15. anova for balanced split-plot experiments among whole-plot-factor marginal means inferences on )

15. ANOVA for Balanced

Split-Plot Experiments

Copyright c©2018 (Iowa State University) 15. Statistics 510 1 / 47

A Traditional Split-Plot Experiment

Field

Block 1

Block 2

Block 3

Block 4 Genotype A Genotype B Genotype C

Genotype A Genotype B Genotype C



0 50 100 150 50 0 100 150 150 0 100 50

150 0 100 50 0 100 50 150 100 0 50 150

100 150 50 0 0 50 100 150 50 0 100 150

0 150 50 100 150 0 100 50 50 0 150 100

Whole Plot or Main Plot

Split Plot or

Sub Plot


A Model for Data from the Traditional Split-Plot

Experiment

Genotype i = 1, 2, 3, Fertilizer j = 1, 2, 3, 4, Block k = 1, 2, 3, 4

yijk = µij + bk + wik + eijk

µij = mean for Genotype i, Fertilizer j

bk = random block effect

wik = random whole-plot exp. unit effect

eijk = random error = random split-plot exp. unit effect


Table of Means and Marginal Means

Fertilizer AmountGenotype 0 50 100 150

A µ11 µ12 µ13 µ14 µ1·

B µ21 µ22 µ23 µ24 µ2·

C µ31 µ32 µ33 µ34 µ3·

µ·1 µ·2 µ·3 µ·4 µ··


Best Linear Unbiased Estimators

Because the experiment is balanced, the GLS estimator is equalto the OLS estimator for any estimable Cβ:

CβΣ = C(X′Σ−1X)−X′Σ−1y = C(X′X)−X′y = Cβ.

Because the elements of E(y) are {µij : i = 1, 2, 3; j = 1, 2, 3, 4},the estimable quantities are all linear combinations of the cellmeans {µij : i = 1, 2, 3; j = 1, 2, 3, 4}.

The BLUE of∑3

i=1

∑4j=1 cijµij is

∑3i=1

∑4j=1 cijyij·.


Table of Best Linear Unbiased Estimates

Fertilizer AmountGenotype 0 50 100 150

A y11· y12· y13· y14· y1··

B y21· y22· y23· y24· y2··

C y31· y32· y33· y34· y3··

y·1· y·2· y·3· y·4· y···


ANOVA Table for the Traditional Split-Plot Design

Source DF

Blocks 4− 1 = 3

Genotypes 3− 1 = 2

Blocks× Geno (4− 1)(3− 1) = 6

Fert 4− 1 = 3

Geno× Fert (3− 1)(4− 1) = 6

Blocks× Fert (4− 1)(4− 1)

+Blocks× Geno× Fert +(4− 1)(3− 1)(4− 1) = 27

C.Total 48− 1 = 47



Source DF

Blocks 4− 1 = 3

Genotypes 3− 1 = 2

Blocks× Geno (4− 1)(3− 1) = 6

Fert 4− 1 = 3

Geno× Fert (3− 1)(4− 1) = 6

Error 3(4− 1)(4− 1) = 27

C.Total 48− 1 = 47


Why does SSBlocks×Fert + SSBlocks×Geno×Fert = SSError?

There are no terms in our model corresponding toBlock × Fert combinations; thus, there is no reason to devotea separate line of our ANOVA table to Block × Fert.

Also, it can be shown that

E(MSBlocks×Fert) = E(MSBlocks×Geno×Fert) = σ2e

Thus, it makes sense to estimate σ2e with an inverse variance

weighted average of independent unbiased estimators:


For this slide only, let

1 = Blocks× Fert and 2 = Blocks× Geno× Fert.

For ` = 1, 2, MS` ∼E(MS`)

df`χ2

df` =⇒ Var(MS`) = 2σ4e/df`.

Var−1(MS1)MS1 + Var−1(MS2)MS2

Var−1(MS1) + Var−1(MS2)=

df12σ4

eMS1 + df2

2σ4eMS2

df12σ4

e+ df2

2σ4e

=df1MS1 + df2MS2

df1 + df2

=SS1 + SS2

df1 + df2


Thus, we combine the Blocks× Fert and Blocks× Geno× Fert

lines of the ANOVA table and label the resulting line as Error.

SSBlocks×Fert + SSBlocks×Geno×Fert = SSError

dfBlocks×Fert + dfBlocks×Geno×Fert = dfError

MSError = SSError/dfError

E(MSError) = σ2e


Now let’s look at the ANOVA table and the analyses that can bedone with it in more detail.

For greater generality, let

w = the number of levels of the whole-plot treatment factor,

s = the number of levels of the split-plot treatment factor, and

b = the number of blocks.



