# 5.3 - Models of Independence and Associations in 3-Way Tables

5.3 - Models of Independence and Associations in 3-Way Tables

For multinomial sampling and a two-dimensional table, only independence of row and column is of interest. With three-dimensional tables, there are at least eight models of interest. In the next sections of this lesson, we are going to look at different models of independence and dependence that can capture relationships between variables in a three-way table. Please note that these concepts extend to any number of categorical variables (e.g., $$k$$-way table), and are not unique to categorical data only. But we will consider their mathematical and graphical representation and interpretations within the context of categorical data and links to odd-ratios, marginal, and partial tables. In the later lessons, we will see different ways of fitting these models (e.g., log-linear models, logistic regression, etc.).

Recall that independence means no association, but there are many types of association that might exist among variables. Here are the types of independence and associations relationships (i.e., models) that we will consider.

1. Mutual independence -- all variables are independent from each other, denoted $$(X,Y,Z)$$ or $$X\perp\!\!\perp Y \perp\!\!\perp Z$$.
2. Joint independence -- two variables are jointly independent of the third, denoted $$(XY,Z)$$ or $$X Y \perp\!\!\perp Z$$.
3. Marginal independence -- two variables are independent while ignoring the third, e.g., $$\theta_{XY}=1$$, denoted $$(X,Y)$$.
4. Conditional independence -- two variables are independent given the third, e.g., $$\theta_{XY(Z=k)}=1$$, for all $$k=1,2,\ldots,K$$, denoted $$(XY, YZ)$$ or $$X \perp\!\!\perp Y | Z$$.
5. Homogeneous associations -- conditional (partial) odds-ratios don't depend on the value of the third variable, denoted $$(XY, XZ, YZ)$$.

Before we look at the details, here is a summary of the relationships among these models:

• Mutual independence implies joint independence, i.e., all variables are independent of each other.
• Joint independence implies marginal independence, i.e., one variable is independent of the other two.
• Marginal independence does NOT imply joint independence.
• Marginal independence does NOT imply conditional independence.
• Conditional independence does NOT imply marginal independence.

It is worth noting that a minimum of three variables is required for all the above types of independence to be defined.

# 5.3.1 - Mutual (Complete) Independence

5.3.1 - Mutual (Complete) Independence

The simplest model that one might propose is that ALL variables are independent of one another. Graphically, if we have three random variables, $$X$$, $$Y$$, and $$Z$$ we can express this model as:

In this graph, the lack of connections between the nodes indicates no relationships exist among $$X$$, $$Y$$, and $$Z{,}$$ or there is mutual independence. In the notation of log-linear models, which will learn about later as well, this model is expressed as ($$X$$, $$Y$$, $$Z$$). We will use this notation from here on. The notation indicates which aspects of the data we consider in the model. In this case, the independence model means we care only about the marginals totals $$n_{i++}, n_{+j+}, n_{++k}$$ for each variable separately, and these pieces of information will be sufficient to fit a model and compute the expected counts. Alternatively, as we will see later on, we could keep track of the number of times some joint outcome of two variables occurs.

In terms of odds ratios, the model ($$X$$, $$Y$$, $$Z$$) implies that if we look at the marginal tables $$X \times Y$$, $$X \times Z$$ , and $$Y\times Z$$ , then all of the odds ratios in these marginal tables are equal to 1. In other words, mutual independence implies marginal independence (i.e., there is independence in the marginal tables). All variables are really independent of one another.

Under this model the following must hold:

$$P(X=i,Y=j,Z=k)=P(X=i)P(Y=j)P(Z=k)$$

for all $$i, j, k$$. That is, the joint probabilities are the product of the marginal probabilities. This is a simple extension from the model of independence in two-way tables where it was assumed:

$$P(X=i,Y=j)=P(X=i)P(Y=j)$$

We denote the marginal probabilities

$$\pi_{i++} = P(X = i), i = 1, 2, \ldots , I$$

$$\pi_{+j+} = P(Y = j), j = 1, 2, \ldots, J$$

$$\pi_{++k} = P(Z = k), k = 1, 2, \ldots , K$$

so that $$\pi_{ijk} = [\pi_{i++}\pi_{+j+}\pi_{++k}$$ for all $$i, j, k$$. Then the unknown parameters of the model of independence are

$$\pi_{i++} = (\pi_{1++}, \pi_{2++},\ldots, \pi_{I++})$$

$$\pi_{+j+} = (\pi_{+1+}, \pi_{+2+},\ldots , \pi_{+J+})$$

$$\pi_{++k} = (\pi_{++1}, \pi_{++2},\ldots , \pi_{++K})$$

Under the assumption that the model of independence is true, once we know the marginal probability values, we have enough information to estimate all unknown cell probabilities. Because each of the marginal probability vectors must add up to one, the number of free parameters in the model is $$(I − 1) + (J − 1) + (K −I )$$. This is exactly like the two-way table, but now one more set of additional parameter(s) needs to be taken care of for the additional random variable.

Consider the Berkeley admissions example, where under the independence model, the number of free parameters is $$(2-1) + (2-1) + (6-1) = 7$$, and the marginal distributions are

$$(n_{1++}, n_{2++}, \ldots , n_{I++}) \sim Mult(n, \pi_{i++})$$

$$(n_{+1+}, n_{+2+}, \ldots , n_{+J+}) \sim Mult(n, \pi_{+j+})$$

$$(n_{++1}, n_{++2}, \ldots , n_{++K}) \sim Mult(n, \pi_{++k})$$

and these three vectors are mutually independent. Thus, the three parameter vectors $$\pi_{i++}, \pi_{+j+}$$, and $$\pi_{++k}$$ can be estimated independently of one another. The ML estimates are the sample proportions in the margins of the table,

$$\hat{\pi}_{i++}=p_{i++}=n_{i++}/n,\quad i=1,2,\ldots,I$$

$$\hat{\pi}_{+j+}=p_{+j+}=n_{+j+}/n,\quad j=1,2,\ldots,J$$

$$\hat{\pi}_{++k}=p_{++k}=n_{++k}/n,\quad k=1,2,\ldots,K$$

It then follows that the estimates of the expected cell counts are

$$E(n_{ijk})=n\hat{\pi}_{i++}\hat{\pi}_{+j+}\hat{\pi}_{++k}=\dfrac{n_{i++}n_{+j+}n_{++k}}{n^2}$$

Compare this to a two-way table, where the expected counts were:

$$E(n_{ij})=n_{i+}n_{+j}/n$$.

For the Berkeley admission example, the marginal tables, i.e., counts are $$X=(2691, 1835)$$ (for sex), $$Y=(1755 2771)$$ (for admission status), and $$Z=(933, 585, 918, 792, 584, 714)$$ (for department). The first (estimated) expected count would be $$E(n_{111})=2691(1755)(933)/4526^2=215.1015$$, for example. Then, we compare these expected counts with the corresponding observed counts. That is, 215.1015 would be compared with $$n_{111}=512$$ and so on.

