anova & sib analysis. basics of anova - revision application to sib analysis intraclass...
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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
- analysis of variance (ANOVA) is a way of comparing the ratio of systematic variance to unsystematic
variance in a study
ANOVA as regression
- 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?
- 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 1
0 4 10
person 2
1 7 9
person 3
1 6 8
person 4
2 3 11
person 5
1 5 7
μN = 1 μL = 5 μLB = 9 μ = 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?
Person Condition
Nothing (N) Lectures (L) Lectures + book(LB)
person 1
0 4 10
person 2
1 7 9
person 3
1 6 8
person 4
2 3 11
person 5
1 5 7
μN = 1 μL = 5 μLB = 9 μ = 52 4 6 8 10 12 14
02
46
810
Offspring
Ph
eno
typ
ic v
alu
e
perso
n
scor
e
- 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?
outcomeij = model + errorij
Person Condition
Nothing (N) Lectures (L) Lectures + book(LB)
person 1
0 4 10
person 2
1 7 9
person 3
1 6 8
person 4
2 3 11
person 5
1 5 7
μN = 1 μL = 5 μLB = 9 μ = 52 4 6 8 10 12 14
02
46
810
Offspring
Ph
eno
typ
ic v
alu
e
perso
n
scor
e
Dummy coding:
Condition Dummy variable
Dummy1 (lec)
Dummy2 (lecbook)
Nothing (N) 0 0
Lectures (L) 1 0
Lectures + book (LB)
0 1
Dummy coding:
outcomeij = model + errorij
Condition Dummy variable
Dummy1 (lec)
Dummy2 (lecbook)
Nothing (N) 0 0
Lectures (L) 1 0
Lectures + book (LB)
0 1
Dummy coding:
outcomeij = model + errorij
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
Condition Dummy variable
Dummy1 (lec)
Dummy2 (lecbook)
Nothing (N) 0 0
Lectures (L) 1 0
Lectures + book (LB)
0 1
Dummy coding:
outcomeij = model + errorij
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
knowledgei1 = b0 + b1*dummy11 + b2*dummy21 + εi1
Condition Dummy variable
Dummy1 (lec)
Dummy2 (lecbook)
Nothing (N) 0 0
Lectures (L) 1 0
Lectures + book (LB)
0 1
Dummy coding:
outcomeij = model + errorij
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
knowledgei1 = b0 + b1*dummy11 + b2*dummy21 + εi1
= b0 + b1*0 + b2*0 + εi1
Condition Dummy variable
Dummy1 (lec)
Dummy2 (lecbook)
Nothing (N) 0 0
Lectures (L) 1 0
Lectures + book (LB)
0 1
Dummy coding:
outcomeij = model + errorij
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
knowledgei1 = b0 + b1*dummy11 + b2*dummy21 + εi1
= b0 + b1*0 + b2*0 + εi1
= b0 + εi1
Condition Dummy variable
Dummy1 (lec)
Dummy2 (lecbook)
Nothing (N) 0 0
Lectures (L) 1 0
Lectures + book (LB)
0 1
Dummy coding:
outcomeij = model + errorij
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
knowledgei1 = b0 + b1*dummy11 + b2*dummy21 + εi1
= b0 + b1*0 + b2*0 + εi1
= b0 + εi1
knowledgei2 = b0 + b1*dummy12 + b2*dummy22 + εi2
Condition Dummy variable
Dummy1 (lec)
Dummy2 (lecbook)
Nothing (N) 0 0
Lectures (L) 1 0
Lectures + book (LB)