Source DF

Blocks b− 1

Genotypes w− 1

Blocks× Geno (b− 1)(w− 1)

Fert s− 1

Geno× Fert (w− 1)(s− 1)

Blocks× Fert (b− 1)(s− 1)

+Blocks× Geno× Fert +(b− 1)(w− 1)(s− 1)

C.Total bws− 1



Source DF

Blocks b− 1

Genotypes w− 1

Blocks× Geno (b− 1)(w− 1)

Fert s− 1

Geno× Fert (w− 1)(s− 1)

Error w(b− 1)(s− 1)

C.Total bws− 1


ANOVA Table Sums of Squares

Source Sum of Squares

Block∑w

i=1

∑sj=1

∑bk=1(y··k − y···)2

Geno∑w

i=1

∑sj=1

∑bk=1(yi·· − y···)2

Block × Geno∑w

i=1

∑sj=1

∑bk=1(yi·k − yi·· − y··k + y···)2

Fert∑w

i=1

∑sj=1

∑bk=1(y·j· − y···)2

Geno× Fert∑w

i=1

∑sj=1

∑bk=1(yij· − yi·· − y·j· + y···)2

Error∑w

i=1

∑sj=1

∑bk=1(yijk − yi·k − yij· + yi··)

2

C.Total∑w

i=1

∑sj=1

∑bk=1(yijk − y···)2


Simplified ANOVA Table Sums of Squares

Source Sum of Squares

Block ws∑b

k=1(y··k − y···)2

Geno sb∑w

i=1(yi·· − y···)2

Block × Geno s∑w

i=1

∑bk=1(yi·k − yi·· − y··k + y···)2

Fert wb∑s

j=1(y·j· − y···)2

Geno× Fert b∑w

i=1

∑sj=1(yij· − yi·· − y·j· + y···)2

Error∑w

i=1

∑sj=1

∑bk=1(yijk − yi·k − yij· + yi··)

2

C.Total∑w

i=1

∑sj=1

∑bk=1(yijk − y···)2


E(MSGeno) =sb

w− 1

w∑i=1

E(yi·· − y···)2

=sb

w− 1

w∑i=1

E(µi· − µ·· + wi· − w·· + ei·· − e···)2

= sb{∑w

i=1(µi· − µ··)2

w− 1+ E

[∑wi=1(wi· − w··)2

w− 1

]+ E

[∑wi=1(ei·· − e···)2

w− 1

]}

= sb∑w

i=1(µi· − µ··)2

w− 1+ sb

σ2w

b+ sb

σ2e

sb

= sb∑w

i=1(µi· − µ··)2

w− 1+ sσ2

w + σ2e


E(MSBlock×Geno) =s

(w− 1)(b− 1)

w∑i=1

b∑k=1

E(yi·k − yi·· − y··k + y···)2

=s

(w− 1)(b− 1)

w∑i=1

b∑k=1

E(wik − wi· − w·k + w·· + ei·k − ei·· − e··k + e···)2

=s

(w− 1)(b− 1)E

[w∑

i=1

b∑k=1

(wik − wi·)2 − 2

w∑i=1

b∑k=1

(wik − wi·)(w·k − w··)

+

w∑i=1

b∑k=1

(w·k − w··)2 + e2 sum

]

=s

(w− 1)(b− 1)E

[w∑

i=1

b∑k=1

(wik − wi·)2 − w

b∑k=1

(w·k − w··)2 + e2 sum

]=

s(w− 1)(b− 1)

[w(b− 1)σ2w − w(b− 1)σ2

w/w + E(e2 sum)]


It can be shown that

E(e2 sum) = E

[w∑

i=1

b∑k=1

(ei·k − ei·· − e··k + e···)2

]

=(w− 1)(b− 1)

sσ2

e .

Putting it all together yields

E(MSBlock×Geno) = sσ2w + σ2

e .


Source Expected Mean Squares

Block

Geno sσ2w + σ2

e + sbw−1

∑wi=1(µi· − µ··)2

Block × Geno sσ2w + σ2

e

Fert

Geno× Fert

Error


The Test for Whole-Plot Factor Main Effects

To test for genotype main effects, i.e.,

H0 : µ1· = · · · = µw· ⇐⇒ H0 :sb

w− 1

w∑i=1

(µi· − µ··)2 = 0,

compare MSGenoMSBlock×Geno

to a central F distribution with w− 1 and(w− 1)(b− 1) degrees of freedom.


Comparison of Whole-Plot Factor Marginal Means

The BLUE of µ1· − µ2· is y1·· − y2··.

Var(y1·· − y2··) = Var(µ1· − µ2· + w1· − w2· + e1·· − e2··)

= 2σ2w

b + 2σ2e

sb

= 2sb(sσ2

w + σ2e ) = 2

sbE(MSBlock×Geno)

Var(y1·· − y2··) =2sb

MSBlock×Geno


We can use

t =y1·· − y2·· − (µ1· − µ2·)√

2sbMSBlock×Geno

∼ t(w−1)(b−1)

to get tests of H0 : µ1· = µ2·

or construct confidence intervals for µ1· − µ2·.


