Epidemiology 9509 - Principles of Biostatistics Chapter 15 - One …publish.uwo.ca › ~jkoval › courses › Epid9509 › chapter15 › oneway.pdf · 2012-11-15 · Principles of

Epidemiology 9509 oneway

Epidemiology 9509Principles of Biostatistics

Chapter 15 - One way analysis of variance

John Koval

Department of Epidemiology and BiostatisticsUniversity of Western Ontario

1


What is being covered

1. oneway analysis of variance

2. posthoc tests

2


What is being covered

1. oneway analysis of variance

2. posthoc testsmultiple comparisons

3


ANOVA

ANalysis Of VAriance

2 or more samples

1. may be naturally occurringas in population studynon-smokers, occasional smokers, current smokers

2. may be experimentalas in Randomized Comparative Trial (RCT)Placebo, Acetaminophen (AC), Acetylsalicylic Acid (ASA)orPlacebo, ASA at 200 mgm, ASA at 400 mgm

4


ANOVA - history

◮ agriculturelevels of fertilizerR.A. Fisher (1925, 1934)

◮ psychology

◮ clinicaldosage of drugs

5


Definitions

◮ factor - measured variableoften discrete

◮ levelsvalues of the factor

◮ treatmentparticular combination of levels of one or more factorshere equivalent to a level of a factor

6


Analysing variability

variability measured by

k∑

i=1

ni∑

j=1

(yij − y)2

k groupsyij is j’th observation in group i

7


Analysing variability II

SST - total sum of squarespartition into two

1. within treatment groups (SSW or SSE)

k∑

i=1

ni∑

j=1

(yij − yi)2

2. between groups (SSB or SSTr)

k∑

i=1

ni∑

j=1

(yi − y)2

k∑

i=1

ni (yi − y)2

8


ANOVA Table

anova table for the oneway design

Source Sum of degrees of Mean Square VarianceSquares freedom ratio

Between SSB k-1 SSBk−1

MSBMSW

Within SSW N-k SSWN−k

Total SST N-1

9


Mean Square

thanks to Manny

10


ANOVA - example

temperature reduction (◦C) for three drugs

placebo AC ASA

0.6 1.0 1.00.4 0.5 1.50.0 0.0 0.5-0.4 0.3 1.2-0.6 0.7 0.8

Ti 0.0 2.5 5.0yi 0.0 0.5 1.0s2i 0.26 0.145 0.145si 0.51 0.38 0.38

11


getting entries for ANOVA table

y =

∑

ni yi∑

ni

=5(0.0) + 5(0.5) + 5(1.0)

5 + 5 + 5

=7.5

15= 0.5

SSB =∑

ni(yi − y)2

= 5(0.0 − 0.5)2 + 5(0.5 − 0.5)2 + 5(1.0 − 0.5)2

= 5(−0.5)2 + 0 + 5(0.5)2

= 1.25 + 0 + 1.25 = 2.5

12


getting entries for ANOVA table (continued)

SSW =∑

(ni − 1)s2i

= 4(0.26) + 4(0.145) + 4(0.145)

= 1.04 + 0.58 + 0.58 = 2.20

13


ANOVA for example

temperature reduction (◦C) for three drugs

Source Sum of degrees of Mean Square VarianceSquares freedom ratio

Between 2.50 2 1.25 6.82

Within 2.20 12 0.183

Total 4.70 14

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Hypothesis testHo : none of the treatments (groups)is different from the others

Under Ho

the variance ratio (vr) ∼ F2,12

in generalvr ∼ Fk−1,N−k

in this case0.025 > Pr(F2,12 > 6.82) > 0.010(Table A.4, page A.7)

At α = 0.05, reject Ho

claimHA: at least one of the treatments (groups)is different from the others

which treatments??15


Post hoc test

three comparisons of interest

1. |y1 − y2|2. |y1 − y3|3. |y2 − y3|

16


Post hoc test II

which test

17


Post hoc test II

which testindependent samples t-test

|y1−y2|√

s2p

(

1n1

+ 1n2

)

18


Multiple comparisons

y1 = 0.0, y2 = 0.5, y3 = 1.0

q12 =|0.0−0.5|

√

0.183( 15+ 1

5)= 1.848

q13 = 3.696, q23 = 1.848

if compare to critical value t12,0.025(= 2.179)LSD (Least Significant Difference)q13 is significant, q12 and q23 are not