0 1
Dummy coding:
outcomeij = model + errorij
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
knowledgei1 = b0 + b1*dummy11 + b2*dummy21 + εi1
= b0 + b1*0 + b2*0 + εi1
= b0 + εi1
knowledgei2 = b0 + b1*dummy12 + b2*dummy22 + εi2
= b0 + b1*1 + b2*0 + εi2
Condition Dummy variable
Dummy1 (lec)
Dummy2 (lecbook)
Nothing (N) 0 0
Lectures (L) 1 0
Lectures + book (LB)
0 1
Dummy coding:
outcomeij = model + errorij
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
knowledgei1 = b0 + b1*dummy11 + b2*dummy21 + εi1
= b0 + b1*0 + b2*0 + εi1
= b0 + εi1
knowledgei2 = b0 + b1*dummy12 + b2*dummy22 + εi2
= b0 + b1*1 + b2*0 + εi2
= b0 + b1 + εi2
Condition Dummy variable
Dummy1 (lec)
Dummy2 (lecbook)
Nothing (N) 0 0
Lectures (L) 1 0
Lectures + book (LB)
0 1
Dummy coding:
outcomeij = model + errorij
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
knowledgei1 = b0 + b1*dummy11 + b2*dummy21 + εi1
= b0 + b1*0 + b2*0 + εi1
= b0 + εi1
knowledgei2 = b0 + b1*dummy12 + b2*dummy22 + εi2
= b0 + b1*1 + b2*0 + εi2
= b0 + b1 + εi2
knowledgei3 = b0 + b1*dummy13 + b2*dummy23 + εi3
Condition Dummy variable
Dummy1 (lec)
Dummy2 (lecbook)
Nothing (N) 0 0
Lectures (L) 1 0
Lectures + book (LB)
0 1
Dummy coding:
outcomeij = model + errorij
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
knowledgei1 = b0 + b1*dummy11 + b2*dummy21 + εi1
= b0 + b1*0 + b2*0 + εi1
= b0 + εi1
knowledgei2 = b0 + b1*dummy12 + b2*dummy22 + εi2
= b0 + b1*1 + b2*0 + εi2
= b0 + b1 + εi2
knowledgei3 = b0 + b1*dummy13 + b2*dummy23 + εi3
= b0 + b1*0 + b2*1 + εi3
= b0 + b2 + εi3
Condition Dummy variable
Dummy1 (lec)
Dummy2 (lecbook)
Nothing (N) 0 0
Lectures (L) 1 0
Lectures + book (LB)
0 1
Dummy coding:
outcomeij = model + errorij
knowledgeij = b0 + b1*dummy1j + b2*dummy2j + εij
knowledgei1 = b0 + b1*dummy11 + b2*dummy21 + εi1
= b0 + b1*0 + b2*0 + εi1
= b0 + εi1
knowledgei2 = b0 + b1*dummy12 + b2*dummy22 + εi2
= b0 + b1*1 + b2*0 + εi2
= b0 + b1 + εi2
knowledgei3 = b0 + b1*dummy13 + b2*dummy23 + εi3
= b0 + b1*0 + b2*1 + εi3
= b0 + b2 + εi3
Condition Dummy variable
Dummy1 (lec)
Dummy2 (lecbook)
Nothing (N) 0 0
Lectures (L) 1 0
Lectures + book (LB)
0 1
Therefore:
Dummy coding:
outcomeij = model + errorij
knowledgeij = b0 + b1*dummy1j + b2*dummy2j + εij
knowledgei1 = b0 + b1*dummy11 + b2*dummy21 + εi1
= b0 + b1*0 + b2*0 + εi1
= b0 + εi1
knowledgei2 = b0 + b1*dummy12 + b2*dummy22 + εi2
= b0 + b1*1 + b2*0 + εi2
= b0 + b1 + εi2
knowledgei3 = b0 + b1*dummy13 + b2*dummy23 + εi3
= b0 + b1*0 + b2*1 + εi3
= b0 + b2 + εi3
Condition Dummy variable
Dummy1 (lec)
Dummy2 (lecbook)
Nothing (N) 0 0
Lectures (L) 1 0
Lectures + book (LB)
0 1
Therefore:
→ μcondition1 = b0 b0 is the mean of condition 1 (N)
Dummy coding:
outcomeij = model + errorij
knowledgeij = b0 + b1*dummy1j + b2*dummy2j + εij
knowledgei1 = b0 + b1*dummy11 + b2*dummy21 + εi1
