stats 330: lecture 28
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© Department of Statistics 2012 STATS 330 Lecture 28: Slide 2
Plan of the dayIn today’s lecture we apply Poisson regression to the analysis of contingency tables having 3 or 4 dimensions.Topics
– Types of independence– Connection between independence and interactions for
3 and 4 dimensional tables– Hierarchical models– Graphical models– Examples
Reference: Coursebook, sections 5.3, 5.3.1
© Department of Statistics 2012 STATS 330 Lecture 28: Slide 3
Example: the Florida murder data
• 326 convicted murderers in Florida, 1976-1977
• Classified by– Death penalty (n/y)– Victims race (black/white)– Defendants race (black/white)
© Department of Statistics 2012 STATS 330 Lecture 28: Slide 4
Contingency table
Victim's race
Black White
Defendant's Death Penalty Death Penalty
Race Yes No Yes No
Black 13 195 23 105
White 1 19 39 265
© Department of Statistics 2012 STATS 330 Lecture 28: Slide 5
Getting the data into Rmurder.df<-data.frame(expand.grid( defendant=c("b","w"),dp = c("y","n"),victim=c("b","w")),counts=c(13,1,195,19,23,39,105,265))
Note use of function expand.gridexpand.grid(defendant=c("b","w"),dp = c("y","n"),victim=c("b","w")) defendant dp victim1 b y b2 w y b3 b n b4 w n b5 b y w6 w y w7 b n w8 w n w
© Department of Statistics 2012 STATS 330 Lecture 28: Slide 6
The data frame
defendant dp victim counts1 b y b 132 w y b 13 b n b 1954 w n b 195 b y w 236 w y w 397 b n w 1058 w n w 265
© Department of Statistics 2012 STATS 330 Lecture 28: Slide 7
Questions
• Is death penalty independent of race?
• What is the role of victim’s race?
• What does “independent” mean when we have 3 factors?
© Department of Statistics 2012 STATS 330 Lecture 28: Slide 8
Types of independence
• Suppose we have 3 criteria (factors) A, B and C
• In a multinomial sampling context, let
ijk = Pr(A=i, B=j, C=k)
• Various forms of independence may be of interest. These can be expressed in terms of the probabilities ijk
© Department of Statistics 2012 STATS 330 Lecture 28: Slide 9
Marginal probabilities
ij
kijk
k
k
kCjBiAPjBiAP
kCjBiAjBiA
),,(),(
},,{},{
© Department of Statistics 2012 STATS 330 Lecture 28: Slide 10
Marginal probabilities (ii)
i
j kijk
j k
kj
kCjBiAPiAP
kCjBiAiA
),,()(
},,{}{,
© Department of Statistics 2012 STATS 330 Lecture 28: Slide 11
All three factors independent
• This can be expressed as
kji
ijk
k)(Cj)(Bi)(A
kCjBiA
PrPrPr
),,Pr(
© Department of Statistics 2012 STATS 330 Lecture 28: Slide 12
One factor independent of the other two
• This can be expressed as
jki
ijk
k)Cj(Bi)(A
kCjBiA
,PrPr
),,Pr(
© Department of Statistics 2012 STATS 330 Lecture 28: Slide 13
Two factors conditionally independent, given a third
This can be expressed as
k
jkkiijk
kCjBkCiA
kCjBiA
toEquivalent
)|Pr()|Pr(
)|,Pr(
© Department of Statistics 2012 STATS 330 Lecture 28: Slide 14
Parameterising tables of Poisson means with main effects and
interactions• Recall (see Slides 22 and 23 of lecture 19) that in
ordinary 3-way ANOVA we split up the table of means into main effects and interactions:
mijkijkijjkik
ijk
• We can do exactly the same thing with the logs of the Poission means in a 3-way table:
Log(mjk
ijkijjkik ijk
© Department of Statistics 2012 STATS 330 Lecture 28: Slide 15
Parameterising tables of probabilities with main effects and
interactions
• Corresponding multinomial probabilities are given by the model
Log(jk /
ijkijjkik
ijk
© Department of Statistics 2012 STATS 330 Lecture 28: Slide 16
Parameterising tables with main effects and
interactions (cont)• The interactions are related to the different forms of
independence: – if all the interactions are zero, then the 3 factors are
mutually independent – If the ABC and AB interactions are zero, then A and B are
independent, given C– If the ABC, AB and AC interactions are zero, then A is
independent of B and C• Using the connection between multinomial and
Poisson sampling, we can test for the various types of independence by fitting a Poisson regression with A, B and C as explanatory variables, and testing for interactions.
