN μLμL μ LB μ Sums of squares SS T = Σ(score ij - μ) 2

32 Person Condition Nothing (N) Lectures (L)Lectures + book (LB) person person 2179 person 3168 person person 5157 μ N = 1μ L = 5μ LB = 9μ = 5 μNμN μLμL μ LB μ Sums of squares SS T = Σ(score ij - μ) 2

33 Person Condition Nothing (N) Lectures (L)Lectures + book (LB) person person 2179 person 3168 person person 5157 μ N = 1μ L = 5μ LB = 9μ = 5 μNμN μLμL μ LB μ Sums of squares SS T = Σ(score ij - μ) 2 SS B = ΣN j (μ j - μ) 2

34 Person Condition Nothing (N) Lectures (L)Lectures + book (LB) person person 2179 person 3168 person person 5157 μ N = 1μ L = 5μ LB = 9μ = 5 μNμN μLμL μ LB μ Sums of squares SS T = Σ(score ij - μ) 2 SS B = ΣN j (μ j - μ) 2

35 Person Condition Nothing (N) Lectures (L)Lectures + book (LB) person person 2179 person 3168 person person 5157 μ N = 1μ L = 5μ LB = 9μ = 5 μNμN μLμL μ LB μ Sums of squares SS T = Σ(score ij - μ) 2 SS B = ΣN j (μ j - μ) 2 SS W = Σ(score ij - μ j ) 2

36 Person Condition Nothing (N) Lectures (L)Lectures + book (LB) person person 2179 person 3168 person person 5157 μ N = 1μ L = 5μ LB = 9μ = 5 μNμN μLμL μ LB μ Sums of squares SS T = Σ(score ij - μ) 2 SS B = ΣN j (μ j - μ) 2 SS W = Σ(score ij - μ j ) 2

37 Person Condition Nothing (N) Lectures (L)Lectures + book (LB) person person 2179 person 3168 person person 5157 μ N = 1μ L = 5μ LB = 9μ = 5 μNμN μLμL μ LB μ Sums of squares SS T = Σ(score ij - μ) 2 SS B = ΣN j (μ j - μ) 2 SS W = Σ(score ij - μ j ) 2 SS T = SS B + SS W

38 Person Condition Nothing (N) Lectures (L)Lectures + book (LB) person person 2179 person 3168 person person 5157 μ N = 1μ L = 5μ LB = 9μ = 5 μNμN μLμL μ LB μ Sums of squares SS T = Σ(score ij - μ) 2 SS B = ΣN j (μ j - μ) 2 SS W = Σ(score ij - μ j ) 2 SS T = SS B + SS W Degrees of freedom df T = MN - 1 df B = M – 1 df W = M(N – 1) N = number of people per condition M = number of conditions

39 Person Condition Nothing (N) Lectures (L)Lectures + book (LB) person person 2179 person 3168 person person 5157 μ N = 1μ L = 5μ LB = 9μ = 5 μNμN μLμL μ LB μ Sums of squares SS T = Σ(score ij - μ) 2 SS B = ΣN j (μ j - μ) 2 SS W = Σ(score ij - μ j ) 2 SS T = SS B + SS W Degrees of freedom df T = MN - 1 df B = M – 1 df W = M(N – 1) Mean squares MS T = SS T /df T MS B = SS B /df B MS W = SS W /df W N = number of people per condition M = number of conditions

40 Person Condition Nothing (N) Lectures (L)Lectures + book (LB) person person 2179 person 3168 person person 5157 μ N = 1μ L = 5μ LB = 9μ = 5 μNμN μLμL μ LB μ Sums of squares SS T = Σ(score ij - μ) 2 SS B = ΣN j (μ j - μ) 2 SS W = Σ(score ij - μ j ) 2 SS T = SS B + SS W Degrees of freedom df T = MN - 1 df B = M – 1 df W = M(N – 1) Mean squares MS T = SS T /df T MS B = SS B /df B MS W = SS W /df W N = number of people per condition M = number of conditions F-ratio F = MS B /MS W = MS model /MS error

41 Sib analysis

42 Sire 1 Sire 2 Sire 3 Sire 4 Sire 5 Sire 6 Sire 7 Sire 8 - number of males (sires) each mated to number of females (dams)