## Chi-Squared Test of Independence

The hypothesis of independence can be tested using the general method described earlier:

$$H_0 \colon$$ the independence model is true vs. $$H_A \colon$$ the saturated model is true

In other words, we can check directly $$H_0\colon \pi_{ijk} = \pi_{i++}\pi_{+j+}\pi_{++k}$$ for all $$i, j, k$$, vs. $$H_A\colon$$ the saturated model

• Estimate the unknown parameters of the independence model, i.e.., the marginal probabilities.
• Calculate estimated cell probabilities and expected cell frequencies $$E_{ijk}=E(n_{ijk})$$ under the model of independence.
• Calculate $$X^2$$ and/or $$G^2$$ by comparing the expected and observed values, and compare them to the appropriate chi-square distribution.

$$X^2=\sum\limits_i \sum\limits_j \sum\limits_k \dfrac{(E_{ijk}-n_{ijk})^2}{E_{ijk}}$$

$$G^2=2\sum\limits_i \sum\limits_j \sum\limits_k n_{ijk} \log\left(\dfrac{n_{ijk}}{E_{ijk}}\right)$$

The degrees of freedom (DF) for this test are $$\nu = (IJK − 1) − [ (I − 1) + (J − 1) + (K − 1)]$$. As before, this is a difference between the number of free parameters for the saturated model $$(IJK-1)$$ and the number of free parameters in the current model of independence, $$(I-1)+(J-1)+(K-1)$$. In the case of the Berkeley data, we have $$23-7 = 16$$ degrees of freedom for the mutual independence test.

Recall that we also said that mutual independence implies marginal independence. So, if we reject marginal independence for any pair of variables, we can immediately reject mutual independence overall. For example, recall the significant result when testing for the marginal association between sex and admission status ($$X^2=92.205$$). Since we rejected marginal independence in that case, it would expect to reject mutual (complete) independence as well.

Stop and Think!
What is your conclusion? Would you reject the model of complete independence? Are these three variables mutually independent?

## Example: Boys Scouts and Juvenile Delinquency

This is a $$2 \times 2 \times 3$$ table. It classifies $$n = 800$$ boys according to socioeconomic status (S), whether they participated in boy scouts (B), and whether they have been labeled as a juvenile delinquent (D):

Socioeconomic status Boy scout Delinquent
Yes No
Low Yes 11 43
No 42 169
Medium Yes 14 104
No 20 132
High Yes 8 196
No 2 59

To fit the mutual independence model, we need to find the marginal totals for B,

$$n_{1++} = 11 + 43 + 14 + 104 + 8 + 196 = 376$$

$$n_{2++} = 42 + 169 + 20 + 132 + 2 + 59 = 424$$

for D,

$$n_{+1+} = 11 + 42 + 14 + 20 + 8 + 2 = 97$$

$$n_{+2+} = 43 + 169 + 104 + 132 + 196 + 59 = 703$$

and for S,

$$n_{++1} = 11 + 43 + 42 + 169 = 265$$

$$n_{++2} = 14 + 104 + 20 + 132 = 270$$

$$n_{++3} = 8 + 196 + 2 + 59 = 265$$

Next, we calculate the expected counts for each cell,

$$E_{ijk}=\dfrac{n_{i++}n_{+j+}n_{++k}}{n^2}$$

and then calculate the chi-square statistic. The degrees of freedom for this test are $$(2\times 2\times 3-1)-[(2-1)+(2-1)+(3-1)]=7$$ so the $$p$$-value can be found as $$P(\chi^2_7 \geq X^2)$$ and $$P(\chi^2_7 \geq G^2)$$.

In SAS, we can use the following lines of code:

data answers;
a1=1-probchi(218.6622,7);
;
run;

In R, we can use 1-pchisq(218.6622, 7).

The p-values are essentially zero, indicating that the mutual independence model does not fit. Remember, in order for the chi-squared approximation to work well, the $$E_{ijk}$$ needs to be sufficiently large. Sufficiently large means that most of them (e.g., about at least 80%) should be at least five, and none should be less than one. We should examine the $$E_{ijk}$$ to see if they are large enough.

Fortunately, both R and SAS will give the expected counts, (in parentheses) and the observed counts:

Socioeconomic status Boy scout Delinquent
Yes No
Low Yes 11
(15.102)
43
(109.448)
No 42
(17.030)
169
(123.420)
Medium Yes 14
(15.387 )

104
(111.513)

No 20
(17.351)
132
(125.749)
High Yes

8
(15.102)

196
(109.448)
No 2
(17.030)

59
(123.420)

Here, the expected counts are sufficiently large for the chi-square approximation to work well, and thus we must conclude that the variables B (boys scout), D (delinquent), and S (socioeconomic status) are not mutually independent.

Note! Most software packages should give you a warning if more than 20% of the expected cells are less than 5, and this may have an influence on large sample approximations.

There is no single function or a call in SAS nor R that will directly test the mutual independence model; see will see how to fit this model via a log-linear model. However, we can test this by relying on our understanding of two-way tables, and of marginal and partial tables and related odds ratios. For mutual independence to hold, all of the tests for independence in marginal tables must hold. Thus, we can do the analysis of all two-way marginal tables using the chi-squared test of independence in each. Alternatively, we can consider the odds-ratios in each two-way table. In this case, for example, the estimated odds ratio for the $$B \times D$$ table, is 0.542, and it is not equal to 1; i.e., 1 is not in covered by the 95% odds-ratio confidence interval, (0.347, 0.845). Therefore, we have sufficient evidence to reject the null hypothesis that boy scout status and delinquent status are independent of one another and thus that $$B$$, $$D$$, and $$S$$ are not mutually independent.

The following SAS or R code supports the above analysis by testing independence of two-way marginal tables. Again, we will see later in the course that this is done more efficiently via log-linear models.

Here we will use the SAS code found in the program boys.sas as shown below.