© Department of Statistics 2012 STATS 330 Lecture 28: Slide 17
Summary of independence models
• All 3 factors independent in multinomial model– Equivalent to all interactions zero in
Poisson model– Poisson Model is count ~ A + B + C
• A independent of B and C– Equivalent to all interactions between A
and the others zero– Poisson Model is count ~ A + B + C + B:C
or count ~ A + B*C
© Department of Statistics 2012 STATS 330 Lecture 28: Slide 18
Summary of independence models (cont)
• Factors A and B conditionally independent given C – Equivalent to all interactions containing both A
and B zero– Poisson Model is
count ~ A + B + C + A:C + B:C– Equivalent to count ~ A*C + B*C
© Department of Statistics 2012 STATS 330 Lecture 28: Slide 19
Analysis strategy: Florida data
• We will fit some models to this data and investigate the pattern of independence/ dependence between the factors.
• We fit a maximal model, and then investigate suitable submodels.
© Department of Statistics 2012 STATS 330 Lecture 28: Slide 20
The analysis> maximal.glm<-glm(counts~victim*defendent*dp, family=poisson, data=murder.df)> anova(maximal.glm, test="Chisq") Df Deviance Resid. Df Resid. Dev P(>|Chi|)NULL 7 774.73 victim 1 64.10 6 710.63 1.183e-15defendent 1 0.22 5 710.42 0.64dp 1 443.51 4 266.90 1.861e-98victim:defendent 1 254.15 3 12.75 3.230e-57victim:dp 1 10.83 2 1.92 9.999e-04defendent:dp 1 1.90 1 0.02 0.17victim:defendent:dp 1 0.02 0 -2.442e-15 0.89>
No interactions between defendants race and death penalty - Seems that model victim*dp + victim*defendant is appropriate “given the victim’s race, defendants race and death penalty are independent” ie the conditional independence model. Model also selected by stepwise, other anovas.
© Department of Statistics 2012 STATS 330 Lecture 28: Slide 21
Testing if the model is adequate
> submodel.glm<-glm(counts~victim*defendant + victim*dp, family=poisson, data=murder.df)
> summary(submodel.glm)Deviance Residuals: 1 2 3 4 5 6 7
8 0.0636 -0.2127 -0.0163 0.0525 1.0389 -0.7138 -0.4453
0.2860 Coefficients: Estimate Std. Error z value Pr(>|z|) (Intercept) 2.5472 0.2680 9.503 < 2e-16 ***victimw 0.3635 0.3057 1.189 0.23449 defendantw -2.3418 0.2341 -10.003 < 2e-16 ***dpn 2.7269 0.2759 9.885 < 2e-16 ***victimw:defendantw 3.2068 0.2567 12.491 < 2e-16 ***victimw:dpn -0.9406 0.3081 -3.053 0.00227 ** • ---
© Department of Statistics 2012 STATS 330 Lecture 28: Slide 22
Testing if the model is adequate
Null deviance: 774.7325 on 7 degrees of freedomResidual deviance: 1.9216 on 2 degrees of freedomAIC: 56.638
Number of Fisher Scoring iterations: 4
> 1-pchisq(1.9216,2) [1] 0.3825867
Model seems OK
Some hint cell 8 not well fitted
© Department of Statistics 2012 STATS 330 Lecture 28: Slide 23
Example: the Copenhagen housing data
• 317 apartment residents in Copenhagen were surveyed on their housing. 3 variables were measured:– sat: satisfaction with housing (Low,
medium,high)– cont: amount of contact with other residents
(Low,high)– infl: influence on management decisions– (Low, medium,high)
© Department of Statistics 2012 STATS 330 Lecture 28: Slide 24
Copenhagen housing data
sat infl cont count Low Low Low 61 Medium Low Low 23 High Low Low 17 Low Medium Low 43 Medium Medium Low 35 High Medium Low 40 Low High Low 26 Medium High Low 18 High High Low 549 more lines…
© Department of Statistics 2012 STATS 330 Lecture 28: Slide 25
Copenhagen housing data
cont
Low High
Low 61 78
sat Medium 23 46
High 17 43
cont
Low High
Low 43 48
sat Medium 35 45
High 40 86
cont
Low High
Low 26 15
sat Medium 18 25
High 54 62
infl = Low infl = Med
infl = High
© Department of Statistics 2012 STATS 330 Lecture 28: Slide 26
The analysis > housing.glm<-glm(count~sat*infl*cont, family=poisson, data=housing.df)> anova(housing.glm, test]”Chisq”)Df Deviance Resid. Df Resid. Dev P(>|Chi|)NULL 17 166.757 sat 2 26.191 15 140.566 2.054e-06infl 2 20.040 13 120.526 4.451e-05cont 1 22.544 12 97.983 2.054e-06sat:infl 4 75.577 8 22.406 1.504e-15sat:cont 2 7.745 6 14.661 0.021infl:cont 2 11.986 4 2.675 0.002sat:infl:cont 4 2.675 0 9.546e-15 0.614
This time, only the 3-factor interaction is insignificant.