43 Sib analysis Sire 1 Sire 2 Sire 3 Sire 4 Sire 5 Sire 6 Sire 7 Sire 8 - number of males (sires) each mated to number of females (dams) - mating and selection of sires and dams → random

44 Sib analysis Sire 1 Sire 2 Sire 3 Sire 4 Sire 5 Sire 6 Sire 7 Sire 8 - number of males (sires) each mated to number of females (dams) - mating and selection of sires and dams → random - thus: population of full sibs (same father, same mother; same cell in table) and half sibs (same father, different mother; same row in table)

45 Sib analysis Sire 1 Sire 2 Sire 3 Sire 4 Sire 5 Sire 6 Sire 7 Sire 8 - number of males (sires) each mated to number of females (dams) - mating and selection of sires and dams → random - thus: population of full sibs (same father, same mother; same cell in table) and half sibs (same father, different mother; same row in table) - data: measurements of all offspring

46 Sire 1 Sire 2 Sire 3 Sire 4 Sire 5 Sire 6 Sire 7 Sire 8 μ sire1 μ dam1sire1 score offspring1dam1sire1 Sib analysis - example with 3 sires:

47 Sib analysis Sire 1 Sire 2 Sire 3 Sire 4 Sire 5 Sire 6 Sire 7 Sire 8 ANOVA: Partitioning the phenotypic variance (V P ):

48 Sib analysis Sire 1 Sire 2 Sire 3 Sire 4 Sire 5 Sire 6 Sire 7 Sire 8 ANOVA: Partitioning the phenotypic variance (V P ): - between-sire component

49 Sib analysis Sire 1 Sire 2 Sire 3 Sire 4 Sire 5 Sire 6 Sire 7 Sire 8 ANOVA: Partitioning the phenotypic variance (V P ): - between-sire component - component attributable to differences between the progeny of different males

50 Sire 1 Sire 2 Sire 3 Sire 4 Sire 5 Sire 6 Sire 7 Sire 8 μ sire1 Sib analysis μ sire2 μ sire3

51 Sire 1 Sire 2 Sire 3 Sire 4 Sire 5 Sire 6 Sire 7 Sire 8 μ sire1 Sib analysis μ sire2 μ sire3

52 Sib analysis Sire 1 Sire 2 Sire 3 Sire 4 Sire 5 Sire 6 Sire 7 Sire 8 ANOVA: Partitioning the phenotypic variance (V P ): - between-sire component

53 Sib analysis Sire 1 Sire 2 Sire 3 Sire 4 Sire 5 Sire 6 Sire 7 Sire 8 ANOVA: Partitioning the phenotypic variance (V P ): - between-sire component - between-dam, within-sire component

54 Sib analysis Sire 1 Sire 2 Sire 3 Sire 4 Sire 5 Sire 6 Sire 7 Sire 8 ANOVA: Partitioning the phenotypic variance (V P ): - between-sire component - between-dam, within-sire component - component attributable to differences between progeny of females mated to same male

55 Sire 1 Sire 2 Sire 3 Sire 4 Sire 5 Sire 6 Sire 7 Sire 8 μ sire1 Sib analysis μ sire2 μ sire3

56 Sire 1 Sire 2 Sire 3 Sire 4 Sire 5 Sire 6 Sire 7 Sire 8 μ sire1 Sib analysis μ sire2 μ sire3

57 Sib analysis Sire 1 Sire 2 Sire 3 Sire 4 Sire 5 Sire 6 Sire 7 Sire 8 ANOVA: Partitioning the phenotypic variance (V P ): - between-sire component - between-dam, within-sire component

58 Sib analysis Sire 1 Sire 2 Sire 3 Sire 4 Sire 5 Sire 6 Sire 7 Sire 8 ANOVA: Partitioning the phenotypic variance (V P ): - between-sire component - between-dam, within-sire component - within-progeny component

59 Sib analysis Sire 1 Sire 2 Sire 3 Sire 4 Sire 5 Sire 6 Sire 7 Sire 8 ANOVA: Partitioning the phenotypic variance (V P ): - between-sire component - between-dam, within-sire component - within-progeny component - component attributable to differences between offspring of the same female