19


Probability of making mistake

General - and for LSD

1. 5% for each of three testsERPC - error rate per comparison (α)=5%

2. expected number of errorsError rate per experimentERPE = np = n(0.05) = 0.15

3. probability of at least one error(αE - experimentwise error rate)experimentwise error rateEWER = 1− (1− α)c = 1− (1− 0.05)3

= 1− (0.95)3 = 1− .857375 = 0.142625≈ ERPE

ERPE and EWER are less if comparisons are correlatedand they are

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10 groups

1. ERPC = 0.05

2. ERPE = np = 10(0.05) = 0.5assumes independence

3. EWER = 1− (1− α)c = 1− (1− 0.05)10

= 1− (0.95)10 = 1− 0.5987 = 0.4013ERPE not such a good approximationand this is upper bound (assumes independence)

21


more conservative procedure - Bonferroni

ERPE and EWER too largedivide α by number of possible comparisons(32

)

= 3

α = 0.05/3 = 0.01667

1. ERPC = α = 0.01667

2. ERPE = n(0.01667) = 0.05

3. EWER = 1− (1− 0.01667)3

= 1− (0.98333)3 = 1− 0.95082 = 0.04918this is upper bound

22


more conservative procedure - Bonferroni - moderate n

(102

)

= 45

α = 0.05/45 = 0.00111

1. ERPC = α = 0.00111

2. ERPE = 45(0.00111) = 0.05

3. EWER = 1− (1− 0.00111)45

= 1− (0.99889)45 = 1− 0.95125 = 0.04875this is upper bound

23


Bonferroni - example

n = 3α = 0.01667t12,0.00833 = 2.779same conclusions as for LSD

24


less conservative? - Scheffe

compare q with√

(k − 1)Fk−1,N−k

critical value is√

2F2,12,0.05=

√

2(3.885) =√7.566 = 2.787

same conclusions as for otherssimilar to Bonferronibut more conservative than Bonferroni

WHY??handles all possible linear contrastsegHo : 1

2(µ1 + µ2) = µ3

use y1+y22 − y3

25


better - Tukey

Tukey(1953) defined Studentized Range Statisticrange of means

√

s2

n

and got its distribution,Qk,N−k

not quite what we wantonly for equal-sample size

compare qij with qk,N−k,0.05/√2

in our case, q3,12,0.05 = 3.783/1.414 = 2.675

same conclusions as for previous tests

26


summary

Critical values multiple comparison tests

Name formula α = 0.05

t-test tN−k,1−(α/2) t12,0.025 = 2.179

(LSD)

Bonferroni tN−k,1−(α/[k(k−1)]) t12,0.00833 = 2.779

Scheffe√

(k − 1)Fk−1,N−k,1−α

√

2F2,12,0.05=

√

2(3.885) = 2.787

Tukey qk,N−k,1−α/√2 q3,12,0.05/

√2

(HSD) = 3.783/1.414 = 2.675

27


other tests

Multiple Range tests:SNK (Student, Newman, Keuls)

◮ put means in ascending ordery[1], ..., y[k]

◮ for |y[1] − y[k]|use Qk,N−k,0.05

◮ for |y[1] − y[k−1]||y[2] − y[k]|use Qk−1,N−k,0.05

◮ etc

◮ or |y[1] − y[2]|, ...,|y[k−1] − y[k]|use Q2,N−k,0.05

like independent samples t-test

28


SNK example

◮ put means in ascending ordery[1], y[2], y[3]

◮ for |y[1] − y[3]|use Q3,N−3,0.05

◮ for |y[1] − y[2]||y[2] − y[3]|use Q2,N−3,0.05

like independent samples t-test

29


conclusion

1. for this data setall tests have the same result

2. t-test (LSD) is anti-conservative

3. Bonferroni and Scheffe are conservative

4. Tukey (HSD) is bestfor equal sample size

30


unequal sample size

Tukey studentised range statistic - equal sample size

q =|y1 − y2|√

s2pn

Kramer(1956) modified for unequal sample size

q =|y1 − y2|

√

s2p2

(

1n1

+ 1n2

)

31


unequal sample size II

SAS uses a different version (sometimes)based on harmonic mean

nh =k

∑

n−1i

so that

q =|y1 − y2|√

s2pnh

32

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Epidemiology 9509 - Principles of Biostatistics Chapter 15 - One …publish.uwo.ca › ~jkoval › courses › Epid9509 › chapter15 › oneway.pdf · 2012-11-15 · Principles of