This is the “homogeneous association model”
© Department of Statistics 2012 STATS 330 Lecture 28: Slide 27
Homogeneous association model
• Association between two factors measured by sets of odds ratios
ji
ij
ijOR
11
11
© Department of Statistics 2012 STATS 330 Lecture 28: Slide 28
Copenhagen housing OR’s
cont
Low High
Low * *
sat Medium * 1.56
High * 1.97
cont
Low High
Low * *
sat Medium * 1.15
High * 1.92
cont
Low High
Low * *
sat Medium * 2.40
High * 1.99
infl = Low infl = Med
infl = High26 15
18 25
54 62
26*25/(18*15)
26*62/(54*15)
© Department of Statistics 2012 STATS 330 Lecture 28: Slide 29
Homogeneous association model (2)
• For the homogeneous association model, the conditional odds ratios for A and B (ie using the conditional distributions of A and B given C=k) do not depend on k. That is, the pattern of association between A and B is the same for all levels of C.
• Common value of the conditional AB Log OR’s are estimated by the AB interactions
© Department of Statistics 2012 STATS 330 Lecture 28: Slide 30
Estimating conditional OR for housing data
• Estimated Sat-cont conditional log ORs are estimated with Sat-cont interactions in homogeneous association model
> homogen.glm<-glm(count~sat*infl*cont-sat:infl:cont, family=poisson, data=housing.df)> summary(homogen.glm)Coefficients: Estimate Std. Error z value Pr(>|z|) satMedium:contHigh 0.4157 0.1948 2.134 0.032818 * satHigh:contHigh 0.6496 0.1823 3.563 0.000367 ***
© Department of Statistics 2012 STATS 330 Lecture 28: Slide 31
Estimating conditional OR for housing data (2)
• For med/High, est for log OR is 0.4157, std error is 0.1948• CI for OR is exp(0.4157 +/- 1.96* 0.1 0.1948)
i.e. (1.034, 2.220)• Estimated OR for high/high is exp(0.6496) = 1.1914
> 0.4157+c(-1,1)*1.96*0.1948 [1] 0.033892 0.797508> exp(0.4157+c(-1,1)*1.96*0.1948) [1] 1.034473 2.220002> exp(0.4157)[1] 1.515431> exp(0.6496)[1] 1.914775
© Department of Statistics 2012 STATS 330 Lecture 28: Slide 32
4 dimensional tables
• Similar results apply for 4-dimensional tables
• For example, models for 4 factors A, B, C and D– A, B, C and D all independent: A + B + C +D – A, B independent of C and D: A*B + C*D– D conditionally independent of C, given A and
B: A*B*C + A*B*D
© Department of Statistics 2012 STATS 330 Lecture 28: Slide 33
Hierarchical models
• We will assume that all models are hierarchical: if the model includes an interaction with factors A1, … Ak, then it includes all main effects and interactions that can be formed from A1, … Ak
• We can represent hierarchical models by listing these “maximal interactions”
© Department of Statistics 2012 STATS 330 Lecture 28: Slide 34
Examples
• 2 factor model A + B + A:B is hierarchical– Hierarchical notation: [AB]
• 3 factor model A + B + C + A:B is hierarchical– Hierarchical notation: [AB][C]
• 3 factor model A + B + C + A:B + A:C is hierarchical– Hierarchical notation: [AB][AC]
© Department of Statistics 2012 STATS 330 Lecture 28: Slide 35
Graphical ModelsA way of visualising independence patterns:• A subset of hierarchical models• Each factor represented by the vertex of an
“association” graph• Two vertices connected by edges if they have a
non-zero interaction• Then
– A vertex not connected to any other vertex is independent of the other vertices
– two vertices not directly connected are conditionally independent, given the connecting vertices
© Department of Statistics 2012 STATS 330 Lecture 28: Slide 36
Examples• 2 factor model A + B + A:B or [AB]
• 3 factor model A + B + C +AB or [AB][C] C independent of A and B
• 3 factor model A + B + C + A:B + A:C +B:C or [AB][AC][BC]
A B
A
B C
A
B C
© Department of Statistics 2012 STATS 330 Lecture 28: Slide 37
More Examples• 3 factor model
A + B + C + AB + AC
[AB][AC]
• 4 factor model
A + B + C + D + A:B + A:C +B:C
[AB][AC][BC][D]
A
B
D
A
B C
C
© Department of Statistics 2012 STATS 330 Lecture 28: Slide 38
Another Example4 factor model
A + B + C + D + A:B + B:C + A:D
[AB][BC][AD]
C and D are conditionally independent given A and B
A
B
D
C
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