© department of statistics 2012 stats 330 lecture 28: slide 1 stats 330: lecture 28

© Department of Statistics 2012 STATS 330 Lecture 28: Slide 1

Stats 330: Lecture 28


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


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)


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


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


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


Questions

• Is death penalty independent of race?

• What is the role of victim’s race?

• What does “independent” mean when we have 3 factors?


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


Marginal probabilities

ij

kijk

k

k

kCjBiAPjBiAP

kCjBiAjBiA

),,(),(

},,{},{


Marginal probabilities (ii)

i

j kijk

j k

kj

kCjBiAPiAP

kCjBiAiA

),,()(

},,{}{,


All three factors independent

• This can be expressed as

kji

ijk

k)(Cj)(Bi)(A

kCjBiA

PrPrPr

),,Pr(


One factor independent of the other two

• This can be expressed as

jki

ijk

k)Cj(Bi)(A

kCjBiA

,PrPr

),,Pr(


Two factors conditionally independent, given a third

This can be expressed as

k

jkkiijk

kCjBkCiA

kCjBiA

toEquivalent

)|Pr()|Pr(

)|,Pr(


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


Parameterising tables of probabilities with main effects and

interactions

• Corresponding multinomial probabilities are given by the model

Log(jk /

ijkijjkik

ijk


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.


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


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


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.


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.


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 ** • ---


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


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)


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…


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


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”


Homogeneous association model

• Association between two factors measured by sets of odds ratios

ji

ij

ijOR

11

11


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)


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


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 ***


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


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


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”


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]


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


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


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


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


Another Example4 factor model

A + B + C + D + A:B:C + A:B:D

[ABC][ABD]

C and D are conditionally independent given A and B

A

B

D

C

© department of statistics 2012 stats 330 lecture 28: slide 1 stats 330: lecture 28

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