data boys;
input SES $scout$ delinquent $count @@; datalines; low yes yes 11 low yes no 43 low no yes 42 low no no 169 med yes yes 14 med yes no 104 med no yes 20 med no no 132 high yes yes 8 high yes no 196 high no yes 2 high no no 59 ; proc freq; weight count; tables SES; tables scout; tables delinquent; tables SES*scout/all nocol nopct; tables SES*delinquent/all nocol nopct; tables scout*delinquent/all nocol nopct; tables SES*scout*delinquent/chisq cmh expected nocol nopct; run;  The output for this program can be found in this file: boys.lst. For R, we will use the R program file boys.R to create contingency tables and carry out the test of mutual independence. #### Test of conditional independence #### Use of oddsratio(), confit() from {vcd} #### Cochran-Mantel-Haenszel test #### Also testing via Log-linear models ############################################ #### Input the table deliquent=c("no","yes") scout=c("no", "yes") SES=c("low", "med","high") table=expand.grid(deliquent=deliquent,scout=scout,SES=SES) count=c(169,42,43,11,132,20,104,14,59,2,196,8) table=cbind(table,count=count) table temp=xtabs(count~deliquent+scout+SES,table) temp #### Create "flat" contigency tables ftable(temp) #### Let's see how we can create various subtables #### One-way Table SES Frequency=as.vector(margin.table(temp,3)) CumFrequency=cumsum(Frequency) cbind(SES,Frequency=Frequency,Percentage=Frequency/sum(Frequency),CumFrequency=CumFrequency,CumPercentage=CumFrequency/sum(Frequency)) #### One-way Table scout Frequency=as.vector(margin.table(temp,2)) CumFrequency=cumsum(Frequency) cbind(scout,Frequency=Frequency,Percentage=Frequency/sum(Frequency),CumFrequency=CumFrequency,CumPercentage=CumFrequency/sum(Frequency)) #### One-way Table deliquent Frequency=as.vector(margin.table(temp,1)) CumFrequency=cumsum(Frequency) cbind(deliquent,Frequency=Frequency,Percentage=Frequency/sum(Frequency),CumFrequency=CumFrequency,CumPercentage=CumFrequency/sum(Frequency)) #### Test the Mutual Independence, step by step #### compute the expected frequencies E=array(NA,dim(temp)) for (i in 1:dim(temp)[1]) { for (j in 1:dim(temp)[2]) { for (k in 1:dim(temp)[3]) { E[i,j,k]=(margin.table(temp,3)[k]*margin.table(temp,2)[j]*margin.table(temp,1))[i]/(sum(temp))^2 }}} E #### compute the X^2, and G^2 df=(length(temp)-1)-(sum(dim(temp))-3) X2=sum((temp-E)^2/E) X2 1-pchisq(X2,df) G2=2*sum(temp*log(temp/E)) G2 1-pchisq(G2,df) #### Test for Mutual indpendence (and other models) by considering analysis of all two-way tables #### Two-way Table SES*scout #### This is a test of Marginal Independence for SES and Scout SES_Scout=margin.table(temp,c(3,2)) result=chisq.test(SES_Scout) result result$expected
result=chisq.test(SES_Scout,correct = FALSE)
result
SES_Scout=list(Frequency=SES_Scout,RowPercentage=prop.table(SES_Scout,2))

#### Two-way Table SES*deliquent
#### temp=xtabs(count~scout+SES+deliquent,table)
#### This is a test of Marginal Independence for SES and deliquent status

SES_deliquent=margin.table(temp,c(3,1))
result=chisq.test(SES_deliquent)
result
result$expected result=chisq.test(SES_deliquent,correct = FALSE) result SES_deliquent=list(Frequency=SES_deliquent,RowPercentage=prop.table(SES_deliquent,2)) #### Two-way Table deliquent*scout #### This is a test of Marginal Independence for Deliquent status and the Scout status deliquent_scout=margin.table(temp,c(1,2)) result=chisq.test(deliquent_scout) result$expected
result=chisq.test(deliquent_scout,correct = FALSE)
result
deliquent_scout1=list(Frequency=deliquent_scout,RowPercentage=prop.table(deliquent_scout,1))

#### compute the log(oddsraio), oddsratio and its 95% CI using {vcd} package

lor=oddsratio(deliquent_scout)
lor
OR=exp(lor)
OR
OR=oddsratio(deliquent_scout, log=FALSE)
OR
CI=exp(confint(lor))
CI
CI=confint(OR)
CI

#### Table of deliquent*scout at different level of SES

temp

#### Test for Joint Independence of (D,BS)
#### creating 6x2 table, BS x D

SESscout_deliquent=ftable(temp, row.vars=c(3,2))
result=chisq.test(SESscout_deliquent)
result

#### Test for Marginal Independence (see above analysis of two-way tables)

#### Test for conditional independence
#### To get partial tables of DB for each level of S

temp[,,1]
chisq.test(temp[,,1], correct=FALSE)
temp[,,2]
chisq.test(temp[,,2], correct=FALSE)
temp[,,3]
chisq.test(temp[,,3], correct=FALSE)
X2=sum(chisq.test(temp[,,1], correct=FALSE)$statistic+chisq.test(temp[,,2], correct=FALSE)$statistic+chisq.test(temp[,,3], correct=FALSE)$statistic) 1-pchisq(X2,df=3) #### Cochran-Mantel-Haenszel test mantelhaen.test(temp) mantelhaen.test(temp,correct=FALSE) #### Breslow-Day test #### make sure to first source/run breslowday.test_.R breslowday.test(temp) ################################################################ #### Log-linear Models #### Let's see how we would do this via Log_linear models #### We will look at the details later in the course #### test of conditional independence #### via loglinear model, but the object has to be a table! is.table(temp) #### to check if table temp<-as.table(temp) #### to save as table temp.condind<-loglin(temp, list(c(1,3), c(2,3)), fit=TRUE, param=TRUE) #### fit the cond.indep. model temp.condind 1-pchisq(temp.condind$lrt, temp.condind$df) #### There is no way to do Breslow-Day stats in R; #### you need to write your own function or fit the homogeneous association model #### to test for identity of odds-ratios #### test of homogenous association model temp.hom<-loglin(temp, list(c(1,3), c(2,3), c(1,2)), fit=TRUE, param=TRUE) temp.hom 1-pchisq(temp.hom$lrt, temp.hom$df) #### Here is how to do the same but with the glm() function in R #### test of conditional independence #### via loglinear mode, but using glm() funcion #### the object now needs to be dataframe temp.data<-as.data.frame(temp) temp.data temp.condind<-glm(temp.data$Freq~temp.data$scout*temp.data$SES+temp.data$deliquent*temp.data$SES, family=poisson())
summary(temp.condind)
fitted(temp.condind)

