Analyses of Variance. Simple Situation Genotype AGenotype B 13534.

Analyses of VarianceAnalyses of Variance

Simple SituationSimple Situation

Genotype A Genotype B135 34

Simple SituationSimple Situation

Genotype A Genotype B135 34115 76102 83110 64

115.5 64.2

Observations and QuestionsObservations and Questions

From the replicated means, genotype A From the replicated means, genotype A is “better” than genotype B.is “better” than genotype B.

What is the probability that this result What is the probability that this result will be repeated if this test were done will be repeated if this test were done say 100 times?say 100 times?

Could this result have occurred by Could this result have occurred by randomrandom chance? chance?

tt-test-test

|x|x11-x-x22| | 2[(2[(11

22++2222)/()/(nn11++nn22)])]

tt = =

ReplicateGenotype

Stephens Lambart1 55 782 66 913 49 974 64 825 70 856 68 77

tt-test-test

ReplicateReplicateGenotypeGenotype

StephensStephens LambartLambart112233445566

555566664949646470706868

787891919797828285857777

xxmean xmean x

3723726262

5105108585



555566664949646470706868

787891919797828285857777

xxmean xmean x

xx22

((x)x)22//nnSS(x)SS(x)

3723726262

23,40223,40223,06423,064

338338

5105108585

43,65243,65243,35043,350

302302



555566664949646470706868

787891919797828285857777

xxmean xmean x

xx22

((x)x)22//nnSS(x)SS(x)

dfdf

3723726262

23,40223,40223,06423,064

33833855

5105108585

43,65243,65243,35043,350

30230255

22 76.676.6 60.460.4

tt-test-test

Stephens = 62 bushelsStephens = 62 bushels

Lambart = 85 bushelsLambart = 85 bushels

Significant?Significant?

tt-test-test

Stephens = 62 bushels Stephens = 62 bushels Lambart = 85 bushelsLambart = 85 bushels

|x|xStephensStephens-x-xLambartLambart| | 2[(2[(StSt

22++LaLa22)/()/(nnStSt++nnLaLa)])]

tt = =

tt = |85-62|/ = |85-62|/2[(60+77)/12] = 4.822[(60+77)/12] = 4.82

cwcw tt10df10df : exceeds 99% table value : exceeds 99% table value

More than two treatmentsMore than two treatments

Rep.Genotype

Brundage Lambert Croft Stephens1 64 78 75 552 72 91 93 663 68 97 78 494 77 82 71 645 56 85 63 706 95 77 76 68

Multiple Multiple tt-tests-tests

Brundage Brundage vv Lambert; Brundage Lambert; Brundage v v Croft; Croft; Brundage Brundage v v Stephens; Lambert Stephens; Lambert v v Croft; Croft; Lambert Lambert v v Stephens; Croft Stephens; Croft v v StephensStephens..

Problems?Problems? If all tests were done at 95% significance If all tests were done at 95% significance

level, and one difference was significant, level, and one difference was significant, we have done 6 tests and would expect 1/20 we have done 6 tests and would expect 1/20 to be significant, at random.to be significant, at random.


#Genotype

TotalBrundage Lambert Croft Stephens

xx

43272

51085

45676

37262

1,770295


#Genotype


xx

43272

51085

45676

37262

1,770295

x2

(x)2/nss(x)

df

31,99431,104

8905

43,65243,350

3025

35,14443,656

4885

23,40223,064

3385

134,192132,174

2,01820


#Genotype


xx

43272

51085

45676

37262

1,770295

x2

(x)2/nss(x)

df

31,99431,104

8905

43,65243,350

3025

35,14443,656

4885

23,40223,064

3385

134,192132,174

2,01820

2 178.0 60.4 97.6 76.6 100.9

Analysis of VarianceAnalysis of Variance

From pooled variance we can estimate From pooled variance we can estimate a pooled SED between any two means a pooled SED between any two means = = (2)(100.9)/6 = (2)(100.9)/6 = ++ 5.80, and use this 5.80, and use this in all in all tt-tests.-tests.

Alternatively an analysis of variance Alternatively an analysis of variance could be carried out.could be carried out.

This form of analysis was first This form of analysis was first proposed by Fisher in the 1920’sproposed by Fisher in the 1920’s


Is an elegant and quicker way to Is an elegant and quicker way to calculate a pooled error term.calculate a pooled error term.

Analysis is simple in simple designs but Analysis is simple in simple designs but can be complicated and lengthy in some can be complicated and lengthy in some designs (i.e. rectangular lattices).designs (i.e. rectangular lattices).

In some experimental designs the In some experimental designs the ANOVA is the only method to estimate ANOVA is the only method to estimate a pooled error term.a pooled error term.


