ANOVA & sib analysis

ANOVA & sib analysis• basics of ANOVA - revision• application to sib analysis

• intraclass correlation coefficient

- 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?

ANOVA as regression

Person ConditionNothing (N) Lectures (L) Lectures +

book(LB)

person 1

0 4 10

person 2

person 3

person 4

2 3 11

person 5

μN = 1 μL = 5 μLB = 9 μ = 5

ANOVA as regression

book(LB)

person 1

0 4 10

person 2

person 3

person 4

2 3 11

person 5

μN = 1 μL = 5 μLB = 9 μ = 52 4 6 8 10 12 14

Offspring

person

ANOVA as regression

outcomeij = model + errorij

book(LB)

person 1

0 4 10

person 2

person 3

person 4

2 3 11

person 5

μN = 1 μL = 5 μLB = 9 μ = 52 4 6 8 10 12 14

Offspring

person

Dummy coding:

Condition Dummy variableDummy1

(lec)Dummy2 (lecbook)

Nothing (N) 0 0Lectures (L) 1 0Lectures + book (LB)

Dummy coding:

knowledgeij = b0 + b1*dummy1j + b2*dummy2j + εij i = 1, … N, N = number of people per

condition = 5 j = 1, … M, M = number of conditions = 3

Dummy coding:

knowledgei1 = b0 + b1*dummy11 + b2*dummy21 + εi1

Dummy coding:

= b0 + b1*0 + b2*0 + εi1

Dummy coding:

= b0 + b1*0 + b2*0 + εi1

= b0 + εi1

Dummy coding:

= b0 + b1*0 + b2*0 + εi1

= b0 + εi1

Dummy coding:

= b0 + b1*0 + b2*0 + εi1

= b0 + εi1

= b0 + b1*1 + b2*0 + εi2

Dummy coding:

= b0 + b1*0 + b2*0 + εi1

= b0 + εi1

= b0 + b1*1 + b2*0 + εi2

= b0 + b1 + εi2

Dummy coding:

= b0 + b1*0 + b2*0 + εi1

= b0 + εi1

= b0 + b1*1 + b2*0 + εi2

= b0 + b1 + εi2

Dummy coding:

= b0 + b1*0 + b2*0 + εi1

= b0 + εi1

= b0 + b1*1 + b2*0 + εi2

= b0 + b1 + εi2

= b0 + b1*0 + b2*1 + εi3

= b0 + b2 + εi3

Dummy coding:

knowledgeij = b0 + b1*dummy1j + b2*dummy2j + εij

= b0 + b1*0 + b2*0 + εi1

= b0 + εi1

= b0 + b1*1 + b2*0 + εi2

= b0 + b1 + εi2

= b0 + b1*0 + b2*1 + εi3

= b0 + b2 + εi3

Therefore:

Dummy coding:

= b0 + b1*0 + b2*0 + εi1

= b0 + εi1

= b0 + b1*1 + b2*0 + εi2

= b0 + b1 + εi2

= b0 + b1*0 + b2*1 + εi3

= b0 + b2 + εi3

Therefore:

→ μcondition1 = b0 b0 is the mean of condition 1 (N)

Dummy coding:

= b0 + b1*0 + b2*0 + εi1

= b0 + εi1

= b0 + b1*1 + b2*0 + εi2

= b0 + b1 + εi2

= b0 + b1*0 + b2*1 + εi3

= b0 + b2 + εi3

Therefore:

→ μcondition2 = b0 + b1 = μcondition1 + b1

μcondition2 - μcondition1 = b1

Dummy coding:

= b0 + b1*0 + b2*0 + εi1

= b0 + εi1

= b0 + b1*1 + b2*0 + εi2

= b0 + b1 + εi2

= b0 + b1*0 + b2*1 + εi3

= b0 + b2 + εi3

Therefore:

→ μcondition2 = b0 + b1 = μcondition1 + b1 b1 is the difference in means of

μcondition2 - μcondition1 = b1 condition 1 (N) and condition 2 (L)

Dummy coding:

= b0 + b1*0 + b2*0 + εi1

= b0 + εi1

= b0 + b1*1 + b2*0 + εi2

= b0 + b1 + εi2

= b0 + b1*0 + b2*1 + εi3

= b0 + b2 + εi3

Therefore:

→ μcondition3 = b0 + b2 = μcondition1 + b2

μcondition3 - μcondition1 = b2

Dummy coding:

= b0 + b1*0 + b2*0 + εi1

= b0 + εi1

= b0 + b1*1 + b2*0 + εi2

= b0 + b1 + εi2

= b0 + b1*0 + b2*1 + εi3

= b0 + b2 + εi3

Therefore:

μcondition3 - μcondition1 = b2 condition 1 (N) and condition 3 (LB)

Dummy coding:

Person ConditionNothing (N) Lectures (L) Lectures + book

(LB)person 1 0 4 10person 2 1 7 9person 3 1 6 8person 4 2 3 11person 5 1 5 7

μN = 1 μL = 5 μLB = 9 μ = 5

2 4 6 8 10 12 14

Offspring

μN = 1 μL = 5 μLB = 9 μ = 5

2 4 6 8 10 12 14

Offspring

μN = 1 μL = 5 μLB = 9 μ = 5

2 4 6 8 10 12 14

Offspring

μN = 1 μL = 5 μLB = 9 μ = 5

2 4 6 8 10 12 14

Offspring

Sums of squares

μN = 1 μL = 5 μLB = 9 μ = 5

2 4 6 8 10 12 14

Offspring

Sums of squares

SST = Σ(scoreij - μ)2

μN = 1 μL = 5 μLB = 9 μ = 5

2 4 6 8 10 12 14

Offspring

Sums of squares

μN = 1 μL = 5 μLB = 9 μ = 5

2 4 6 8 10 12 14

Offspring

Sums of squares

SSB = ΣNj(μj - μ)2

μN = 1 μL = 5 μLB = 9 μ = 5

2 4 6 8 10 12 14

Offspring

Sums of squares

μN = 1 μL = 5 μLB = 9 μ = 5

2 4 6 8 10 12 14

Offspring

Sums of squares

SSW = Σ(scoreij - μj)2

μN = 1 μL = 5 μLB = 9 μ = 5

2 4 6 8 10 12 14

Offspring

Sums of squares

μN = 1 μL = 5 μLB = 9 μ = 5

2 4 6 8 10 12 14

Offspring

Sums of squares

SST = SSB + SSW

μN = 1 μL = 5 μLB = 9 μ = 5

2 4 6 8 10 12 14

Offspring

Sums of squares

SST = SSB + SSW

Degrees of freedom

dfT = MN - 1dfB = M – 1dfW = M(N – 1)

N = number of people per conditionM = number of conditions

μN = 1 μL = 5 μLB = 9 μ = 5

2 4 6 8 10 12 14

Offspring

Sums of squares

SST = SSB + SSW

Degrees of freedom

dfT = MN - 1dfB = M – 1dfW = M(N – 1)

Mean squares

MST = SST/dfT

MSB = SSB/dfB

MSW = SSW/dfW

μN = 1 μL = 5 μLB = 9 μ = 5

2 4 6 8 10 12 14

Offspring

Sums of squares

SST = SSB + SSW

Degrees of freedom

dfT = MN - 1dfB = M – 1dfW = M(N – 1)

Mean squares

MST = SST/dfT

MSB = SSB/dfB

MSW = SSW/dfW

F-ratio

F = MSB/MSW

= MSmodel/MSerror

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)

Sib analysis

Sire 1

Sire 2

Sire 3

Sire 4

Sire 5

Sire 6

Sire 7

Sire 8

- mating and selection of sires and dams → random

Sib analysis

Sire 1

Sire 2

Sire 3

Sire 4

Sire 5

Sire 6

Sire 7

Sire 8

- thus: population of full sibs (same father, same mother; same cell in table) and half sibs (same father, different mother; same row in table)