#### test of homogenous association model

temp.hom<-glm(temp.data$Freq~temp.data$scout*temp.data$SES+temp.data$deliquent*temp.data$SES+temp.data$scout*temp.data$deliquent, family=poisson()) summary(temp.hom) fitted(temp.hom) The output file, boys.out contains the results for these procedures. Stop and Think! If two or more variables in a $$k$$-way table are not independent, then where is this difference coming from? That is, what are some other possible relationships that hold? What are some other models that can capture this data? Maybe $$S$$ and $$B$$ are jointly independent of $$D$$? # 5.3.2 - Joint Independence 5.3.2 - Joint Independence The joint independence model implies that two variables are jointly independent of a third. For example, let's suggest that C is jointly independent of $$X$$ and $$Y$$. In the log-linear notation, this model is denoted as ($$XY$$, $$Z{)}$$. Here is a graphical representation of this model: which indicates that $$X$$ and $$Y$$ are jointly independent of $$Z$$. The line linking $$X$$ and $$Y$$ indicates that $$X$$ and $$Y$$ are possibly related, but not necessarily so. Therefore, the model of complete independence is a special case of this one. For three variables, three different joint independence models may be considered. If the model of complete independence ($$X$$, $$Y$$, $$Z$$) fits a data set, then the model ($$XY$$, $$Z$$) will also fit, as will ($$XZ$$, $$Y$$) and ($$YZ$$, $$X$$). In that case, we will prefer to use ($$X$$, $$Y$$, $$Z$$) because it is more parsimonious; our goal is to find the simplest model that fits the data. Note that joint independence does not imply mutual independence. This is one of the reasons why we start with the overall model of mutual independence before we collapse and look at models of joint independence. Assuming that the ($$XY$$, $$Z$$) model holds, the cell probabilities would be equal to the product of the marginal probabilities from the $$XY$$ margin and the $$Z$$ margin: $$\pi_{ijk} = P(X = i,Y= j) P(Z = k) = \pi_{ij}+ \pi_{++k},$$ where $$\sum_i \sum_j \pi_{ij+} = 1$$ and $$\sum_k \pi_{++k} = 1$$. Thus if we know the counts in the $$XY$$ table and the $$Z$$ table, we can compute the expected counts in the $$XYZ$$ table. The number of free parameters, that is the number of unknown parameters that we need to estimate is $$(IJ − 1) + (K − 1)$$, and their ML estimates are $$\hat{\pi}_{ij+}=n_{ij+}/n$$, and $$\hat{\pi}_{++k}=n_{++k}/n$$. The estimated expected cell frequencies are: $$\hat{E}_{ijk}=\dfrac{n_{ij+}n_{++k}}{n}$$ (1) Notice the similarity between this formula and the one for the model of independence in a two-way table, $$\hat{E}_{ij}=n_{i+}n_{+j}/n$$. If we view $$X$$ and $$Y$$ as a single categorical variable with $$IJ$$ levels, then the goodness-of-fit test for ($$XY$$, $$Z$$) is equivalent to the test of independence between the combined variable $$XY$$ and $$Z$$. The implication of this is that you can re-write the 3-way table as a 2-way table and do its analysis. To determine the degrees of freedom in order to evaluate this model and to construct the chi-squared and deviance statistics, we need to find the number of free parameters in the saturated model and subtract from this the number of free parameters in the new model. We take the difference of the number of parameters we fit in the saturated model and the number of parameters we fit in the assumed model: DF = $$(IJK-1)-[(IJ-1)+(K-1)]$$, and this is equal to $$(IJ-1)(K-1)$$, which ties in the statement above that this is equivalent to the test of independence in a two-way table with a variable ($$XY$$) vs. variable ($$Z$$). Stop and Think! Can you write this two-way table for the Berkeley admissions example? What do you think, is the $$X=$$ sex, $$Y=$$ admission status, jointly independent of $$Z=$$ department? Answer by conducting the chi-square test of independence for the $$XY\times Z$$ table. ## Example: Boy Scouts and Juvenile Delinquency Going back to the boy scout example, let’s think of juvenile delinquency D as the response variable, and the other two — boy scout status B and socioeconomic status S—as explanatory variables or predictors. Therefore, it would be useful to test the null hypothesis that D is independent of B and S, or (D, BS). Because we used $$i$$, $$j$$, and $$k$$ to index B, D, and S, respectively, the estimated expected cell counts under the model "B and S are independent of D" are a slight rearrangement of (1) above, $$E_{ijk}=\dfrac{n_{i+k}n_{+j+}}{n}$$ We can calculate the $$E_{ijk}$$s by this formula or in SAS or R by entering the data as a two-way table with one dimension corresponding to D and the other dimension corresponding to B $$\times$$ S. In fact, looking at the original table, we see that the data are already arranged in the correct fashion; the two columns correspond to the two levels of D, and the six rows correspond to the six levels of a new combined variable B $$\times$$ S. To test the hypothesis "D is independent of B and S," we simply enter the data as a $$6\times 2$$ matrix, where we have combined the socioeconomic status (SES) with scout status, essentially collapsing the 3-way table into a 2-way table. The SAS code for this example is: boys.sas. Here is the SAS code shown below: data boys; input SES$scout $delinquent$ count @@;
datalines;
low yes yes 11 low yes no 43
low no yes 42 low no no 169
med yes yes 14 med yes no 104
med no yes 20 med no no 132
high yes yes 8 high yes no 196
high no yes 2 high no no 59
;
proc freq;
weight count;
tables SES;
tables scout;
tables delinquent;
tables SES*scout/all nocol nopct;
tables SES*delinquent/all nocol nopct;
tables scout*delinquent/all nocol nopct;
tables SES*scout*delinquent/chisq cmh expected nocol nopct;
run;

Expected counts are printed below in the SAS output as the second entry in each cell:

The SAS System

The FREQ Procedure

Frequency
Expected
Row Pct

Table of SES_scout by delinquent
SES_scout delinquent
no yes Total
high_no
59
53.604
96.72
2
7.3963
3.28
61

high_yes
196
179.27
96.08
8
24.735
3.92
204

low_no
169
185.42
80.09
42
25.584
19.91
211

low_yes
43
47.453
79.63
11
6.5475
20.37
54

med_no
132
133.57
86.84
20
18.43
13.16
152

med_yes
104
103.69
88.14
14
14.308
11.86
118

Total
703
97
800

Statistics for Table of SES_scout by delinquent

Statistic DF Value Prob
Chi-Square 5 32.9576 <.0001
Likelihood Ratio Chi-Square 5 36.4147 <.0001
Mantel-Haenszel Chi-Square 1 5.5204 0.0188
Phi Coefficient   0.2030
Contingency Coefficient   0.1989
Cramer's V   0.2030

The SAS output from this program is essentially evaluating a 2-way table, which we have already done before.

We will use the R code boys.R and look for the "Joint Independence" comment heading, and use of the ftable() function as shown below.

#### Test for Joint Independence of (D,BS)
#### creating 6x2 table, BS X D

SESscout_delinquent=ftable(temp, row.vars=c(3,2))
results=chisq.test(SESscout_delinquent)
result

Here is the output that R gives us:

Pearson's Chi-squared test

data: SESscout_deliquent
X-squared = 32.958, df = 5, p-value = 3.837e-06

Our conclusions are based on the following evidence: the chi-squared statistics values (e.g., $$X^2, G^2$$) are very large (e.g., $$X^2=32.9576, df=5$$), and the p-value is essentially zero, indicating that B and S are not independent of D. The expected cell counts are all greater than five, so this p-value is reliable and our hypothesis does not hold - we reject this model of joint independence.

Notice that we can also test for the other joint independence models, e.g., (BD, S), that is, the scout status and delinquency are jointly independent of the SES, and this involves analyzing a $$4\times3$$ table, and (SD,B), that is the SES and Delinquency are jointly independent of the Boy's Scout status, and this involves analyzing a $$6\times2$$ table. These are just additional analyses of two-way tables.

It is worth noting that while many different varieties of models are possible, one must keep in mind the interpretation of the models. For example, assuming delinquency to be the response, the model (D, BS) would have a natural interpretation; if the model holds, it means that B and S cannot predict D. But if the model does not hold, either B or S may be associated with D. However, (BD, S) and (SD, B) may not render themselves to easy interpretation, although statistically, we can perform the tests of independence.

# 5.3.3 - Marginal Independence

5.3.3 - Marginal Independence

Marginal independence has already been discussed before. Variables $$X$$ and $$Y$$ are marginally independent if:

$$\pi_{ij+}=\pi_{i++}\pi_{+j+}$$

They are marginally independent if they are independent within the marginal table. In other words, we consider the relationship between $$X$$ and $$Y$$ only and completely ignore $$Z$$. Controlling for, or adjusting for different levels of $$Z$$ would involve looking at the partial tables.