It can provide an It can provide an FF-test to tests specific -test to tests specific hypotheses. (i.e. to test general hypotheses. (i.e. to test general differences between different differences between different treatments).treatments).

Can be an invaluable Can be an invaluable initial initial contributioncontribution to interpretation of to interpretation of experiments.experiments.

Theory of Analysis of VarianceTheory of Analysis of Variance

Consider a simple CRB design.Consider a simple CRB design.Four treatments (Four treatments (nn = 4). = 4).With all treatments replicated 5 times With all treatments replicated 5 times

(k(k = 5) = 5)..The total experiment would be The total experiment would be nn x x kk = =

20 experimental units.20 experimental units.


Rep.Treatment

A B C D1 x11 x21 x31 x41

2 x12 x22 x32 x42

3 x13 x23 x33 x43

4 x14 x24 x34 x44

5 x15 x25 x35 x45


Rep.Treatment

MeanA B C D

1 x11 x21 x31 x41 x.1

2 x12 x22 x32 x42 x.2

3 x13 x23 x33 x43 x.3

4 x14 x24 x34 x44 x.4

5 x15 x25 x35 x45 x.5


Rep.Treatment

MeanA B C D

1 x11 x21 x31 x41 x.1

2 x12 x22 x32 x42 x.2

3 x13 x23 x33 x43 x.3

4 x14 x24 x34 x44 x.4

5 x15 x25 x35 x45 x.5

Mean x1. x2. x3. x4. x..


TSS = TSS = iijj(x(xijij-x-x....))22

TMS = TMS = iijj(x(xijij-x-x....))22/(/(nknk-1)-1)

iijj(x(xijij-x-x....))2 2 = = iijj[(x[(xijij-x-xi.i.) + (x) + (xi.i.-x-x....)])]2 2

iijj[(x[(xijij-x-xi.i.))22+2(x+2(xijij-x-xi.i.)(x)(xi.i.-x-x....)+(x)+(xi.i.-x-x....))22] ]


iijj[(x[(xijij-x-xi.i.))22+2(x+2(xijij-x-xi.i.)(x)(xi.i.-x-x....)+(x)+(xi.i.-x-x....))22] ]

22iijj(x(xijij-x-xi.i.)(x)(xi.i.-x-x....))

ii[2[2nn (x (xi.i.-x-x....) ) jj(x(xijij-x-xi.i.)])]

But!But! jj(x(xijij-x-xi.i.) = 0) = 0

22iijj(x(xijij-x-xi.i.)(x)(xi.i.-x-x....) = 0) = 0


iijj[(x[(xijij-x-xi.i.))22++2(x2(xijij-x-xi.i.)(x)(xi.i.-x-x....))+(x+(xi.i.-x-x....))22] ]

iijj[(x[(xijij-x-xi.i.))22 + (x + (xi.i.-x-x....))22] ]

iijj(x(xijij-x-x....))22 = = iijj(x(xijij-x-xi.i.))22++kkii(x(xi.i.-x-x....))22] ]

kkii(x(xi.i.-x-x....))22 = Between Treatment SS = Between Treatment SS

iijj(x(xijij-x-xi.i.))2 2 = Within Treatment SS= Within Treatment SS


df [WTSS] = df [WTSS] = nknk--n n : df [BTSS] = : df [BTSS] = nn-1-1

MS = SS/dfMS = SS/df

WTMS ~WTMS ~ 22nk-n dfnk-n df : B : BTMS ~TMS ~ 22

n-1 dfn-1 df

kkii(x(xi.i.-x-x....))22 = Between Treatment SS = Between Treatment SS

iijj(x(xijij-x-xi.i.))2 2 = Within Treatment SS= Within Treatment SS


WTMS ~WTMS ~ 22nk-n dfnk-n df : B : BTMS ~TMS ~ 22

n-1 dfn-1 df

YYijij = = + g + gii + e + eijij

ggii = BTMS : e = BTMS : eij ij = WTMS= WTMS

Assumption is homogeneity of error Assumption is homogeneity of error variance between treatments.variance between treatments.