Furthermore, suppose C is a matrix whose rows are contrastvectors so that C1 = 0. Then

Var

C

y1··...

yw··

= Var

C

b· + w1· + e1··

...

b· + ww· + ew··

= Var

C1b· + C

w1· + e1··

...

ww· + ew··

= C Var

w1· + e1··...

ww· + ew··

C′

= C(σ2

w

b+σ2

e

sb

)IC′ =

(σ2

w

b+σ2

e

sb

)CC′ =

E(MSBlock×Geno)

sbCC′


An F statistic, with q and (w− 1)(b− 1) degrees of freedom, fortesting

H0 : C

µ1·...µw·

= 0, is

F =

C

y1·...

yw·

′ [

MSBlock×Genosb CC′

]−1

C

y1··...

yw··

q,

where q is the number of rows of C (which must have full rowrank to ensure that the hypothesis is testable).


Inference for the Split-Plot Factor

E(MSFert) =wb

s− 1

s∑j=1

E(y·j· − y···)2

=wb

s− 1

s∑j=1

E(µ·j − µ·· + e·j· − e···)2

=wb

s− 1

s∑j=1

(µ·j − µ··)2 + σ2e .

Likewise, it can be shown that

E(MSError) = σ2e .



Block

Geno sσ2w + σ2

e + sbw−1

∑wi=1(µi· − µ··)2


e

Fert σ2e + wb

s−1

∑sj=1(µ·j − µ··)2

Geno× Fert

Error σ2e


The Test for Split-Plot Factor Main Effects

To test for fertilizer main effects, i.e.,

H0 : µ·1 = · · · = µ·s ⇐⇒ H0 :wb

s− 1

s∑j=1

(µ·j − µ··)2 = 0,

compare MSFertMSError

to a central F distribution with s− 1 andw(s− 1)(b− 1) degrees of freedom.


Comparison of Split-Plot Factor Marginal Means

The BLUE of µ·1 − µ·2 is y·1· − y·2·.

y·1· − y·2· = (µ·1 + b· + w·· + e·1·)− (µ·2 + b· + w·· + e·2·)

Var(y·1· − y·2·) = Var(µ·1 − µ·2 + e·1· − e·2·)

= 2wbσ

2e = 2

wbE(MSError)

Var(y·1· − y·2·) = 2wbMSError


We can use

t =y·1· − y·2· − (µ·1 − µ·2)√

2wbMSError

∼ tw(s−1)(b−1)

to get tests of H0 : µ·1 = µ·2

or to construct confidence intervals for µ·1 − µ·2.


Furthermore, suppose C is a matrix with rows that are contrastvectors so that C1 = 0. Then

Var

C

y·1·...

y·s·

= Var

C

b· + w·· + e·1·

...b· + w·· + e·s·

= Var

C1b· + C1w·· + C

e·1·...

e·s·

= C Var

e·1·...

e·s·

C′

= C(σ2

e

wb

)IC′ =

E(MSError)

wbCC′


An F statistic, with q and w(s− 1)(b− 1) degrees of freedom, fortesting

H0 : C

µ·1...µ·s

= 0, is

F =

C

y·1·...

y·s·

′ [

MSErrorwb CC′

]−1

C

y·1·...

y·s·

q

where q is the number of rows of C (which must have full rowrank to ensure that the hypothesis is testable).


Inference for Interactions

E(MSGeno×Fert) =b

(w− 1)(s− 1)

w∑i=1

s∑j=1

E(yij· − yi·· − y·j· + y···)2

=b

(w− 1)(s− 1)

w∑i=1

s∑j=1

E(µij − µi· − µ·j + µ·· + eij· − ei·· − e·j· + e···)2

= · · ·

=b

(w− 1)(s− 1)

w∑i=1

s∑j=1

(µij − µi· − µ·j + µ··)2 + σ2

e .


It can be shown that

µij − µi· − µ·j + µ·· = 0 ∀ i, j

is equivalent to

µij − µij∗ − µi∗j + µi∗j∗ = 0 ∀ i 6= i∗, j 6= j∗.

Thus,b

(w− 1)(s− 1)

w∑i=1

s∑j=1

(µij − µi· − µ·j + µ··)2 = 0

is equivalent to no interactions between genotypes andfertilizers.