= b0 + b1*0 + b2*0 + εi1
= b0 + εi1
knowledgei2 = b0 + b1*dummy12 + b2*dummy22 + εi2
= b0 + b1*1 + b2*0 + εi2
= b0 + b1 + εi2
knowledgei3 = b0 + b1*dummy13 + b2*dummy23 + εi3
= b0 + b1*0 + b2*1 + εi3
= b0 + b2 + εi3
Condition Dummy variable
Dummy1 (lec)
Dummy2 (lecbook)
Nothing (N) 0 0
Lectures (L) 1 0
Lectures + book (LB)
0 1
Therefore:
→ μcondition1 = b0 b0 is the mean of condition 1 (N)
→ μcondition2 = b0 + b1
= μcondition1 + b1
μcondition2 - μcondition1 = b1
Dummy coding:
outcomeij = model + errorij
knowledgeij = b0 + b1*dummy1j + b2*dummy2j + εij
knowledgei1 = b0 + b1*dummy11 + b2*dummy21 + εi1
= b0 + b1*0 + b2*0 + εi1
= b0 + εi1
knowledgei2 = b0 + b1*dummy12 + b2*dummy22 + εi2
= b0 + b1*1 + b2*0 + εi2
= b0 + b1 + εi2
knowledgei3 = b0 + b1*dummy13 + b2*dummy23 + εi3
= b0 + b1*0 + b2*1 + εi3
= b0 + b2 + εi3
Condition Dummy variable
Dummy1 (lec)
Dummy2 (lecbook)
Nothing (N) 0 0
Lectures (L) 1 0
Lectures + book (LB)
0 1
Therefore:
→ μcondition1 = b0 b0 is the mean of condition 1 (N)
→ μcondition2 = b0 + b1
= μcondition1 + b1 b1 is the difference in means of
μcondition2 - μcondition1 = b1 condition 1 (N) and condition 2 (L)
Dummy coding:
outcomeij = model + errorij
knowledgeij = b0 + b1*dummy1j + b2*dummy2j + εij
knowledgei1 = b0 + b1*dummy11 + b2*dummy21 + εi1
= b0 + b1*0 + b2*0 + εi1
= b0 + εi1
knowledgei2 = b0 + b1*dummy12 + b2*dummy22 + εi2
= b0 + b1*1 + b2*0 + εi2
= b0 + b1 + εi2
knowledgei3 = b0 + b1*dummy13 + b2*dummy23 + εi3
= b0 + b1*0 + b2*1 + εi3
= b0 + b2 + εi3
Condition Dummy variable
Dummy1 (lec)
Dummy2 (lecbook)
Nothing (N) 0 0
Lectures (L) 1 0
Lectures + book (LB)
0 1
Therefore:
→ μcondition1 = b0 b0 is the mean of condition 1 (N)
→ μcondition2 = b0 + b1
= μcondition1 + b1 b1 is the difference in means of
μcondition2 - μcondition1 = b1 condition 1 (N) and condition 2 (L)
→ μcondition3 = b0 + b2
= μcondition1 + b2
μcondition3 - μcondition1 = b2
Dummy coding:
outcomeij = model + errorij
knowledgeij = b0 + b1*dummy1j + b2*dummy2j + εij
knowledgei1 = b0 + b1*dummy11 + b2*dummy21 + εi1
= b0 + b1*0 + b2*0 + εi1
= b0 + εi1
knowledgei2 = b0 + b1*dummy12 + b2*dummy22 + εi2
= b0 + b1*1 + b2*0 + εi2
= b0 + b1 + εi2
knowledgei3 = b0 + b1*dummy13 + b2*dummy23 + εi3
= b0 + b1*0 + b2*1 + εi3
= b0 + b2 + εi3
Condition Dummy variable
Dummy1 (lec)
Dummy2 (lecbook)
Nothing (N) 0 0
Lectures (L) 1 0
Lectures + book (LB)
0 1
Therefore:
→ μcondition1 = b0 b0 is the mean of condition 1 (N)
→ μcondition2 = b0 + b1
= μcondition1 + b1 b1 is the difference in means of
μcondition2 - μcondition1 = b1 condition 1 (N) and condition 2 (L)
→ μcondition3 = b0 + b2
= μcondition1 + b2 b2 is the difference in means of
μcondition3 - μcondition1 = b2 condition 1 (N) and condition 3
(LB)
Dummy coding:
Person Condition
Nothing (N) Lectures (L) Lectures + book(LB)
person 1 0 4 10
person 2 1 7 9
person 3 1 6 8