Source of variation df EMS

Between treatments n-1 e2 + kt

2

Within treatments nk-n e2

Total nk-1

[e2 + kt

2]/e2 = 1, if kt

2 = 0

Analysis of Variance of CRBAnalysis of Variance of CRB

Source df SS

Between treatments k-1 [G1

2/n1 + G22/n2 … Gk

2/nk] - CF

Within treatments jk-k By difference

Total jk-1 [x112 + x12

2 + … + xjk2] - CF

CF = [CF = [xxijij]]22//jkjk


Rep.Genotype

Brundage Lambert Croft Stephens1 64 78 75 552 72 91 93 663 68 97 78 494 77 82 71 645 56 85 63 706 95 77 76 68


#Genotype


xx

43272

51085

45676

37262

1,770295

TSS = TSS = ∑∑(64(6422 + 72 + 7222 + 68 + 6822 + .... + 68 + .... + 6822) - CF) - CF

CF = CF = ∑∑(64 + 72 + 68 + .... + 68)(64 + 72 + 68 + .... + 68)22/24/24

BSS = BSS = ∑∑(432(43222/6 + 510/6 + 51022/6 + 456/6 + 45622/6 + 372/6 + 37222/6) - CF/6) - CF

WSS = TSS - BSSWSS = TSS - BSS


#Genotype


xx

43272

51085

45676

37262

1,770295

TSS = TSS = ∑∑(64(6422 + 72 + 7222 + 68 + 6822 + .... + 68 + .... + 6822) - CF) - CF

CF = CF = ∑∑(64 + 72 + 68 + .... + 68)(64 + 72 + 68 + .... + 68)22/24/24

BSS = BSS = ∑∑(432(43222 + 510 + 51022 + 456 + 45622 + 372 + 37222)/6 - CF)/6 - CF



#Genotype


xx

43272

51085

45676

37262

1,770295

TSS = TSS = ∑∑(64(6422 + 72 + 7222 + 68 + 6822 + .... + 68 + .... + 6822) - CF) - CF

CF = CF = ∑∑(64 + 72 + 68 + .... + 68)(64 + 72 + 68 + .... + 68)22/24/24

BSS = BSS = ∑∑(432(43222/6 + 510/6 + 51022/6 + 456/6 + 45622/6 + 372/6 + 37222/6) - CF/6) - CF


Example of Analysis of VarianceExample of Analysis of Variance

Source df SS MS F

Between genotypes 3 1636.5 545.5 5.41**

Within genotypes 20 2018.0 100.9

Total 23 3654.5

** = 0.01 > P > 0.001

Rep.Treatment

A B C D1 x11 x21 x31 x41

2 x12 x22 x32 x42

3 x13 x23 x33 x43

4 - x24 x34 x44

5 - - x35 -


Rep.Treatment

A B C D1 x11 x21 x31 x41

2 x12 x22 x32 x42

3 x13 x23 x33 x43

4 - x24 x34 x44

5 - - x35 -Total G1 G2 G3 G4



Source df SS

Between treatments k-1 [G1

2/n1 + G22/n2 … Gk

2/nk] - CF

Within treatments jk-k By difference

Total jk-1 [x112 + x12

2 + … + xjk2] - CF

CF = [CF = [xxijij]]22//jkjk

Assumptions behind the ANOVAAssumptions behind the ANOVA

Assumption of data being normally Assumption of data being normally distributed.distributed.

Homogeneity of error variance.Homogeneity of error variance.Additivity of variance effects.Additivity of variance effects.Data collected from a properly Data collected from a properly

randomized experiment.randomized experiment.

Dealing with Wrongful DataDealing with Wrongful Data

It is usually assumed that the data It is usually assumed that the data collected is collected is correctcorrect!.!.

Why would data not be Why would data not be correct?correct?Mis-recording, mis-classification, Mis-recording, mis-classification,

transcription errors, errors in data transcription errors, errors in data entry.entry.

Outliers.Outliers.


What things can help?What things can help?Keep detailed records, on each Keep detailed records, on each

experimental unit.experimental unit.Decide beforehand what values Decide beforehand what values

would arouse suspision.would arouse suspision.


What do you do with suspicios data?What do you do with suspicios data?If correct, and it is discarded, then If correct, and it is discarded, then

valuable information is lost. This valuable information is lost. This will bias the results.will bias the results.

If wrong and included, will bias If wrong and included, will bias results and may have extreme results and may have extreme consequences.consequences.

Checking ANOVA AccurucyChecking ANOVA Accurucy

Coefficient of variation: [Coefficient of variation: [ee//]x100.]x100. CV=(CV=(√100.9/73.75)*100=13.6%√100.9/73.75)*100=13.6%

RR22 value = {[TSS-ESS]/TSS}x100. value = {[TSS-ESS]/TSS}x100.RR22 = (1654/3654)*100 = 44.7%. = (1654/3654)*100 = 44.7%.

Compare the effect of blocking or Compare the effect of blocking or sub-blocking (discussed later).sub-blocking (discussed later).

Next ClassNext ClassANOVA of RCB ANOVA of RCB

DesignsDesigns

ANOVA of Latin ANOVA of Latin Square DesignsSquare Designs

Analyses of Variance. Simple Situation Genotype AGenotype B 13534.

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