60 Sire 1 Sire 2 Sire 3 Sire 4 Sire 5 Sire 6 Sire 7 Sire 8 μ sire1 Sib analysis μ sire2 μ sire3

61 Sire 1 Sire 2 Sire 3 Sire 4 Sire 5 Sire 6 Sire 7 Sire 8 μ sire1 Sib analysis μ sire2 μ sire3

62 Sib analysis Sire 1 Sire 2 Sire 3 Sire 4 Sire 5 Sire 6 Sire 7 Sire 8 ANOVA: Partitioning the phenotypic variance (V P ): - between-sire component - between-dam, within-sire component - within-progeny component

63 Sib analysis Sire 1 Sire 2 Sire 3 Sire 4 Sire 5 Sire 6 Sire 7 Sire 8 ANOVA: Partitioning the phenotypic variance (V P ): - between-sire component - between-dam, within-sire component - within-progeny component

64 Sib analysis Sire 1 Sire 2 Sire 3 Sire 4 Sire 5 Sire 6 Sire 7 Sire 8 ANOVA: Partitioning the phenotypic variance (V P ): - between-sire component (σ 2 S ) - between-dam, within-sire component (σ 2 D ) - within-progeny component (σ 2 W )

65 Sib analysis Sire 1 Sire 2 Sire 3 Sire 4 Sire 5 Sire 6 Sire 7 Sire 8 ANOVA: Partitioning the phenotypic variance (V P ): σ 2 T = σ 2 S + σ 2 D + σ 2 W - between-sire component (σ 2 S ) - between-dam, within-sire component (σ 2 D ) - within-progeny component (σ 2 W )

66 Sib analysis Sire 1 Sire 2 Sire 3 Sire 4 Sire 5 Sire 6 Sire 7 Sire 8 ANOVA: Partitioning the phenotypic variance (V P ): - between-sire component (σ 2 S ) = variance between means of half-sib families = cov HS = ¼V A - between-dam, within-sire component (σ 2 D ) - within-progeny component (σ 2 W ) σ 2 T = σ 2 S + σ 2 D + σ 2 W

67 Sib analysis Sire 1 Sire 2 Sire 3 Sire 4 Sire 5 Sire 6 Sire 7 Sire 8

68 Sib analysis Sire 1 Sire 2 Sire 3 Sire 4 Sire 5 Sire 6 Sire 7 Sire 8 ANOVA: Partitioning the phenotypic variance (V P ): - between-sire component (σ 2 S ) = variance between means of half-sib families = cov HS = ¼V A - between-dam, within-sire component (σ 2 D ) - within-progeny component (σ 2 W ) σ 2 T = σ 2 S + σ 2 D + σ 2 W

69 Sib analysis Sire 1 Sire 2 Sire 3 Sire 4 Sire 5 Sire 6 Sire 7 Sire 8 ANOVA: Partitioning the phenotypic variance (V P ): - between-sire component (σ 2 S ) = variance between means of half-sib families = cov HS = ¼V A - between-dam, within-sire component (σ 2 D ) - within-progeny component (σ 2 W ) = total variance minus variance between groups = V P – cov FS = ½V A + ¾V D + V Ew σ 2 T = σ 2 S + σ 2 D + σ 2 W

70 Sib analysis Sire 1 Sire 2 Sire 3 Sire 4 Sire 5 Sire 6 Sire 7 Sire 8 ANOVA: Partitioning the phenotypic variance (V P ): - between-sire component (σ 2 S ) = variance between means of half-sib families = cov HS = ¼V A - between-dam, within-sire component (σ 2 D ) = σ 2 T - σ 2 S - σ 2 W = cov FS – cov HS = ¼V A + ¼V D + V Ec - within-progeny component (σ 2 W ) = total variance minus variance between groups = V P – cov FS = ½V A + ¾V D + V Ew σ 2 T = σ 2 S + σ 2 D + σ 2 W

71 Question: Why is any between group variance component equal to the covariance of the members of the groups?

72 Question: Why is any between group variance component equal to the covariance of the members of the groups? Conceptually: If all offspring in a group have relatively high values, the mean value for that group will also be relatively high. Conversely, when all members of a group have relatively low values, the mean for that group will be relatively low.

73 Question: Why is any between group variance component equal to the covariance of the members of the groups? Conceptually: If all offspring in a group have relatively high values, the mean value for that group will also be relatively high. Conversely, when all members of a group have relatively low values, the mean for that group will be relatively low. Hence, the greater the correlation, the greater will be the variability among the means of the groups (i.e., between-groups variability) as a proportion of the total variability, and the smaller will be the proportion of total variability inside the groups (i.e., within-groups variability).

