There are different ways to do this depending on the format of the data. As before, for details, you need to refer to SAS or R help. Here are some general guidelines to keep in mind. Please note that we make a distinction about the way the data are entered into SAS (or R).

If data come in a tabular form, i.e., the response pattern is with counts (as seen in the previous example).

model y/n = x1 x2 /link=logit dist=binomial;

model y/n = x1 x2;[enter code here]

• PROC GENMOD: We need a variable that specifies the number of cases that equals marginal frequency counts or number of trials (e.g. n), and the number of events (y)
• PROC LOGISTIC: We do need a variable that specifies the number of cases that equals marginal frequency counts

If data come in a matrix form, i.e., subject \(\times\) variables matrix with one line for each subject, like a database

model y/n = x1 x2 / link = logit dist = binomial;

model y = x1 x2;

• PROC GENMOD: We need a variable that specifies the number of cases that equals marginal frequency counts or number of trials (e.g. n), and the number of events (y)
• PROC LOGISTIC: We do NOT need a variable that specifies the number of cases that equals marginal frequency counts

For both GENMOD and LOGISTIC, as before, include interaction terms with *, and make sure to include all lower-order terms.

We will follow both the SAS output through to explain the different parts of model fitting. The outputs from R will be essentially the same.

Once an appropriate model is fitted, the success probabilities need to be estimated using the model parameters. Note that success probabilities are now NOT simply the ratio of observed number of successes and the number of trials. A model fit introduces a structure on the success probabilities. The estimates will now be functions of model parameters. What are the parameter estimates? What is the fitted model?

The fitted model is

\(\mbox{logit}(\hat{\pi_i})=\log\dfrac{\hat{\pi_i}}{1-\hat{\pi_i}}=\hat{\beta_0}+\hat{\beta_1} X_i=-1.837+0.459\mbox{smoke}\)

where smoke is a dummy variable (e.g. design variable) that takes value 1 if at least one parent smokes and 0 if neither smokes as discussed above.

Estimated \(\beta_0=-1.827\) with standard error 0.078 is significant and it says that log-odds of a child smoking versus not smoking if neither parent is smoking (the baseline level) is \(-1.827\) and it's statistically significant.

Estimated \(\beta_1=0.459\) with a standard error of 0.088 is significant and it says that log-odds-ratio of a child smoking versus not smoking if at least one parent smokes versus neither parents smokes (the baseline level) is 0.459 and it's statistically significant. \(\exp(0.459)=1.58\) are the estimated odds-ratios. Thus there is a strong association between parent and child's smoking behavior, and the model fits well.

And the estimated probability of a student smoking given the explanatory variable (e.g. parent smoking behavior) is

\(\hat{\pi}_i=\dfrac{\exp(-1.826+0.4592X_i)}{1+\exp(-1.826+0.4592x_i)}\)

In particular, the predicted probability of a student smoking given that neither parent smokes is

\(P(Y_i=1|x_i=0)=\dfrac{\exp(-1.826+0.4592\times 0)}{1+\exp(-1.826+0.4592\times 0)}=0.14\)

and the predicted probability of a student being a smoker if at least one parent smokes is

\(P(Y_i=1|x_i=1)=\dfrac{\exp(-1.826+0.4592(X_i=1)}{1+\exp(-1.826+0.4592(x_i=1))}=0.20\)

By invoking the following option in the MODEL statement, output out=predict pred=prob, PROC LOGISTIC will print the predicted probabilities in the output:

So, we do not need to do this calculation by hand, but it may be useful to try it out to be sure of what's going on. In this model,\(\beta_0\) is the log-odds of children smoking for no-smoking parents /(x_i = 0\). Thus \(\exp(−1.8266)\) are the odds that a student smokes when the parents do not smoke.

Looking at the \(2\times2\) table, the estimated log-odds for nonsmokers is

\(\log\left(\dfrac{188}{1168}\right)=\log(0.161)=-1.8266\)

which agrees with \(\hat{\beta}_0\) from the logistic model.

From the analysis of this example as a \(2\times2\) table, \(\exp(\beta_1)\) should be the estimated odds ratio. The estimated coefficient of the dummy variable,

\(\hat{\beta}_1=0.4592\)

agrees exactly with the log-odds ratio from the \(2\times2\) table (i.e., \(\log(1.58) = (816\times1168) / (188\times 3203) = 0.459\)). That is exp(0.459) = 1.58, which is the estimated odds ratio of a student smoking.

To relate this to the interpretation of the coefficients in a linear regression, we could say that for every one-unit increase in the explanatory variable \(x_1\) (e.g., changing from no smoking parents to smoking parents), the odds of "success" \(\pi_i / (1 − \pi_i)\) will be multiplied by \(\exp(\beta_1)\), given that all the other variables are held constant.

This is not surprising, because in the logistic regression model \(\beta_1\) is the difference in the log-odds of children smoking as we move from "nosmoke" (neither parent smokes) /(x_i = 0/) to "smoke" (at least one parent smokes) \(x_i = 1\), and the difference in log-odds is a log-odds ratio.