Sib analysis

Sire 1

Sire 2

Sire 3

Sire 4

Sire 5

Sire 6

Sire 7

Sire 8

- 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

Sire 1

Sire 2

Sire 3

Sire 4

Sire 5

Sire 6

Sire 7

Sire 8

0 5 10 15

Offspring

μsire1

μdam1sire1

scoreoffspring1dam1sire1

Sib analysis

- example with 3 sires:

Sib analysis

Sire 1

Sire 2

Sire 3

Sire 4

Sire 5

Sire 6

Sire 7

Sire 8

ANOVA:

Partitioning the phenotypic variance (VP):

Sib analysis

Sire 1

Sire 2

Sire 3

Sire 4

Sire 5

Sire 6

Sire 7

Sire 8

ANOVA:

- between-sire component

Sib analysis

Sire 1

Sire 2

Sire 3

Sire 4

Sire 5

Sire 6

Sire 7

Sire 8

ANOVA:

- between-sire component - component attributable to differences between the progeny of different males

Sire 1

Sire 2

Sire 3

Sire 4

Sire 5

Sire 6

Sire 7

Sire 8

0 5 10 15

Offspring

μsire1

Sib analysis

μsire2

μsire3

Sire 1

Sire 2

Sire 3

Sire 4

Sire 5

Sire 6

Sire 7

Sire 8

0 5 10 15

Offspring

μsire1

Sib analysis

μsire2

μsire3

Sib analysis

Sire 1

Sire 2

Sire 3

Sire 4

Sire 5

Sire 6

Sire 7

Sire 8

ANOVA:

- between-sire component

Sib analysis

Sire 1

Sire 2

Sire 3

Sire 4

Sire 5

Sire 6

Sire 7

Sire 8

ANOVA:

- between-sire component- between-dam, within-sire component

Sib analysis

Sire 1

Sire 2

Sire 3

Sire 4

Sire 5

Sire 6

Sire 7

Sire 8

ANOVA:

- between-sire component- between-dam, within-sire component - component attributable to differences between progeny of females mated to same male

Sire 1

Sire 2

Sire 3

Sire 4

Sire 5

Sire 6

Sire 7

Sire 8

0 5 10 15

Offspring

μsire1

Sib analysis

μsire2

μsire3

Sire 1

Sire 2

Sire 3

Sire 4

Sire 5

Sire 6

Sire 7

Sire 8

0 5 10 15

Offspring

μsire1

Sib analysis

μsire2

μsire3

Sib analysis

Sire 1

Sire 2

Sire 3

Sire 4

Sire 5

Sire 6

Sire 7

Sire 8

ANOVA:

- between-sire component- between-dam, within-sire component

Sib analysis

Sire 1

Sire 2

Sire 3

Sire 4

Sire 5

Sire 6

Sire 7

Sire 8

ANOVA:

- between-sire component- between-dam, within-sire component- within-progeny component

Sib analysis

Sire 1

Sire 2

Sire 3

Sire 4

Sire 5

Sire 6

Sire 7

Sire 8

ANOVA:

- between-sire component- between-dam, within-sire component- within-progeny component - component attributable to differences between offspring of the same female

Sire 1

Sire 2

Sire 3

Sire 4

Sire 5

Sire 6

Sire 7

Sire 8

0 5 10 15

Offspring

μsire1

Sib analysis

μsire2

μsire3

Sire 1

Sire 2

Sire 3

Sire 4

Sire 5

Sire 6

Sire 7

Sire 8

0 5 10 15

Offspring

μsire1

Sib analysis

μsire2

μsire3

Sib analysis

Sire 1

Sire 2

Sire 3

Sire 4

Sire 5

Sire 6

Sire 7

Sire 8

ANOVA:

- between-sire component- between-dam, within-sire component- within-progeny component

Sib analysis

Sire 1

Sire 2

Sire 3

Sire 4

Sire 5

Sire 6

Sire 7

Sire 8

ANOVA:

- between-sire component - between-dam, within-sire component - within-progeny component

Sib analysis

Sire 1

Sire 2

Sire 3

Sire 4

Sire 5

Sire 6

Sire 7

Sire 8

ANOVA:

- between-sire component (σ2S)

- between-dam, within-sire component (σ2D)

- within-progeny component (σ2W)