person 4 2 3 11
person 5 1 5 7
μN = 1 μL = 5 μLB = 9 μ = 5
2 4 6 8 10 12 14
02
46
810
Offspring
Ph
eno
typ
ic v
alu
e
μN
μL
μLB
μ
Person Condition
Nothing (N) Lectures (L) Lectures + book(LB)
person 1 0 4 10
person 2 1 7 9
person 3 1 6 8
person 4 2 3 11
person 5 1 5 7
μN = 1 μL = 5 μLB = 9 μ = 5
2 4 6 8 10 12 14
02
46
810
Offspring
Ph
eno
typ
ic v
alu
e
μN
μL
μLB
μ
b0
b1
b2
Person Condition
Nothing (N) Lectures (L) Lectures + book(LB)
person 1 0 4 10
person 2 1 7 9
person 3 1 6 8
person 4 2 3 11
person 5 1 5 7
μN = 1 μL = 5 μLB = 9 μ = 5
2 4 6 8 10 12 14
02
46
810
Offspring
Ph
eno
typ
ic v
alu
e
μN
μL
μLB
μ
Person Condition
Nothing (N) Lectures (L) Lectures + book(LB)
person 1 0 4 10
person 2 1 7 9
person 3 1 6 8
person 4 2 3 11
person 5 1 5 7
μN = 1 μL = 5 μLB = 9 μ = 5
2 4 6 8 10 12 14
02
46
810
Offspring
Ph
eno
typ
ic v
alu
e
μN
μL
μLB
μ
Sums of squares
Person Condition
Nothing (N) Lectures (L) Lectures + book(LB)
person 1 0 4 10
person 2 1 7 9
person 3 1 6 8
person 4 2 3 11
person 5 1 5 7
μN = 1 μL = 5 μLB = 9 μ = 5
2 4 6 8 10 12 14
02
46
810
Offspring
Ph
eno
typ
ic v
alu
e
μN
μL
μLB
μ
Sums of squares
SST = Σ(scoreij - μ)2
Person Condition
Nothing (N) Lectures (L) Lectures + book(LB)
person 1 0 4 10
person 2 1 7 9
person 3 1 6 8
person 4 2 3 11
person 5 1 5 7
μN = 1 μL = 5 μLB = 9 μ = 5
2 4 6 8 10 12 14
02
46
810
Offspring
Ph
eno
typ
ic v
alu
e
μN
μL
μLB
μ
Sums of squares
SST = Σ(scoreij - μ)2
Person Condition
Nothing (N) Lectures (L) Lectures + book(LB)
person 1 0 4 10
person 2 1 7 9
person 3 1 6 8
person 4 2 3 11
person 5 1 5 7
μN = 1 μL = 5 μLB = 9 μ = 5
2 4 6 8 10 12 14
02
46
810
Offspring
Ph
eno
typ
ic v
alu
e
μN
μL
μLB
μ
Sums of squares
SST = Σ(scoreij - μ)2
SSB = ΣNj(μj - μ)2
Person Condition
Nothing (N) Lectures (L) Lectures + book(LB)
person 1 0 4 10
person 2 1 7 9
person 3 1 6 8
person 4 2 3 11
person 5 1 5 7
μN = 1 μL = 5 μLB = 9 μ = 5
2 4 6 8 10 12 14
02
46
810
Offspring
Ph
eno
typ
ic v
alu
e
μN
μL
μLB
μ
Sums of squares
SST = Σ(scoreij - μ)2
SSB = ΣNj(μj - μ)2
Person Condition
Nothing (N) Lectures (L) Lectures + book(LB)
person 1 0 4 10
person 2 1 7 9
person 3 1 6 8
person 4 2 3 11
person 5 1 5 7
μN = 1 μL = 5 μLB = 9 μ = 5
2 4 6 8 10 12 14
02
46
810
Offspring
Ph
eno
typ
ic v
alu
e
μN
μL
μLB
μ
Sums of squares
SST = Σ(scoreij - μ)2
SSB = ΣNj(μj - μ)2
SSW = Σ(scoreij - μj)2
Person Condition
Nothing (N) Lectures (L) Lectures + book(LB)
person 1 0 4 10
person 2 1 7 9
person 3 1 6 8
person 4 2 3 11
person 5 1 5 7
μN = 1 μL = 5 μLB = 9 μ = 5
2 4 6 8 10 12 14
02
46
810
Offspring
Ph
eno
typ
ic v
alu
e
μN
μL
μLB
μ
Sums of squares
SST = Σ(scoreij - μ)2
SSB = ΣNj(μj - μ)2
SSW = Σ(scoreij - μj)2
Person Condition
Nothing (N) Lectures (L) Lectures + book(LB)
person 1 0 4 10
person 2 1 7 9
person 3 1 6 8
person 4 2 3 11
person 5 1 5 7
μN = 1 μL = 5 μLB = 9 μ = 5
2 4 6 8 10 12 14
02
46