Testing the hypothesis that the probability of the characteristic depends on the value of the \(j\)th variable.

Testing \(H_0\colon \beta_j = 0\) versus \(H_A\colon \beta_j \ne 0\).

The Wald chi-squared statistics \(z^2=(\hat{\beta}_j/\mbox{SE}(\hat{\beta}_k))^2\) for these tests are displayed along with the estimated coefficients in the "Analysis of Maximum Likelihood Estimates" section.

The standard error for \(\hat{\beta}_1\), 0.0878, agrees exactly with the standard error that we can calculate from the corresponding \(2\times2\) table.

The values indicate the significant relationship between the logit of the odds of student smoking in parents' smoking behavior. Or, we can use the information from the "Type III Analysis of Effects" section.

Again, this information indicates that parents' smoking behavior is a significant factor in the model. We could compare \(z^2\) to a chi-square with one degree of freedom; the p-value would then be the area to the right of \(z^2\) under the \(\chi^2_1\) density curve. A value of \(z^2\) (Wald statistic) bigger than 3.84 indicates that we can reject the null hypothesis \(\beta_j = 0\) at the .05-level.

\(\beta_1:\left(\dfrac{0.4592}{0.0878}\right)^2=27.354\)

An approximate \((1 − \alpha) 100\%\) confidence interval for \(\beta_j\) is given by

\(\hat{\beta}_j \pm z_{(1-\alpha/2)}SE(\hat{\beta}_j)\)

For example, a 95% confidence interval for \(\beta_1\)

\(0.4592 \pm 1.96 (0.0878)= (0.287112, 0.63128)\)

where 0.0878 is the "Standard Error"' from the "Analysis of Maximum Likelihood Estimates" section. Then, the 95% confidence interval for the odds-ratio of a child smoking if one parent is smoking in comparison to neither smoking is

\((\exp(0.287112), \exp(0.63128)) = (1.3325, 1.880)\)

Compare this with the output we get from PROC LOGISTIC:

When fitting logistic regression, we need to evaluate the overall fit of the model, the significance of individual parameter estimates and consider their interpretation. For assessing the fit of the model, we also need to consider the analysis of residuals. Definition of Pearson, deviance, and adjusted residuals is as before, and you should be able to carry this analysis.

If we include the statement...

output out=predict pred=phat reschi=pearson resdev=deviance;

in the PROC LOGISTIC call, then SAS creates a new dataset called "results" that includes all of the variables in the original dataset, the predicted probabilities \(\hat{\pi}_i\), the Pearson residuals, and the deviance residuals. Then we can add some code to calculate and print out the estimated expected number of successes \(\hat{\mu}_i=n_i\hat{\pi}_i\) and failures \(n_i-\hat{\mu}_i=n_i(1-\hat{\pi}_i)\).

There are different ways to run logistic regression depending on the format of the data. Here are some general guidelines to keep in mind with a simple example outlined in dataformats.R where we created two binary random variables with \(n\) number of trials, e.g., \(n=100\).

##create two binary vectors of length 100
x=sample(c(0,1),100, replace=T)
y=sample(c(0,1),100, replace=T)

If the data come in a tabular form, i.e., response pattern is with counts (as seen in the previous example), the data are said to be "grouped".

> ##create a 2x2 table with counts
> xytab=table(x,y)
> xytab
y
x 0 1
0 24 29
1 26 21

glm(): Let \(Y\) be the response variable capturing the number of events with the number of success (\(Y = 1\)) and failures (\(Y = 0\)). We need to create a response table that has a count for both the "success" and "failure" out of \(n\) trials in its columns. Notice that the count table below could be also the number of success \(Y = 1\), and then a column computed as \(n-Y\)

> count=cbind(xytab[,2],xytab[,1])
> count
[,1] [,2]
0 29 24
1 21 26

We also need a categorical predictor variable.

> xfactor=factor(c("0","1"))
> xfactor
[1] 0 1
Levels: 0 1

We need to specify the response distribution and a link, which in this case is done by specifying family=binomial("logit").

> tmp3=glm(count~xfactor, family=binomial("logit"))
> tmp3
Call: glm(formula = count ~ xfactor, family = binomial("logit"))
Coefficients:
(Intercept) xfactor1
0.1892 -0.4028
Degrees of Freedom: 1 Total (i.e. Null); 0 Residual
Null Deviance: 1.005
Residual Deviance: -4.441e-15 AIC: 12.72

If data come in a matrix form, i.e., subject \(\times\) variables matrix with one line for each subject, like a database, where data are "ungrouped". 