We saw this already with the Berkeley admission example. Recall from the marginal table between sex and admission status, where the estimated odds-ratio was 1.84. We also had significant evidence that the corresponding odds-ratio in the population was different from 1, which indicates a marginal relationship between sex and admission status.

Question: How would you test the model of marginal independence between scouting and SES status in the boy-scout example? See the files boys.sas (boys.lst) or boys.R (boys.out) to answer this question. Hint: look for the chi-square statistic $$X^2=172.2$$.

Recall, that joint independence implies marginal, but not the other way around. For example, if the model ($$XY$$, $$Z$$) holds, it will imply $$X$$ independent of $$Z$$, and $$Y$$ independent of $$Z$$. But if $$X$$ is independent of $$Z$$ , and $$Y$$ is independent of $$Z$$, this will NOT necessarily imply that $$X$$ and $$Y$$ are jointly independent of $$Z$$.

Stop and Think!
Can you see why this is the case? Can you construct a simple example?

If $$X$$ is independent of $$Z$$ and $$Y$$ is independent of $$Z$$ we can’t tell if there is an association between $$X$$ and $$Y$$ or not; it could go either way, that is in our graphical representation there could be a link between $$X$$ and $$Y$$ but there may not be one either.

For example, let’s consider a $$2\times2\times2$$ table.

The following $$XZ$$ table has $$X$$ independent of $$Z$$ because each cell count is equal to the product of the corresponding margins divided by the total (e.g., $$n_{11}=2=8(5/20)$$, or equivalently, OR = 1)

C=1 C=2 Total
A=1 2 6 8
A=2 3 9 12
Total 5 15 20

The following $$YZ$$ table has $$Y$$ independent of $$Z$$ because each cell count is equal to the product of the corresponding margins divided by the total (e.g., $$n_{11}=2=8(5/20)$$).

C=1

C=2

Total

B=1

2

6

8

B=2

3

9

12

Total

5

15

20

The following $$XY$$ table is consistent with the above two tables, but here $$X$$ and $$Y$$ are NOT independent because all cell counts are not equal to the product of the corresponding margins divided by the total (e.g., $$n_{11}=2$$), whereas $$8(8/20)=16/5$$). Or you can see this via the odds ratio, which is not equal to 1.

B=1 B=2 Total
A=1 2 6 8
A=2 6 6 12
Total 8 12 20

Furthermore, the following $$XY\times Z$$ table is consistent with the above tables but here $$X$$ and $$Y$$ are NOT jointly independent of $$Z$$ because all cell counts are not equal to the product of the corresponding margins divided by the total (e.g., $$n_{11}=1$$, whereas $$2(5/20)=0.5$$).

C=1 C=2 Total
A=1, B=1 1 1 2
A=1, B=2 1 5 6
A=2, B=1 1 5 6
A=2, B=2 2 4 6
Total 5 15 20
So this example shows, that we could have (1) $$X$$ independent of $$Z$$, and (2) $$Y$$ independent of $$Z$$, but (3) $$X$$ and $$Y$$ are not jointly independent of $$Z$$.

Other marginal independence models in a three-way table would be, $$(X,Y)$$ while ignoring $$Z$$, and $$(Y,Z)$$, ignoring any information on $$X$$.

# 5.3.4 - Conditional Independence

5.3.4 - Conditional Independence

The concept of conditional independence is very important and it is the basis for many statistical models (e.g., latent class models, factor analysis, item response models, graphical models, etc.).

There are three possible conditional independence models with three random variables: ($$XY$$, $$XZ$$), ($$XY$$, $$YZ$$), and ($$XZ$$, $$YZ$$). Consider the model ($$XY$$, $$XZ$$),

which means that $$Y$$ and $$Z$$ are conditionally independent given $$X$$. In mathematical terms, the model ($$XY$$, $$XZ$$) means that the conditional probability of $$Y$$ and $$Z$$, given $$X$$ equals the product of conditional probabilities of $$Y$$ given $$X$$ and $$Z$$, given $$X$$:

$$P(Y=j,Z=k|X=i)=P(Y=j|X=i) \times P(Z=k|X=i)$$

In terms of odds-ratios, this model implies that if we look at the partial tables, that is $$Y\times Z$$ tables at each level of $$X = 1, \ldots , I$$, that the odds-ratios in these tables should not significantly differ from 1. Tying this back to 2-way tables, we can test in each of the partial $$Y\times Z$$ tables at each level of $$X$$ to see if independence holds.

$$H_0\colon \theta_{YZ\left(X=i\right)} = 1$$ for all i

vs.

$$H_0\colon$$ at least one $$\theta_{YZ\left(X=i\right)} \ne 1$$

Stop and Think!
It is straightforward to show that the model $$(XY, Z)$$, $$(XZ, Y)$$ and $$(X,Y,Z)$$ are special cases of this model. Therefore, if any of these simpler models fit, then $$(XY, XZ)$$ will also fit. Can you see this? What is a solution for when (X,Y,Z) holds?
1. Recall, $$(XY, XZ)$$ implies that $$P(Y,Z| X)=P(Y|X)P(Z|X)$$.
2. If $$(X,Y,Z)$$ holds then we know that $$P(X, Y, Z)= P(X) P(Y) P(Z)$$, and we also know that the mutual independence implies that $$P(Y|X)=P(Y)$$ and $$P(Z|X)=P(Z)$$.
3. Thus from the probability properties, $$P(Y,Z|X)=P(X,Y,Z)/P(X)$$ and from (2) above, $$P(Y,Z|X)=P(X,Y,Z)/P(X) = P(X)P(Y)P(Z)/P(X)= P(Y)P(Z)=P(Y|X)P(Z|X)$$, which equals the expression in (1)

Intuitively, $$(XY, XZ)$$ means that any relationship that may exist between $$Y$$ and $$Z$$ can be explained by $$X$$ . In other words, $$Y$$ and $$Z$$ may appear to be related if $$X$$ is not considered (e.g. only look at the marginal table $$Y\times Z$$, but if one could control for $$X$$ by holding it constant (i.e. by looking at subsets of the data having identical values of $$X$$, that is looking at partial tables $$Y\times Z$$ for each level of $$X$$), then any apparent relationship between $$Y$$ and $$Z$$ would disappear (remember Simpson's paradox?). Marginal and conditional associations can be different!

Under the conditional independence model, the cell probabilities can be written as

\begin{align} \pi_{ijk} &= P(X=i) P(Y=j,Z=k|X=i)\\ &= P(X=i)P(Y=j|X=i)P(Z=k|X=i)\\ &= \pi_{i++}\pi_{j|i}\pi_{k|i}\\ \end{align}

where $$\sum_i \pi_{i++} = 1, \sum_j \pi_{j | i} = 1$$ for each i, and $$\sum_k \pi_{k | i} = 1$$ for each $$i$$. The number of free parameters is $$(I − 1) + I (J − 1) + I (K − 1)$$.