Sib analysis

Sire 1

Sire 2

Sire 3

Sire 4

Sire 5

Sire 6

Sire 7

Sire 8

ANOVA:

σ2T = σ2

S + σ2D +

- between-sire component (σ2S)

Sib analysis

Sire 1

Sire 2

Sire 3

Sire 4

Sire 5

Sire 6

Sire 7

Sire 8

ANOVA:

- between-sire component (σ2S) = variance between means of half-sib families = covHS = ¼VA

σ2T = σ2

S + σ2D +

0 5 10 15

Offspring

Sib analysis

Sire 1

Sire 2

Sire 3

Sire 4

Sire 5

Sire 6

Sire 7

Sire 8

Sib analysis

Sire 1

Sire 2

Sire 3

Sire 4

Sire 5

Sire 6

Sire 7

Sire 8

ANOVA:

σ2T = σ2

S + σ2D +

Sib analysis

Sire 1

Sire 2

Sire 3

Sire 4

Sire 5

Sire 6

Sire 7

Sire 8

ANOVA:

- within-progeny component (σ2W) = total variance minus variance between groups = VP – covFS = ½VA +

¾VD + VEw

σ2T = σ2

S + σ2D +

Sib analysis

Sire 1

Sire 2

Sire 3

Sire 4

Sire 5

Sire 6

Sire 7

Sire 8

ANOVA:

- between-dam, within-sire component (σ2D) = σ2

T - σ2S - σ2

W = covFS – covHS = ¼VA + ¼VD + VEc

- within-progeny component (σ2W) = total variance minus variance between groups = VP – covFS = ½VA +

¾VD + VEw

σ2T = σ2

S + σ2D +

Question:

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

Question:

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.

Question:

Conceptually:

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).

Question:

Conceptually:

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).

Measures of phenotype

Offspring 1 Offspring 2 Offspring 3 Offspring 4 Mean

Sire 1 1 2 3 4 μs1 = 2.5

Sire 2 5 6 7 8 μs2 = 6.5

Sire 3 9 10 11 12 μs3 = 10.5

μ = 6.5

- not a nested design -> each sire mated to only 1 dam -> families of full sibs

Sire 1 1 2 3 4 μs1 = 2.5

Sire 2 5 6 7 8 μs2 = 6.5

Sire 3 9 10 11 12 μs3 = 10.5

μ = 6.5

2 4 6 8 10 12

Offspring

Sire 1 1 2 3 4 μs1 = 2.5

Sire 2 5 6 7 8 μs2 = 6.5

Sire 3 9 10 11 12 μs3 = 10.5

μ = 6.5

2 4 6 8 10 12

Offspring

σ2s = Σ(μSi - μ)2/dfs

= [N1(μs1 - μ)2 + N2(μs2 - μ)2 + N3(μs3 - μ)2]/(3-1) = [4*(2.5 – 6.5)2 + 4*(6.5 – 6.5)2 + 4*(10.5 – 6.5)2]/2 = (4*42 + 4*0 + 4*42)/2 = 64

- this is the variance between the means of 3 groups, or the between-group variance component

2 4 6 8 10 12

Offspring

Sire 1 1 2 3 4 μs1 = 2.5

Sire 2 5 6 7 8 μs2 = 6.5

Sire 3 9 10 11 12 μs3 = 10.5

μ = 6.5

= [N1(μs1 - μ)2 + N2(μs2 - μ)2 + N3(μs3 - μ)2]/(3-1) = [4*(2.5 – 6.5)2 + 4*(6.5 – 6.5)2 + 4*(10.5 – 6.5)2]/2 = (4*42 + 4*0 + 4*42)/2 = 64

- how does the magnitude of this variance component relate to the covariance within the groups?