74 Question: Why is any between group variance component equal to the covariance of the members of the groups? Conceptually: If all offspring in a group have relatively high values, the mean value for that group will also be relatively high. Conversely, when all members of a group have relatively low values, the mean for that group will be relatively low. Hence, the greater the correlation, the greater will be the variability among the means of the groups (i.e., between-groups variability) as a proportion of the total variability, and the smaller will be the proportion of total variability inside the groups (i.e., within-groups variability). Computationally: We can illustrate this using the intraclass correlation coefficient (ICC).

75 Measures of phenotype Offspring 1Offspring 2Offspring 3Offspring 4Mean Sire 11234μ s1 = 2.5 Sire 25678μ s2 = 6.5 Sire μ s3 = 10.5 μ = not a nested design -> each sire mated to only 1 dam -> families of full sibs

76 Measures of phenotype Offspring 1Offspring 2Offspring 3Offspring 4Mean Sire 11234μ s1 = 2.5 Sire 25678μ s2 = 6.5 Sire μ s3 = 10.5 μ = 6.5 μ s1 μ s2 μ s3 μ

77 Measures of phenotype Offspring 1Offspring 2Offspring 3Offspring 4Mean Sire 11234μ s1 = 2.5 Sire 25678μ s2 = 6.5 Sire μ s3 = 10.5 μ = 6.5 μ s1 μ s2 μ s3 μ σ 2 s = Σ(μ Si - μ) 2 /df s = [N 1 (μ s1 - μ) 2 + N 2 (μ s2 - μ) 2 + N 3 (μ s3 - μ) 2 ]/(3-1) = [4*(2.5 – 6.5) 2 + 4*(6.5 – 6.5) 2 + 4*(10.5 – 6.5) 2 ]/2 = (4* *0 + 4*4 2 )/2 = 64

78 σ 2 s = Σ(μ Si - μ) 2 /df s = [N 1 (μ s1 - μ) 2 + N 2 (μ s2 - μ) 2 + N 3 (μ s3 - μ) 2 ]/(3-1) = [4*(2.5 – 6.5) 2 + 4*(6.5 – 6.5) 2 + 4*(10.5 – 6.5) 2 ]/2 = (4* *0 + 4*4 2 )/2 = 64 - this is the variance between the means of 3 groups, or the between-group variance component μ s1 μ s2 μ s3 μ Measures of phenotype Offspring 1Offspring 2Offspring 3Offspring 4Mean Sire 11234μ s1 = 2.5 Sire 25678μ s2 = 6.5 Sire μ s3 = 10.5 μ = 6.5

79 σ 2 s = Σ(μ Si - μ) 2 /df s = [N 1 (μ s1 - μ) 2 + N 2 (μ s2 - μ) 2 + N 3 (μ s3 - μ) 2 ]/(3-1) = [4*(2.5 – 6.5) 2 + 4*(6.5 – 6.5) 2 + 4*(10.5 – 6.5) 2 ]/2 = (4* *0 + 4*4 2 )/2 = 64 - this is the variance between the means of 3 groups, or the between-group variance component - how does the magnitude of this variance component relate to the covariance within the groups? Measures of phenotype Offspring 1Offspring 2Offspring 3Offspring 4Mean Sire 11234μ s1 = 2.5 Sire 25678μ s2 = 6.5 Sire μ s3 = 10.5 μ = 6.5 μ s1 μ s2 μ s3 μ

80 Families of sibs σ 2 s = Σ(μ Si - μ) 2 /df s = [N 1 (μ s1 - μ) 2 + N 2 (μ s2 - μ) 2 + N 3 (μ s3 - μ) 2 ]/(3-1) = [4*(2.5 – 6.5) 2 + 4*(6.5 – 6.5) 2 + 4*(10.5 – 6.5) 2 ]/2 = (4* *0 + 4*4 2 )/2 = 64 - this is the variance between the means of 3 groups, or the between-group variance component - how does the magnitude of this variance component relate to the covariance within the groups? Measures of phenotype Offspring 1Offspring 2Offspring 3Offspring 4Mean Sire 11234μ s1 = 2.5 Sire 25678μ s2 = 6.5 Sire μ s3 = 10.5 μ = 6.5 μ s1 μ s2 μ s3 μ

