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

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