Block

Geno sσ2w + σ2

e + sbw−1

∑wi=1(µi· − µ··)2


e

Fert σ2e + wb

s−1

∑sj=1(µ·j − µ··)2

Geno× Fert σ2e + b

(w−1)(s−1)

∑wi=1

∑sj=1(µij − µi· − µ·j + µ··)

2

Error σ2e


The Test for Whole × Split Interaction Effects

To test for genotype × fertilizer interaction effects, i.e.,

H0 : µij − µi· − µ·j + µ·· = 0 ∀ i, j⇐⇒

H0 :b

(w− 1)(s− 1)

w∑i=1

s∑j=1

(µij − µi· − µ·j + µ··)2 = 0,

compare MSGeno×FertMSError

to a central F distribution with (w− 1)(s− 1)

and w(s− 1)(b− 1) degrees of freedom.


Inference for Simple Effects

Consider the difference between two fertilizer means within agenotype, e.g., µ11 − µ12 whose BLUE is y11· − y12·.

Var(y11· − y12·) = Var(µ11 − µ12 + b· − b· + w1· − w1· + e11· − e12·)

= 2bσ

2e

Var(y11· − y12·) = 2bMSError


We can use

t =y11· − y12· − (µ11 − µ12)√

2bMSError

∼ tw(s−1)(b−1)

to get tests of H0 : µ11 = µ12

or construct confidence intervals for µ11 − µ12.


Now consider the difference between two genotype meanswithin a fertilizer, e.g., µ11 − µ21 whose BLUE is y11· − y21·.

Var(y11· − y21·) = Var(µ11 − µ21 + w1· − w2· + e11· − e21·)

= 2σ2w

b + 2σ2e

b

= 2b(σ2

w + σ2e ).

This variance is not a constant times any expected mean squarefrom our ANOVA table.


We need an estimator of σ2w + σ2

e . We have

E(MSBlock×Geno) = sσ2w + σ2

e , E(MSError) = σ2e , and

E(

1s

MSBlock×Geno +s− 1

sMSError

)= σ2

w +σ2

e

s+

(s− 1)σ2e

s

= σ2w + σ2

e .

Thus,1s

MSBlock×Geno +s− 1

sMSError

is an unbiased estimator of σ2w + σ2

e .


It follows that

Var(y11· − y21·) ≡2sb

MSBlock×Geno +2(s− 1)

sbMSError

is an unbiased estimator of Var(y11· − y21·).

We can use

y11· − y21· − (µ11 − µ21)√Var(y11· − y21·)

·∼ td, with d degrees of freedom

computed by Cochran-Satterthwaite to get approximate tests ofH0 : µ11 = µ21 or to construct approximate confidence intervalsfor µ11 − µ21.


Full Table of Expected Mean Squares


Block wsσ2b + sσ2

w + σ2e

Geno sσ2w + σ2

e + sbw−1

∑wi=1(µi· − µ··)2


e

Fert σ2e + wb

s−1

∑sj=1(µ·j − µ··)2

Geno× Fert σ2e + b

(w−1)(s−1)

∑wi=1

∑sj=1(µij − µi· − µ·j + µ··)

2

Error σ2e


Inferences for Cell Mean µij

Var(yij·) = Var(µij + b· + wi· + eij·)

=σ2

b

b+σ2

w

b+σ2

e

b

We can construct the unbiased estimator

Var(yij·) =1

wbs[MSBlock + (w− 1) MSBlock×Geno + w(s− 1) MSError]

with approximate degrees of freedom fromCochran-Satterthwaite.


Inferences for Whole-Plot-Factor Means µi·

Var(yi··) = Var(µi· + b· + wi· + ei··)

=σ2

b

b+σ2

w

b+σ2

e

sb

This can be estimated with a linear combination of meansquares.


If block effects are considered fixed rather than random,

Var(yi··) = Var(µi· + b· + wi· + ei··)

=σ2

w

b+σ2

e

sb

=1sb

(sσ2

w + σ2e

)We can estimate this variance by 1

sbMSBlock×Geno with(w− 1)(b− 1) degrees of freedom.


Inferences for Split-Plot-Factor Means µ·j

If block effects are considered random,

Var(y·j·) = Var(µ·j + b· + w·· + e·j·)

=σ2

b

b+σ2

w

wb+σ2

e

wb

If block effects are considered fixed,

Var(y·j·) = Var(µ·j + b· + w·· + e·j·)

=σ2

w

wb+σ2

e

wb.

Both can be estimated by linear combinations of mean squares.


Summary of ANOVA for a Balanced Split-Plot

Use whole-plot-error mean square for inferences oncontrasts among whole-plot-factor marginal means

Use split-plot-error mean square for inferences oncontrasts among split-plot-factor marginal means

whole× split interactions

a simple effect within a whole-plot treatment

Construct a linear combination of mean squares forinferences on

a simple effect within a split-plot treatment

a comparison within neither whole-plot nor split-plot treatments

(e.g., µ11 − µ22)

most means


15. anova for balanced split-plot experiments among whole-plot-factor marginal means inferences on )

Documents