81 Measures of phenotype Offspring 1Offspring 2Offspring 3Offspring 4Mean Sire 11234μ s1 = 2.5 Sire 25678μ s2 = 6.5 Sire μ s3 = 10.5 μ = 6.5 μ s1 μ s2 μ s3 μ σ 2 s = Σ(μ Si - μ) 2 /df s = [N 1 (μ s1 - μ) 2 + N 2 (μ s2 - μ) 2 + N 3 (μ s3 - μ) 2 ]/(3-1) = [4*(2.5 – 6.5) 2 + 4*(6.5 – 6.5) 2 + 4*(10.5 – 6.5) 2 ]/2 = (4* *0 + 4*4 2 )/2 = 64 - this is the variance between the means of 3 groups, or the between-group variance component - how does the magnitude of this variance component relate to the covariance within the groups?

82 Measures of phenotype Offspring 1Offspring 2Offspring 3Offspring 4Mean Sire 11234μ s1 = 2.5 Sire 25678μ s2 = 6.5 Sire μ s3 = 10.5 μ = 6.5 σ 2 s = Σ(μ Si - μ) 2 /df s = [N 1 (μ s1 - μ) 2 + N 2 (μ s2 - μ) 2 + N 3 (μ s3 - μ) 2 ]/(3-1) = [4*(2.5 – 6.5) 2 + 4*(6.5 – 6.5) 2 + 4*(10.5 – 6.5) 2 ]/2 = (4* *0 + 4*4 2 )/2 = 64 - this is the variance between the means of 3 groups, or the between-group variance component - how does the magnitude of this variance component relate to the covariance within the groups? How to summarize the correlations between these 4 variables?

83 Measures of phenotype Offspring 1Offspring 2Offspring 3Offspring 4Mean Sire 11234μ s1 = 2.5 Sire 25678μ s2 = 6.5 Sire μ s3 = 10.5 μ = 6.5 σ 2 s = Σ(μ Si - μ) 2 /df s = [N 1 (μ s1 - μ) 2 + N 2 (μ s2 - μ) 2 + N 3 (μ s3 - μ) 2 ]/(3-1) = [4*(2.5 – 6.5) 2 + 4*(6.5 – 6.5) 2 + 4*(10.5 – 6.5) 2 ]/2 = (4* *0 + 4*4 2 )/2 = 64 - this is the variance between the means of 3 groups, or the between-group variance component - how does the magnitude of this variance component relate to the covariance within the groups? How to summarize the correlations between these 4 variables? - use Pearson r (bivariately) to obtain a correlation matrix?

84 Measures of phenotype Offspring 1Offspring 2Offspring 3Offspring 4Mean Sire 11234μ s1 = 2.5 Sire 25678μ s2 = 6.5 Sire μ s3 = 10.5 μ = 6.5 σ 2 s = Σ(μ Si - μ) 2 /df s = [N 1 (μ s1 - μ) 2 + N 2 (μ s2 - μ) 2 + N 3 (μ s3 - μ) 2 ]/(3-1) = [4*(2.5 – 6.5) 2 + 4*(6.5 – 6.5) 2 + 4*(10.5 – 6.5) 2 ]/2 = (4* *0 + 4*4 2 )/2 = 64 - this is the variance between the means of 3 groups, or the between-group variance component - how does the magnitude of this variance component relate to the covariance within the groups? How to summarize the correlations between these 4 variables? - use Pearson r (bivariately) to obtain a correlation matrix? - no, because a) we need a single measure of relationship

85 Measures of phenotype Offspring 1Offspring 2Offspring 3Offspring 4Mean Sire 11234μ s1 = 2.5 Sire 25678μ s2 = 6.5 Sire μ s3 = 10.5 μ = 6.5 σ 2 s = Σ(μ Si - μ) 2 /df s = [N 1 (μ s1 - μ) 2 + N 2 (μ s2 - μ) 2 + N 3 (μ s3 - μ) 2 ]/(3-1) = [4*(2.5 – 6.5) 2 + 4*(6.5 – 6.5) 2 + 4*(10.5 – 6.5) 2 ]/2 = (4* *0 + 4*4 2 )/2 = 64 - this is the variance between the means of 3 groups, or the between-group variance component - how does the magnitude of this variance component relate to the covariance within the groups? How to summarize the correlations between these 4 variables? - use Pearson r (bivariately) to obtain a correlation matrix? - no, because a) we need a single measure of relationship b) r sensitive to reshuffling data in rows (thus, if we reshuffle data in rows, the row means [μ s ] and between-group variance component [σ 2 s ] would stay the same, while r would change)