Sire 1 1 2 3 4 μs1 = 2.5

Sire 2 5 6 7 8 μs2 = 6.5

Sire 3 9 10 11 12 μs3 = 10.5

μ = 6.5

2 4 6 8 10 12

Offspring

Families of sibs

= [N1(μs1 - μ)2 + N2(μs2 - μ)2 + N3(μs3 - μ)2]/(3-1) = [4*(2.5 – 6.5)2 + 4*(6.5 – 6.5)2 + 4*(10.5 – 6.5)2]/2 = (4*42 + 4*0 + 4*42)/2 = 64

Sire 1 1 2 3 4 μs1 = 2.5

Sire 2 5 6 7 8 μs2 = 6.5

Sire 3 9 10 11 12 μs3 = 10.5

μ = 6.5

2 4 6 8 10 12

Offspring

Sire 1 1 2 3 4 μs1 = 2.5

Sire 2 5 6 7 8 μs2 = 6.5

Sire 3 9 10 11 12 μs3 = 10.5

μ = 6.5

2 4 6 8 10 12

Offspring

= [N1(μs1 - μ)2 + N2(μs2 - μ)2 + N3(μs3 - μ)2]/(3-1) = [4*(2.5 – 6.5)2 + 4*(6.5 – 6.5)2 + 4*(10.5 – 6.5)2]/2 = (4*42 + 4*0 + 4*42)/2 = 64

Sire 1 1 2 3 4 μs1 = 2.5

Sire 2 5 6 7 8 μs2 = 6.5

Sire 3 9 10 11 12 μs3 = 10.5

μ = 6.5

= [N1(μs1 - μ)2 + N2(μs2 - μ)2 + N3(μs3 - μ)2]/(3-1) = [4*(2.5 – 6.5)2 + 4*(6.5 – 6.5)2 + 4*(10.5 – 6.5)2]/2 = (4*42 + 4*0 + 4*42)/2 = 64

How to summarize the correlations between these 4 variables?

Sire 1 1 2 3 4 μs1 = 2.5

Sire 2 5 6 7 8 μs2 = 6.5

Sire 3 9 10 11 12 μs3 = 10.5

μ = 6.5

= [N1(μs1 - μ)2 + N2(μs2 - μ)2 + N3(μs3 - μ)2]/(3-1) = [4*(2.5 – 6.5)2 + 4*(6.5 – 6.5)2 + 4*(10.5 – 6.5)2]/2 = (4*42 + 4*0 + 4*42)/2 = 64

- use Pearson r (bivariately) to obtain a correlation matrix?

Sire 1 1 2 3 4 μs1 = 2.5

Sire 2 5 6 7 8 μs2 = 6.5

Sire 3 9 10 11 12 μs3 = 10.5

μ = 6.5

= [N1(μs1 - μ)2 + N2(μs2 - μ)2 + N3(μs3 - μ)2]/(3-1) = [4*(2.5 – 6.5)2 + 4*(6.5 – 6.5)2 + 4*(10.5 – 6.5)2]/2 = (4*42 + 4*0 + 4*42)/2 = 64

- use Pearson r (bivariately) to obtain a correlation matrix?- no, because a) we need a single measure of relationship

Sire 1 1 2 3 4 μs1 = 2.5

Sire 2 5 6 7 8 μs2 = 6.5

Sire 3 9 10 11 12 μs3 = 10.5

μ = 6.5

= [N1(μs1 - μ)2 + N2(μs2 - μ)2 + N3(μs3 - μ)2]/(3-1) = [4*(2.5 – 6.5)2 + 4*(6.5 – 6.5)2 + 4*(10.5 – 6.5)2]/2 = (4*42 + 4*0 + 4*42)/2 = 64

- 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 [μs] and between-group variance component [σ2

s] would stay the same, while r would change)

Sire 1 1 2 3 4 μs1 = 2.5

Sire 2 5 6 7 8 μs2 = 6.5

Sire 3 9 10 11 12 μs3 = 10.5

μ = 6.5

= [N1(μs1 - μ)2 + N2(μs2 - μ)2 + N3(μs3 - μ)2]/(3-1) = [4*(2.5 – 6.5)2 + 4*(6.5 – 6.5)2 + 4*(10.5 – 6.5)2]/2 = (4*42 + 4*0 + 4*42)/2 = 64

- solution: ICC (intraclass correlation coefficient):