The ML estimates of these parameters are

$$\hat{\pi}_{i++}=n_{i++}/n$$

$$\hat{\pi}_{j|i}=n_{ij+}/n_{i++}$$

$$\hat{\pi}_{k|i}=n_{i+k}/n_{i++}$$

and the estimated expected frequencies are

$$\hat{E}_{ijk}=\dfrac{n_{ij+}n_{i+k}}{n_{i++}}.$$

Notice again the similarity to the formula for independence in a two-way table.

The test for conditional independence of $$Y$$ and $$Z$$ given $$X$$ is equivalent to separating the table by levels of $$X = 1, \ldots , I$$ and testing for independence within each level.

There are two ways we can test for conditional independence:

1. The overall $$X^2$$ or $$G^2$$statistics can be found by summing the individual test statistics for $$YZ$$ independence across the levels of $$X$$. The total degrees of freedom for this test must be $$I (J − 1)(K − 1)$$. See the example below, and we’ll see more on this again when we look at log-linear models. Note, if we can reject independence in one of the partial tables, then we can reject the conditional independence and don't need to run the full analysis.
2. Cochran-Mantel-Haenszel Test (using option CMH in PROC FREQ/ TABLES/ in SAS and mantelhaen.test in R). This test produces the Mantel-Haenszel statistic also known as the "average partial association" statistic.

## Example: Boy Scouts and Juvenile Delinquency

Let us return to the table that classifies $$n = 800$$ boys by scout status B, juvenile delinquency D, and socioeconomic status S. We already found that the models of mutual independence (D, B, S) and joint independence (D, BS) did not fit. Thus we know that either B or S (or both) are related to D. Let us temporarily ignore S and see whether B and D are related (marginal independence). Ignoring S means that we classify individuals only by the variables B and D; in other words, we form a two-way table for B $$\times$$ D, the same table that we would get by collapsing (i.e. adding) over the levels of S.

Boy scout Delinquent
Yes No
Yes 33 343
No 64 360

The $$X^2$$ test for this marginal independence demonstrates that a relationship between B and D does exist. Expected counts are printed below the observed counts:

Delinquent=Yes Delinquent=No Total
Boy Scout=Yes 33
45.59
343
330.41
376
Boy Scout=No 64
51.41
360
372.59
424
Total 97 703 800

$$X^2 = 3.477 + 0.480 + 3.083 + 0.425 = 7.465$$, where each value in the sum is a contribution (squared Pearson residual) of each cell to the overall Pearson $$X^2$$ statistic. With df = 1, the p-value=1- PROBCHI(7.465,1)=0.006 in SAS or in R p-value=1-pchisq(7.465,1)=0.006, rejecting the marginal independence of B and D. This would also be consistent with the chi-square test of independence in the $$2\times2$$ table.

The odds ratio of $$(33 \cdot360)/(64 \cdot 343) = 0.54$$ indicates a strong negative relationship between boy scout status and delinquency; it appears that boy scouts are 46% less likely (on the odds scale) to be delinquent than non-boy scouts.

To a proponent of scouting, this result might suggest that being a boy scout has substantial benefits in reducing the rates of juvenile delinquency. But boy scouts tend to differ from non-scouts on a wide variety of characteristics. Could one of these characteristics—say, socioeconomic status—explain the apparent relationship between B and D?

Let’s now test the hypothesis that B and D are conditionally independent, given S. To do this, we enter the data for each $$2 \times 2$$ table of B $$\times$$ D corresponding to different levels of, S = 1, S = 2, and S = 3, respectively, then perform independence tests on these tables, and add up the $$X^2$$ statistics (or run the CMH test -- as in the next section).

To do this in SAS you can run the following command in boys.sas:

tables SES*scouts*delinquent / chisq;

Notice that the order is important; SAS will create partial tables for each level of the first variable; see boys.lst

The individual chi-square statistics from the output after each partial table are given below. To test the conditional independence of (BS, DS) we can add these up to get the overall chi-squared statistic:

0.053+0.006 + 0.101 = 0.160.

Each of the individual tests has 1 degree of freedom, so the total number of degrees of freedom is 3. The p-value is $$P(\chi^2_3 \geq 0.1600)=0.984$$, indicating that the conditional independence model fits extremely well. As a result, we will not reject this model here. However, the p-value is so high - doesn't it make you wonder what is going on here?

The apparent relationship between B and D can be explained by S; after the systematic differences in social class among scouts and non-scouts are accounted for, there is no additional evidence that scout membership has any effect on delinquency. The fact that the p-value is so close to 1 suggests that the model fit is too good to be true; it suggests that the data may have been fabricated. (It’s true; some of this data was created in order to illustrate this point!)

In the next section, we will see how to use the CMH option in SAS.

In R, in boys.R for example

temp[,,1]

will give us the B $$\times$$ D partial table for the first level of S, and similarly for the levels 2 and 3, where temp was the name of our 3-way table this code; see boys.out.

The individual chi-square statistics from the output after each partial table are given below.


> chisq.test(temp[,,1], correct=FALSE)
Pearson's Chi-squared test
data:  temp[, , 1]
X-squared = 0.0058, df = 1, p-value = 0.9392
> temp[,,2]
scout
deliquent  no yes
no  132 104
yes  20  14
> chisq.test(temp[,,2], correct=FALSE)
Pearson's Chi-squared test
data:  temp[, , 2]
X-squared = 0.101, df = 1, p-value = 0.7507
> temp[,,3]
scout
deliquent no yes
no  59 196
yes  2   8
> chisq.test(temp[,,3], correct=FALSE)
Pearson's Chi-squared test
data:  temp[, , 3]
X-squared = 0.0534, df = 1, p-value = 0.



To test the conditional independence of (BS, DS) we can add these up to get the overall chi-squared statistic:

0.006 + 0.101 + 0.053 = 0.160.

Each of the individual tests has 1 degree of freedom, so the total number of degrees of freedom is 3. The p-value is $$P(\chi^2_3 \geq 0.1600)=0.984$$, indicating that the conditional independence model fits extremely well. As a result, we will not reject this model here. However, the p-value is so high - doesn't it make you wonder what is going on here?

The apparent relationship between B and D can be explained by S; after the systematic differences in social class among scouts and non-scouts are accounted for, there is no additional evidence that scout membership has any effect on delinquency. The fact that the p-value is so close to 1 suggests that the model fit is too good to be true; it suggests that the data may have been fabricated. (It’s true; some of this data was created in order to illustrate this point!)

In the next section, we will see how to use the mantelhean.test in R.

Spurious Relationship

To see how the spurious relationship between B and D could have been induced, it is worthwhile to examine the B $$\times$$ S and D $$\times$$ S marginal tables.

The B $$\times$$ S marginal table is shown below.