86 Measures of phenotype Offspring 1Offspring 2Offspring 3Offspring 4Mean Sire 11234μ s1 = 2.5 Sire 25678μ s2 = 6.5 Sire μ s3 = 10.5 μ = 6.5 σ 2 s = Σ(μ Si - μ) 2 /df s = [N 1 (μ s1 - μ) 2 + N 2 (μ s2 - μ) 2 + N 3 (μ s3 - μ) 2 ]/(3-1) = [4*(2.5 – 6.5) 2 + 4*(6.5 – 6.5) 2 + 4*(10.5 – 6.5) 2 ]/2 = (4* *0 + 4*4 2 )/2 = 64 - this is the variance between the means of 3 groups, or the between-group variance component - how does the magnitude of this variance component relate to the covariance within the groups? How to summarize the correlations between these 4 variables? - use Pearson r (bivariately) to obtain a correlation matrix? - no, because a) we need a single measure of relationship b) r sensitive to reshuffling data in rows (thus, if we reshuffle data in rows, the row means [μ s ] and between-group variance component [σ 2 s ] would stay the same, while r would change) - solution: ICC (intraclass correlation coefficient):

87 Measures of phenotype Offspring 1Offspring 2Offspring 3Offspring 4Mean Sire 11234μ s1 = 2.5 Sire 25678μ s2 = 6.5 Sire μ s3 = 10.5 μ = 6.5 σ 2 s = Σ(μ Si - μ) 2 /df s = [N 1 (μ s1 - μ) 2 + N 2 (μ s2 - μ) 2 + N 3 (μ s3 - μ) 2 ]/(3-1) = [4*(2.5 – 6.5) 2 + 4*(6.5 – 6.5) 2 + 4*(10.5 – 6.5) 2 ]/2 = (4* *0 + 4*4 2 )/2 = 64 - this is the variance between the means of 3 groups, or the between-group variance component - how does the magnitude of this variance component relate to the covariance within the groups? How to summarize the correlations between these 4 variables? - use Pearson r (bivariately) to obtain a correlation matrix? - no, because a) we need a single measure of relationship b) r sensitive to reshuffling data in rows (thus, if we reshuffle data in rows, the row means [μ s ] and between-group variance component [σ 2 s ] would stay the same, while r would change) - solution: ICC (intraclass correlation coefficient): a) a single measure b) insensitive to reshuffling data in rows

88 Measures of phenotype Offspring 1Offspring 2Offspring 3Offspring 4Mean Sire 11234μ s1 = 2.5 Sire 25678μ s2 = 6.5 Sire μ s3 = 10.5 μ = 6.5 σ 2 s = Σ(μ Si - μ) 2 /df s = [N 1 (μ s1 - μ) 2 + N 2 (μ s2 - μ) 2 + N 3 (μ s3 - μ) 2 ]/(3-1) = [4*(2.5 – 6.5) 2 + 4*(6.5 – 6.5) 2 + 4*(10.5 – 6.5) 2 ]/2 = (4* *0 + 4*4 2 )/2 = 64 ICC = σ 2 s /(σ 2 s + σ 2 w ) How to summarize the correlations between these 4 variables? - use Pearson r (bivariately) to obtain a correlation matrix? - no, because a) we need a single measure of relationship b) r sensitive to reshuffling data in rows (thus, if we reshuffle data in rows, the row means [μ s ] and between-group variance component [σ 2 s ] would stay the same, while r would change) - solution: ICC (intraclass correlation coefficient): a) a single measure b) insensitive to reshuffling data in rows

