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Single Factor Experiments

Topic:

Comparison of more than 2 groups

One-Way Analysis of Variance

F test

Learning Aims:Understand model parametrization

Carry out an anova

Reason: Multiple t tests wont do!

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Potatoe scab

widespread disease

causes economic loss

known factors: variety, soil condition

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Experiment with different treatments

Compare 7 treatments for effectiveness inreducing scab

Field with 32 plots, 100 potatoes are randomlysampled from each plot

For each potatoe the percentage of the surfacearea affected was recorded. Response variable isthe average of the 100 percentages.

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1-Factor Design

Plots, subjects

Randomisation

Group 1 Group 2 . . . Group I

. . .

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Complete Randomisation

a) number the plots 1, . . ., 32.

b) construct a vector with 8 replicates of 1 and 4replicates of 2 to 7.

c) choose a random permutation and apply it to thevector in b).

in R:

> treatment=factor(c(rep(1,8),rep(2:7,each=4)))

> sample(treatment)

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Exploratory data analysis

Group y y

1 12 10 24 29 30 18 32 26 22.625

2 9 9 16 4 9.5

3 16 10 18 18 15.5

4 10 4 4 5 5.755 30 7 21 9 16.75

6 18 24 12 19 18.25

7 17 7 16 17 14.25

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Graphical display

1 2 3 4 5 6 7

5

10

15

20

25

30

Treatment

y

1 2 3 4 5 6 7

5

10

15

20

25

30

Treatment

y

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Two sample t tests

Group 1 Group 2 : H0:1 =2Group 1 Group 3 : H0:1 =3

Group 1 Group 4 : H0:1 =4Group 1 Group 5 : H0:1 =5Group 1 Group 6 : H0:1 =6

Group 1 Group 7 : H0:1 =7. . .

= 5%,P(Test not significant |H0) = 95%7 groups, 21 independent tests:P(none of the tests sign. |H0) = 0.9521 = 0.34P(at least one test sign. |H0) = 0.66 (more realistic: 0.42)

1 (1 )n

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Bonferroni correction

ChooseTsuch that

1 (1 T)n

=E= 5%

(T =testwise,E= experimentwise)

Since1 (1 n)n , the significance level for asingle test has to be divided by the number of tests.

Overcorrection, not very efficient.

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Analysis of variance

Comparison of more than 2 groups

for more complex designs

global F test

Idea:

total variability

in data=

source of

variation 1+

source of

variation 2+ . . .

Comparison of components

total = treatment + experimental error

total = variability of plots with + variability of plots with

different treatments the same treatment

2

+treatment effect 2

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Definitions

Factor: categorical, explanatory variableLevel: value of a factor

Ex 1: Factor= soil treatment, 7 levels 1 7.= One-way analysis of varianceEx 2: 3 varieties with 4 quantities of fertilizer

= Two-way analysis of varianceTreatment: combination of factor levels

Plot, experimental unit: smallest unit to which a

treatment can be appliedEx: feeding (chicken, chicken-houses), dentalmedicine (families, people, teeth)

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One-way analysis of variance

Model:

response = treatment + error (Plot)

Yij = + Ai + ij(1)

i=1,...,I;j=1,...,Ji

=overall meanAi=ith treatment effectij =random error, N(0, 2)iid.

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Illustration of model (1)

0.

05

0.

10

0.

15

0.

20

0.

25

0.

30

o ooo o oo o oo o o oo ooo oo o oo o ooooo oo o o ooo oo oo o

+A1 +A2 +A3 +A4

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Necessary constraint

Model (1) is overparametrized, a restriction is needed.

usual constraint:JiAi= 0,

Ai= 0ifJi=Jfor alli

Ai denotes the deviation from overall mean.

A1= 0, resp.AI= 0First (or last) group is reference group.

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Decomposition of the deviation of a

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Decomposition of the deviation of a

response from the overall mean

yij y..= yi. y..

+ yij yi.

deviation of deviation from

the group mean the group mean

yi.= 1Ji

jyij mean of groupi,

y..= 1N

i

jyij overall mean,N=

Ji.

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Analysis of variance identity

i

j

(yij y..)2

total variability

=i

j

(yi. y..)2

variability between groups

+i

j

(yij yi.)2

variability within groups

total sum = treatment sum + residual sum

of squares of squares of squares

SStot = SStreat + SSres

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Mean squares

Total mean square: M Stot=SStot/(N 1)Residual mean square: M Sres =SSres/(N I)

SSresN I

=

i(Ji 1)S

2i

i(Ji 1) , S2i =

j(yij yi.)

2

Ji 1

M Sres = 2 = V ar(Yij), E(M Sres) =

2

Treatment mean square: M Streat=SStreat/(I 1)

E(M Streat) =2 +

JiA

2i /(I 1)

dftot=dftreat+ dfres, N 1 =I 1 + N I

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F test

H0: allAi= 0

HA: at least oneAi= 0

Sinceij N(0, 2),F =MStreatMSres

has underH0 anF

distribution withI 1andN Idegrees of freedom.

one-sided test:rejectH0 ifF > F95%,I1,NI

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Chisquare and t distribution

LetZ1, . . . , Z n N(0, 1),iid. Then

X=Z21 + Z

22 + + Z

2n

has a2 distribution withndf,X 2n

LetZ N(0, 1)andX 2

n be independentrandom variables. The distribution of

T = Z

X/nis called thetdistribution withndf,T tn

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F distribution

LetX1 2n andX2 2m be independent random

variables. The distribution of

F = X1/n

X2/m

is called theFdistribution withnandmdf,F Fn,m

Properties: F1,m=t2mE(Fn,m) =

mm2

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R: anova table

> mod1=aov(ytreatment,data=scab)

> summary(mod1)

Df Sum Sq Mean Sq F value Pr(>F)

treatment 6 972.34 162.06 3.608 0.0103 *

Residuals 25 1122.88 44.92

F test is significant, there are significant treatmentdifferences.

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