 Socioeconomic status Boy scout Yes No Low 54 211 Medium 118 152 High 204 61

The test of independence for this table yields $$X^2 = 172.2$$ with 2 degrees of freedom, which gives a p-value of essentially zero. There is a highly significant relationship between B and S. To see what the relationship is, we can estimate the conditional probabilities of B = 1 for S = 1, S = 2, and S = 3:

$$P(B=1|S=1)=54/(54 + 211) = .204$$

$$P(B=1|S=2)=118/(118 + 152) = .437$$

$$P(B=1|S=3)=204/(204 + 61) = .769$$

The probability of being a boy scout rises dramatically as socioeconomic status goes up.

Now let’s examine the D $$\times$$ S marginal table.

 Socioeconomic status Delinquent Yes No Low 53 212 Medium 34 236 High 10 255

The test for independence here yields $$X^2 = 32.8$$ with 2 degrees of freedom, p-value $$\approx$$ 0. The estimated conditional probabilities of D = 1 for S = 1, S = 2, and S = 3 are shown below.

$$P(D=1|S=1)=53/(53 + 212) = .200$$

$$P(D=1|S=2)=34/(34 + 236) = .126$$

$$P(D=1|S=3=10/(10 + 255) = .038$$

The rate of delinquency drops as socioeconomic status goes up. Now we see how S induces a spurious relationship between B and D. Boy scouts tend to be of higher social class than non-scouts, and boys in higher social class have a smaller chance of being delinquent. The apparent effect of scouting is really an effect of social class.

In the next section, we study how to test for conditional independence via the CMH statistic.

# 5.3.5 - Cochran-Mantel-Haenszel Test

5.3.5 - Cochran-Mantel-Haenszel Test

This is another way to test for conditional independence, by exploring associations in partial tables for $$2 \times 2 \times K$$ tables. Recall, the null hypothesis of conditional independence is equivalent to the statement that all conditional odds ratios given the levels $$k$$  are equal to 1, i.e.,

$$H_0 : \theta_{XY(1)} = \theta_{XY(2)} = \cdots = \theta_{XY(K)} = 1$$

The Cochran-Mantel-Haenszel (CMH) test statistic is

$$M^2=\dfrac{[\sum_k(n_{11k}-\mu_{11k})]^2}{\sum_k Var(n_{11k})}$$

where $$\mu_{11k}=E(n_{11})=\frac{n_{1+k}n_{+1k}}{n_{++k}}$$ is the expected frequency of the first cell in the $$k$$th partial table assuming the conditional independence holds, and the variance of cell (1, 1) is

$$Var(n_{11k})=\dfrac{n_{1+k}n_{2+k}n_{+1k}n_{+2k}}{n^2_{++k}(n_{++k}-1)}$$.

## Properties of the CMH statistic

• For large samples, when $$H_0$$ is true, the CMH statistic has a chi-squared distribution with df = 1.
• If all $$\theta_{XY(k)} = 1$$, then the CMH statistic is close to zero
• If some or all $$\theta_{XY(k)} > 1$$, then the CMH statistic is large
• If some or all $$\theta_{XY(k)} < 1$$, then the CMH statistic is large
• If some $$\theta_{XY(k)} < 1$$ and others $$\theta_{XY(k)} > 1$$, then the CMH statistic is not as effective; that is, the test works better if the conditional odds ratios are in the same direction and comparable in size.
• The CMH test can be generalized to $$I \times J \times K$$ tables, but this generalization varies depending on the nature of the variables:
• the general association statistic treats both variables as nominal and thus has df $$= (I −1)\times(J −1)$$.
• the row mean scores differ statistic treats the row variable as nominal and column variable as ordinal, and has df $$= I − 1$$.
• the nonzero correlation statistic treats both variables as ordinal, and df = 1.

## Common odds-ratio estimate

As we have seen before, it’s always informative to have a summary estimate of strength of association (rather than just a hypothesis test). If the associations are similar across the partial tables, we can summarize them with a single value: an estimate of the common odds ratio for a $$2 \times2 \times K$$ table is

$$\hat{\theta}_{MH}=\dfrac{\sum_k(n_{11k}n_{22k})/n_{++k}}{\sum_k(n_{12k}n_{21k})/n_{++k}}$$

This is a useful summary statistic especially if the model of homogeneous associations holds, as we will see in the next section.

## Example - Boy Scouts and Juvenile Delinquency

For the boy scout data based on the first method of doing individual chi-squared tests in each conditional table we concluded that B and D are independent given S. Here we repeat our analysis using the CMH test.

In the SAS program file boys.sas, the cmh option (e.g., tables SES*scouts*delinquent / chisq cmh) gives the following summary statistics output where the CMH statistics are:

Summary Statistics for scout by delinquent
Controlling for SES

Cochran-Mantel-Haenszel Statistics (Based on Table Scores)
Statistic Alternative Hypothesis DF Value Prob
1 Nonzero Correlation 1 0.0080 0.9287
2 Row Mean Scores Differ 1 0.0080 0.9287
3 General Association 1 0.0080 0.9287

The small value of the general association statistic, CMH = 0.0080 which is very close to zero indicates that conditional independence model is a good fit for this data; i.e., we cannot reject the null hypothesis.

The hypothesis of conditional independence is tenable, thus $$\theta_{BD(\text{high})} = \theta_{BD(\text{mid})} = \theta_{BD(\text{low})} = 1$$, is also tenable. Below, we can see that the association can be summarized with the common odds ratio value of 0.978, with a 95% CI (0.597, 1.601).

Common Odds Ratio and Relative Risks
Statistic Method Value 95% Confidence Limits
Odds Ratio Mantel-Haenszel 0.9777 0.5970 1.6010
Logit 0.9770 0.5959 1.6020
Relative Risk (Column 1) Mantel-Haenszel 0.9974 0.9426 1.0553
Logit 1.0015 0.9581 1.0468
Relative Risk (Column 2) Mantel-Haenszel 1.0193 0.6706 1.5495
Logit 1.0195 0.6712 1.5484

Since $$\theta_{BD(\text{high})} \approx \theta_{BD(\text{mid})} \approx \theta_{BD(\text{low})}$$, the CMH is typically a more powerful statistic than the Pearson chi-squared statistic we calculated in the previous section, $$X^2 = 0.160$$.

The option in R is mantelhaen.test() and used in the file boys.R as shown below:

Here is the output:

It gives the same value as SAS (e.g., Mantel-Haenszel $$X^2= 0.008$$, df = 1, p-value = 0.9287), and it only computes the general association version of the CMH statistic which treats both variables as nominal, which is very close to zero and indicates that conditional independence model is a good fit for this data; i.e., we cannot reject the null hypothesis.

The hypothesis of conditional independence is tenable, thus $$\theta_{BD(\text{high})} = \theta_{BD(\text{mid})} = \theta_{BD(\text{low})} = 1$$, is also tenable. Above, we can see that the association can be summarized with the common odds ratio value of 0.978, with a 95% CI (0.597, 1.601).

Since $$\theta_{BD(\text{high})} \approx \theta_{BD(\text{mid})} \approx \theta_{BD(\text{low})}$$, the CMH is typically a more powerful statistic than the Pearson chi-squared statistic we calculated in the previous section, $$X^2 = 0.160$$.