89 Measures of phenotype Offspring 1Offspring 2Offspring 3Offspring 4Mean Sire 11234μ s1 = 2.5 Sire 25678μ s2 = 6.5 Sire μ s3 = 10.5 μ = 6.5 σ 2 s = Σ(μ Si - μ) 2 /df s = [N 1 (μ s1 - μ) 2 + N 2 (μ s2 - μ) 2 + N 3 (μ s3 - μ) 2 ]/(3-1) = [4*(2.5 – 6.5) 2 + 4*(6.5 – 6.5) 2 + 4*(10.5 – 6.5) 2 ]/2 = (4* *0 + 4*4 2 )/2 = 64 ICC = σ 2 s /(σ 2 s + σ 2 w ) σ 2 w = Σ(p ij - μ Si ) 2 /df w How to summarize the correlations between these 4 variables? - use Pearson r (bivariately) to obtain a correlation matrix? - no, because a) we need a single measure of relationship b) r sensitive to reshuffling data in rows (thus, if we reshuffle data in rows, the row means [μ s ] and between-group variance component [σ 2 s ] would stay the same, while r would change) - solution: ICC (intraclass correlation coefficient): a) a single measure b) insensitive to reshuffling data in rows

90 Measures of phenotype Offspring 1Offspring 2Offspring 3Offspring 4Mean Sire 11234μ s1 = 2.5 Sire 25678μ s2 = 6.5 Sire μ s3 = 10.5 μ = 6.5 σ 2 s = Σ(μ Si - μ) 2 /df s = [N 1 (μ s1 - μ) 2 + N 2 (μ s2 - μ) 2 + N 3 (μ s3 - μ) 2 ]/(3-1) = [4*(2.5 – 6.5) 2 + 4*(6.5 – 6.5) 2 + 4*(10.5 – 6.5) 2 ]/2 = (4* *0 + 4*4 2 )/2 = 64 ICC = σ 2 s /(σ 2 s + σ 2 w ) σ 2 w = Σ(p ij - μ Si ) 2 /df w = [(p 11 - μ s1 ) 2 + (p 12 - μ s1 ) 2 + … + (p 21 – μ s2 ) 2 + … + (p 34 – μ s3 ) 2 ]/3(4-1) = 15/9 = 1.67 How to summarize the correlations between these 4 variables? - use Pearson r (bivariately) to obtain a correlation matrix? - no, because a) we need a single measure of relationship b) r sensitive to reshuffling data in rows (thus, if we reshuffle data in rows, the row means [μ s ] and between-group variance component [σ 2 s ] would stay the same, while r would change) - solution: ICC (intraclass correlation coefficient): a) a single measure b) insensitive to reshuffling data in rows

91 Measures of phenotype Offspring 1Offspring 2Offspring 3Offspring 4Mean Sire 11234μ s1 = 2.5 Sire 25678μ s2 = 6.5 Sire μ s3 = 10.5 μ = 6.5 σ 2 s = Σ(μ Si - μ) 2 /df s = [N 1 (μ s1 - μ) 2 + N 2 (μ s2 - μ) 2 + N 3 (μ s3 - μ) 2 ]/(3-1) = [4*(2.5 – 6.5) 2 + 4*(6.5 – 6.5) 2 + 4*(10.5 – 6.5) 2 ]/2 = (4* *0 + 4*4 2 )/2 = 64 ICC = σ 2 s /(σ 2 s + σ 2 w ) = 64/( ) = 0.97 σ 2 w = Σ(p ij - μ Si ) 2 /df w = [(p 11 - μ s1 ) 2 + (p 12 - μ s1 ) 2 + … + (p 21 – μ s2 ) 2 + … + (p 34 – μ s3 ) 2 ]/3(4-1) = 15/9 = 1.67 How to summarize the correlations between these 4 variables? - use Pearson r (bivariately) to obtain a correlation matrix? - no, because a) we need a single measure of relationship b) r sensitive to reshuffling data in rows (thus, if we reshuffle data in rows, the row means [μ s ] and between-group variance component [σ 2 s ] would stay the same, while r would change) - solution: ICC (intraclass correlation coefficient): a) a single measure b) insensitive to reshuffling data in rows

92 Measures of phenotype Offspring 1Offspring 2Offspring 3Offspring 4Mean Sire 11234μ s1 = 2.5 Sire 25678μ s2 = 6.5 Sire μ s3 = 10.5 μ = 6.5 ICC = 0.97

93 ICC = 0.10 Measures of phenotype Offspring 1Offspring 2Offspring 3Offspring 4Mean Sire μ s1 = Sire μ s2 = Sire μ s3 = μ =

94 Measures of phenotype Offspring 1Offspring 2Offspring 3Offspring 4Mean Sire μ s1 = 6.5 Sire 25678μ s2 = 6.5 Sire μ s3 = 6.5 μ = 6.5 ICC = 0


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