810
Offspring
Ph
eno
typ
ic v
alu
e
μN
μL
μLB
μ
Sums of squares
SST = Σ(scoreij - μ)2
SSB = ΣNj(μj - μ)2
SSW = Σ(scoreij - μj)2
SST = SSB + SSW
Person Condition
Nothing (N) Lectures (L) Lectures + book(LB)
person 1 0 4 10
person 2 1 7 9
person 3 1 6 8
person 4 2 3 11
person 5 1 5 7
μN = 1 μL = 5 μLB = 9 μ = 5
2 4 6 8 10 12 14
02
46
810
Offspring
Ph
eno
typ
ic v
alu
e
μN
μL
μLB
μ
Sums of squares
SST = Σ(scoreij - μ)2
SSB = ΣNj(μj - μ)2
SSW = Σ(scoreij - μj)2
SST = SSB + SSW
Degrees of
freedom
dfT = MN - 1
dfB = M – 1
dfW = M(N – 1)N = number of people per conditionM = number of conditions
Person Condition
Nothing (N) Lectures (L) Lectures + book(LB)
person 1 0 4 10
person 2 1 7 9
person 3 1 6 8
person 4 2 3 11
person 5 1 5 7
μN = 1 μL = 5 μLB = 9 μ = 5
2 4 6 8 10 12 14
02
46
810
Offspring
Ph
eno
typ
ic v
alu
e
μN
μL
μLB
μ
Sums of squares
SST = Σ(scoreij - μ)2
SSB = ΣNj(μj - μ)2
SSW = Σ(scoreij - μj)2
SST = SSB + SSW
Degrees of
freedom
dfT = MN - 1
dfB = M – 1
dfW = M(N – 1)
Mean squares
MST = SST/dfT
MSB = SSB/dfB
MSW = SSW/dfW
N = number of people per conditionM = number of conditions
Person Condition
Nothing (N) Lectures (L) Lectures + book(LB)
person 1 0 4 10
person 2 1 7 9
person 3 1 6 8
person 4 2 3 11
person 5 1 5 7
μN = 1 μL = 5 μLB = 9 μ = 5
2 4 6 8 10 12 14
02
46
810
Offspring
Ph
eno
typ
ic v
alu
e
μN
μL
μLB
μ
Sums of squares
SST = Σ(scoreij - μ)2
SSB = ΣNj(μj - μ)2
SSW = Σ(scoreij - μj)2
SST = SSB + SSW
Degrees of
freedom
dfT = MN - 1
dfB = M – 1
dfW = M(N – 1)
Mean squares
MST = SST/dfT
MSB = SSB/dfB
MSW = SSW/dfW
N = number of people per conditionM = number of conditions
F-ratio
F = MSB/MSW
= MSmodel/MSerror
Sib analysis
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
- number of males (sires) each mated
to number of females (dams)
- 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
- 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)
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
Sire 1
Sire 2
Sire 3
Sire 4
Sire 5
Sire 6
Sire 7
Sire 8
0 5 10 15
02
46
810
Offspring
Ph
en
oty
pic
va
lue
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:
Partitioning the phenotypic variance
(VP):
- between-sire component
Sib analysis
Sire 1
Sire 2
Sire 3
Sire 4
Sire 5
Sire 6
Sire 7
Sire 8
ANOVA:
Partitioning the phenotypic variance
(VP):
- 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
02
46
810
Offspring
Ph
en
oty
pic
va
lue
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
02
46
810
Offspring
Ph
en
oty
pic
va
lue
μsire1
Sib analysis
μsire2
μsire3
Sib analysis
Sire 1
Sire 2
Sire 3
Sire 4
Sire 5
Sire 6
Sire 7
Sire 8
ANOVA:
Partitioning the phenotypic variance
(VP):
- between-sire component
Sib analysis
Sire 1
Sire 2
Sire 3
Sire 4
Sire 5
Sire 6
Sire 7
Sire 8
ANOVA:
Partitioning the phenotypic variance
(VP):
- 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:
Partitioning the phenotypic variance
(VP):
- 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
02
46
810
Offspring
Ph
en
oty
pic
va
lue
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
02
46
810
Offspring
Ph
en
oty
pic
va
lue
μsire1
Sib analysis
μsire2
μsire3
Sib analysis
Sire 1
Sire 2
Sire 3
Sire 4
Sire 5
Sire 6
Sire 7
Sire 8
ANOVA:
Partitioning the phenotypic variance
(VP):
- 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:
Partitioning the phenotypic variance
(VP):
- 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:
Partitioning the phenotypic variance
(VP):
- 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
02
46
810
Offspring
Ph
en
oty
pic
va
lue
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
02
46
810
Offspring
Ph
en
oty
pic
va
lue
μsire1
Sib analysis
μsire2
μsire3
Sib analysis
Sire 1
Sire 2
Sire 3
Sire 4
Sire 5
Sire 6
Sire 7
Sire 8
ANOVA:
Partitioning the phenotypic variance
(VP):
- 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:
Partitioning the phenotypic variance
(VP):
- 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:
Partitioning the phenotypic variance
(VP):
- 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:
Partitioning the phenotypic variance
(VP):
σ2T = σ2
S + σ2D +
σ2W
- 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:
Partitioning the phenotypic variance
(VP):
- between-sire component (σ2S) = variance between means of half-sib families = covHS = ¼VA
- between-dam, within-sire component (σ2D)
- within-progeny component (σ2W)
σ2T = σ2
S + σ2D +
σ2W
0 5 10 15
02
46
810
Offspring
Ph
en
oty
pic
va
lue
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:
Partitioning the phenotypic variance
(VP):
- between-sire component (σ2S) = variance between means of half-sib families = covHS = ¼VA
- between-dam, within-sire component (σ2D)
- within-progeny component (σ2W)
σ2T = σ2
S + σ2D +
σ2W
Sib analysis
Sire 1
Sire 2
Sire 3
Sire 4
Sire 5
Sire 6
Sire 7
Sire 8
ANOVA:
Partitioning the phenotypic variance
(VP):
- between-sire component (σ2S) = variance between means of half-sib families = covHS = ¼VA
- between-dam, within-sire component (σ2D)
- within-progeny component (σ2W) = total variance minus variance between groups = VP – covFS = ½VA +
¾VD + VEw
σ2T = σ2
S + σ2D +
σ2W
Sib analysis
Sire 1
Sire 2
Sire 3
Sire 4
Sire 5
Sire 6
Sire 7
Sire 8
ANOVA:
Partitioning the phenotypic variance
(VP):
- between-sire component (σ2S) = variance between means of half-sib families = covHS = ¼VA
- 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 +
σ2W
Question:
Why is any between group variance component equal to the covariance of the members of the groups?
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.
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).
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).