# 5.3.6 - Homogeneous Association

5.3.6 - Homogeneous Association

Homogeneous association implies that the conditional relationship between any pair of variables given the third one is the same at each level of the third variable. This model is also known as a no three-factor interactions model or no second-order interactions model.

There is really no graphical representation for this model, but the log-linear notation is $$(XY, YZ, XZ)$$, indicating that if we know all two-way tables, we have sufficient information to compute the expected counts under this model. In the log-linear notation, the saturated model or the three-factor interaction model is $$(XYZ)$$. The homogeneous association model is "intermediate" in complexity, between the conditional independence model, $$(XY, XZ)$$ and the saturated model, $$(XYZ)$$.

If we take the conditional independence model $$(XY, XZ)$$:

and add a direct link between $$Y$$ and $$Z$$, we obtain the saturated model, $$(XYZ)$$:

• The saturated model, $$(XYZ)$$, allows the $$YZ$$ odds ratios at each level of $$X = 1,\ldots , I$$ to be unique.
• The conditional independence model, $$(XY, XZ)$$ requires the $$YZ$$ odds ratios at each level of $$X = 1, \ldots , I$$ to be equal to 1.
• The homogeneous association model, $$(XY, XZ, YZ)$$, requires the $$YZ$$ odds ratios at each level of $$X$$ to be identical, but not necessarily equal to 1.

Under the model of homogeneous association, there are no closed-form estimates for the expected cell probabilities like we have derived for the previous models. ML estimates must be computed by an iterative procedure. The most popular methods are

• iterative proportional fitting (IPF), and
• Newton Raphson (NR).

We will be able to fit this model later using software for logistic regression or log-linear models. For now, we will consider testing for the homogeneity of the odds-ratios via the Breslow-Day statistic.

## Breslow-Day Statistic

To test the hypothesis that the odds ratio between $$X$$ and $$Y$$ is the same at each level of $$Z$$

$$H_0 \colon \theta_{XY(1)} = \theta_{XY(2)} = \dots = \theta_{XY(k)}$$

there is a non-model based statistic, Breslow-Day statistic that is like Pearson $$X^2$$ statistic

$$X^2=\sum\limits_i\sum\limits_j\sum\limits_k \dfrac{(O_{ijk}-E_{ijk})^2}{E_{ijk}}$$

where the $$E_{ijk}$$ are calculated assuming the above $$H_0$$ is true, that is there is a common odds ratio across the level of the third variable.

The Breslow-Day statistic

• has an approximate chi-squared distribution with df $$= K − 1$$, given a large sample size under $$H_0$$
• does not work well for a small sample size, while the CMH test could still work fine
• has not been generalized for $$I \times J \times K$$ tables, while there is such a generalization for the CMH test

If we reject the conditional independence with the CMH test, we should still test for homogeneous associations.

## Example - Boy Scouts and Juvenile Delinquency

We should expect that the homogeneous association model fits well for the boys scout example, since we already concluded that the conditional independence model fits well, and the conditional independence model is a special case of the homogeneous association model. Let's see how we compute the Breslow-Day statistic in SAS or R.

In SAS, the cmh option will produce the Breslow-Day statistic; boys.sas

For the boy scout example, the Breslow-Day statistic is 0.15 with df = 2, p-value = 0.93. We do NOT have sufficient evidence to reject the model of homogeneous associations. Furthermore, the evidence is strong that associations are very similar across different levels of socioeconomic status.

Breslow-Day Test for
Homogeneity of Odds Ratios
Chi-Square 0.1518
DF 2
Pr > ChiSq 0.9269

Total Sample Size = 800

The expected odds ratio for each table are: $$\hat{\theta}_{BD(high)}=1.20\approx \hat{\theta}_{BD(mild)}=0.89\approx \hat{\theta}_{BD(low)}=1.02$$

In this case, the common odds estimate from the CMH test is a good estimate of the above values, i.e., common OR=0.978 with 95% confidence interval (0.597, 1.601).

Of course, this was to be expected for this example, since we already concluded that the conditional independence model fits well, and the conditional independence model is a special case of the homogeneous association model.

There is not a single built-in function in R that will compute the Breslow-Day statistic. We can still use a log-linear models, (e.g. loglin() or glm() in R) to fit the homogeneous association model to test the above hypothesis, or we can use our own function breslowday.test() provided in the file breslowday.test_.R. This is being called in the R code file boys.R below.

#### Breslow-Day test
#### make sure to first source/run breslowday.test_.R
#source(file.choose()) # choose breslowday.test_.R
breslowday.test(temp)

Here is the resulting output:

For the boy scout example, the Breslow-Day statistic is 0.15 with df = 2, p-value = 0.93. We do NOT have sufficient evidence to reject the model of homogeneous associations. Furthermore, the evidence is strong that associations are very similar across different levels of socioeconomic status.

The expected odds ratio for each table are: $$\hat{\theta}_{BD(high)}=1.20\approx \hat{\theta}_{BD(mild)}=0.89\approx \hat{\theta}_{BD(low)}=1.02$$

In this case, the common odds estimate from CMH test is a good estimate of the above values, i.e., common OR=0.978 with 95% confidence interval (0.597, 1.601).

Of course, this was to be expected for this example, since we already concluded that the conditional independence model fits well, and the conditional independence model is a special case of the homogeneous association model.

We will see more about these models and functions in both R and SAS in the upcoming lessons.

Next, let's see how this works for the Berkeley admissions data. Recall that we set $$X=$$ sex, $$Y=$$ admission status, and $$Z=$$ department. The question of bias in admission can be approached with two tests characterized by the following null hypotheses: 1) sex is marginally independent of admission, and 2) sex and admission are conditionally independent, given department.

See berkeley.sas that performs this test, and the output in berkeley.lst.
See berkeley.R that performs this test. This program uses a dataset already in R, and the outputs can be found in berkeley.out.

For the test of marginal independence of sex and admission, the Pearson test statistic is $$X^2 = 92.205$$ with df = 1 and p-value approximately zero. All the expected values are greater than five, so we can rely on the large sample chi-square approximation to conclude that sex and admission are significantly related. More specifically, the estimated odds ratio, 0.5423, with 95% confidence interval (0.4785, 0.6147) indicates that the odds of acceptance for males are about two times as high as that for females.

What about this relationship viewed within a particular department? The CMH test statistic of 1.5246 with df = 1 and p-value = 0.2169 indicates that sex and admission are not (significantly) conditionally related, given department. The Mantel-Haenszel estimate of the common odds ratio is $$0.9047=1/1.1053$$ with 95% CI $$(0.7719, 1.0603)$$. However, the Breslow-Day statistic testing for the homogeneity of the odds ratio is 18.83 with df = 5 and p-value = 0.002!

Stop and Think!
What is going on? Why might the CMH test and the Breslow-Day test report such different results?
Recall from the properties of the CMH test that it is not as effective if the odds ratios are not in the same direction. In this case, some departments have a higher acceptance rate for males, and other departments have a higher acceptance rate for females. This is why it's important to check the properties and assumptions of a test. As a follow-up to this, what happens if we drop Department A?

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