Sire 1 1 2 3 4 μs1 = 2.5

Sire 2 5 6 7 8 μs2 = 6.5

Sire 3 9 10 11 12 μs3 = 10.5

μ = 6.5

= [N1(μs1 - μ)2 + N2(μs2 - μ)2 + N3(μs3 - μ)2]/(3-1) = [4*(2.5 – 6.5)2 + 4*(6.5 – 6.5)2 + 4*(10.5 – 6.5)2]/2 = (4*42 + 4*0 + 4*42)/2 = 64

- solution: ICC (intraclass correlation coefficient): a) a single measure b) insensitive to reshuffling data in rows

Sire 1 1 2 3 4 μs1 = 2.5

Sire 2 5 6 7 8 μs2 = 6.5

Sire 3 9 10 11 12 μs3 = 10.5

μ = 6.5

= [N1(μs1 - μ)2 + N2(μs2 - μ)2 + N3(μs3 - μ)2]/(3-1) = [4*(2.5 – 6.5)2 + 4*(6.5 – 6.5)2 + 4*(10.5 – 6.5)2]/2 = (4*42 + 4*0 + 4*42)/2 = 64

ICC = σ2s/(σ2

s + σ2w)

Sire 1 1 2 3 4 μs1 = 2.5

Sire 2 5 6 7 8 μs2 = 6.5

Sire 3 9 10 11 12 μs3 = 10.5

μ = 6.5

= [N1(μs1 - μ)2 + N2(μs2 - μ)2 + N3(μs3 - μ)2]/(3-1) = [4*(2.5 – 6.5)2 + 4*(6.5 – 6.5)2 + 4*(10.5 – 6.5)2]/2 = (4*42 + 4*0 + 4*42)/2 = 64

ICC = σ2s/(σ2

s + σ2w)

σ2w = Σ(pij - μSi)2/dfw

Sire 1 1 2 3 4 μs1 = 2.5

Sire 2 5 6 7 8 μs2 = 6.5

Sire 3 9 10 11 12 μs3 = 10.5

μ = 6.5

= [N1(μs1 - μ)2 + N2(μs2 - μ)2 + N3(μs3 - μ)2]/(3-1) = [4*(2.5 – 6.5)2 + 4*(6.5 – 6.5)2 + 4*(10.5 – 6.5)2]/2 = (4*42 + 4*0 + 4*42)/2 = 64

ICC = σ2s/(σ2

s + σ2w)

= [(p11 - μs1)2 + (p12 - μs1)2 + … + (p21 – μs2)2 + … + (p34 – μs3)2]/3(4-1) = 15/9 = 1.67

Sire 1 1 2 3 4 μs1 = 2.5

Sire 2 5 6 7 8 μs2 = 6.5

Sire 3 9 10 11 12 μs3 = 10.5

μ = 6.5

= [N1(μs1 - μ)2 + N2(μs2 - μ)2 + N3(μs3 - μ)2]/(3-1) = [4*(2.5 – 6.5)2 + 4*(6.5 – 6.5)2 + 4*(10.5 – 6.5)2]/2 = (4*42 + 4*0 + 4*42)/2 = 64

ICC = σ2s/(σ2

s + σ2w)

= 64/(64+1.67) = 0.97

= [(p11 - μs1)2 + (p12 - μs1)2 + … + (p21 – μs2)2 + … + (p34 – μs3)2]/3(4-1) = 15/9 = 1.67

Sire 1 1 2 3 4 μs1 = 2.5

Sire 2 5 6 7 8 μs2 = 6.5

Sire 3 9 10 11 12 μs3 = 10.5

μ = 6.5

ICC = 0.97

2 4 6 8 10 12

Offspring

2 4 6 8 10 12

Offspring

ICC = 0.10

Sire 1 1 9 4 11 μs1 =

Sire 2 7 2 12 8 μs2 =

Sire 3 6 3 10 5 μs3 =

2 4 6 8 10 12

Offspring

Sire 1 3 5 8 10 μs1 = 6.5

Sire 2 5 6 7 8 μs2 = 6.5

Sire 3 2 4 9 11 μs3 = 6.5

μ = 6.5

ICC = 0

ANOVA & sib analysis

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