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
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
2 4 6 8 10 12
24
68
1012
Offspring
Ph
eno
typ
ic v
alu
e
μs1
μs2
μs3
μ
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
2 4 6 8 10 12
24
68
1012
Offspring
Ph
eno
typ
ic v
alu
e
μs1
μs2
μs3
μ
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
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 component2 4 6 8 10 12
24
68
1012
Offspring
Ph
eno
typ
ic v
alu
e
μs1
μs2
μs3
μ
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
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
- how does the magnitude of this variance
component
relate to the covariance within the groups?
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
2 4 6 8 10 12
24
68
1012
Offspring
Ph
eno
typ
ic v
alu
e
μs1
μs2
μs3
μ
Families of sibs
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
- how does the magnitude of this variance
component
relate to the covariance within the groups?
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
2 4 6 8 10 12
24
68
1012
Offspring
Ph
eno
typ
ic v
alu
e
μs1
μs2
μs3
μ
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
2 4 6 8 10 12
24
68
1012
Offspring
Ph
eno
typ
ic v
alu
e
μs1
μs2
μs3
μ
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
- how does the magnitude of this variance
component
relate to the covariance within the groups?
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
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
- how does the magnitude of this variance
component
relate to the covariance within the groups?
How to summarize the correlations between these 4
variables?
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
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
- 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?
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
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
- 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
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
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
- 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 [μs] and between-group variance
component [σ2s] would stay the same,
while
r would change)
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
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
- 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 [μs] and between-group variance
component [σ2s] would stay the same,
while
r would change)
- solution: ICC (intraclass correlation coefficient):
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
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
- 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 [μs] and between-group variance
component [σ2s] would stay the same,
while
r would change)
- solution: ICC (intraclass correlation coefficient):
a) a single measure
b) insensitive to reshuffling data in rows
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
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
ICC = σ2s/(σ2
s + σ2w)
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 [μs] and between-group variance
component [σ2s] would stay the same,
while
r would change)
- solution: ICC (intraclass correlation coefficient):
a) a single measure
b) insensitive to reshuffling data in rows
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
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
ICC = σ2s/(σ2
s + σ2w)
σ2w = Σ(pij - μSi)2/dfw
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 [μs] and between-group variance
component [σ2s] would stay the same,
while
r would change)
- solution: ICC (intraclass correlation coefficient):
a) a single measure
b) insensitive to reshuffling data in rows
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
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
ICC = σ2s/(σ2
s + σ2w)
σ2w = Σ(pij - μSi)2/dfw
= [(p11 - μs1)2 + (p12 - μs1)2 + … + (p21 – μs2)2 + …
+
(p34 – μs3)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 [μs] and between-group variance
component [σ2s] would stay the same,
while
r would change)
- solution: ICC (intraclass correlation coefficient):
a) a single measure
b) insensitive to reshuffling data in rows
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
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
ICC = σ2s/(σ2
s + σ2w)
= 64/(64+1.67) = 0.97
σ2w = Σ(pij - μSi)2/dfw
= [(p11 - μs1)2 + (p12 - μs1)2 + … + (p21 – μs2)2 + …
+
(p34 – μs3)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 [μs] and between-group variance
component [σ2s] would stay the same,
while
r would change)
- solution: ICC (intraclass correlation coefficient):
a) a single measure
b) insensitive to reshuffling data in rows
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
ICC = 0.97
2 4 6 8 10 12
24
68
1012
Offspring
Ph
eno
typ
ic v
alu
e
2 4 6 8 10 12
24
68
1012
Offspring
Ph
eno
typ
ic v
alu
e
ICC = 0.10
Measures of phenotype
Offspring 1 Offspring 2 Offspring 3 Offspring 4 Mean
Sire 1 1 9 4 11 μs1 =
Sire 2 7 2 12 8 μs2 =
Sire 3 6 3 10 5 μs3 =
μ =
2 4 6 8 10 12
24
68
10
Offspring
Ph
eno
typ
ic v
alu
e
Measures of phenotype
Offspring 1 Offspring 2 Offspring 3 Offspring 4 Mean
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