> xydata=cbind(x,y)

> xydata ## 100 rows, we are showing first 7
x y
[1,] 0 1
[2,] 0 1
[3,] 0 0
[4,] 0 1
[5,] 1 0
[6,] 0 1
[7,] 0 0.....

glm(): We need a binary response variable \(Y\) and a predictor variable \(x\), which in this case was also binary. We need to specify the response distribution and a link, which in this case is done by specifying family=binomial("logit")

> tmp1=glm(y~x, family=binomial("logit"))
> tmp1
Call: glm(formula = y ~ x, family = binomial("logit"))
Coefficients:
(Intercept) x
0.1892 -0.4028
Degrees of Freedom: 99 Total (i.e. Null); 98 Residual
Null Deviance: 138.6
Residual Deviance: 137.6 AIC: 141.6

We will follow the R output through to explain the different parts of model fitting. The output from SAS (or from many other software) will be essentially the same.

Testing the hypothesis that the probability of the characteristic depends on the value of the \(j\)th variable.

Testing \(H_0\colon\beta_j=0\) versus \(H_A\colon\beta_j\ne0\)

The Wald chi-squared statistics \(z^2=(\hat{\beta}_j/\text{SE}(\hat{\beta}_k))^2\) for these tests are displayed along with the estimated coefficients in the "Coefficients" section.

The standard error for \(\hat{\beta}_1\), 0.0878, agrees exactly with the standard error that we can calculate from the corresponding \(2\times2\) table. The values indicate the significant relationship between the logit of the odds of student smoking in parents' smoking behavior. This information indicates that parents' smoking behavior is a significant factor in the model. We could compare \(z^2\) to a chi-square with one degree of freedom; the p-value would then be the area to the right of \(z^2\) under the \(\chi^2_1\) density curve.

A value of \(z^2\) (Wald statistic) bigger than 3.84 indicates that we can reject the null hypothesis \(\beta_j=0\) at the .05-level.

\(\beta_1:\left(\dfrac{0.4592}{0.0878}\right)^2=27.354\)

An approximate \((1 − \alpha) 100\%\) confidence interval for \( \beta_j \) is given by

\(\hat{\beta}_j \pm z_{(1-\alpha/2)} \times SE(\hat{\beta}_j)\)

For example, a 95% confidence interval for \(\beta_1\)

\(0.4592\pm 1.96 (0.0878) = (0.287112, 0.63128)\)

where 0.0878 is the "Standard Error"' from the "Coefficients" section. Then, the 95% confidence interval for the odds-ratio of a child smoking if one parent is smoking in comparison to neither smoking is

\((\exp(0.287112), \exp(0.63128)) = (1.3325, 1.880)\)

Thus, there is a strong association between parent's and children smoking behavior. But does this model fit?

Overall goodness-of-fit testing

Test: \(H_0\colon\) current model vs. \(H_A\colon\) saturated model

Residual deviance: -3.7170e-13 on 0 degrees of freedom
AIC: 19.242

The goodness-of-fit statistics, deviance, \(G^2\) from this model is zero, because the model is saturated. If we want to know the fit of the intercept only model that is provided by

Test: \(H_0\colon\) intercept only model vs. \(H_A\colon\) saturated model

Null deviance: 2.9121e+01 on 1 degrees of freedom

Suppose that we fit the intercept-only model. This is accomplished by removing the predictor from the model statement, like this:

glm(response~1, family=binomial(link=logit)

The goodness-of-fit statistics are shown below.

ull deviance: 29.121 on 1 degrees of freedom

Residual deviance: 29.121 on 1 degrees of freedom
AIC: 46.362
N

The deviance \(G^2 = 29.1207\) is precisely equal to the \(G^2\) for testing independence in this \(2\times2\)  table. Thus by the assumption, the intercept-only model or the null logistic regression model states that student's smoking is unrelated to parents' smoking (e.g., assumes independence, or odds-ratio=1). But clearly, based on the values of the calculated statistics, this model (i.e., independence) does NOT fit well.

Analysis of deviance table

In R, we can test factors’ effects with the anova function to give an analysis of deviance table. We only include one factor in this model. So R dropped this factor (parentsmoke) and fit the intercept-only model to get the same statistics as above, i.e., the deviance \(G^2 = 29.121\).

> anova(smoke.logistic)
Analysis of Deviance Table
Model: binomial, link: logit
Response: response
Terms added sequentially (first to last)
            Df Deviance Resid. Df Resid. Dev
NULL                            1     29.121
parentsmoke  1   29.121         0      0.000

This example shows that analyzing a \(2\times2\) table for association is equivalent to logistic regression with a single dummy variable. We can further compare these tests to the loglinear model of independence in Lesson 10.

The goodness-of-fit statistics, \(X^2\) and \(G^2\), are defined as before in the tests of independence and loglinear models (e.g. compare observed and fitted values). For the \(chi^2\) approximation to work well, we need the \(n_i\)s to be sufficiently large so that the expected values, \(\hat{\mu}_i \ge 5\), and \(n_i − \hat{\mu}_i\ge 5\) for most of the rows. We can afford to have about 20% of these values less than 5, but none of them should fall below 1.

With real data, we often find that the \(n_i\)s are not big enough for an accurate test, and there is no single best solution to handle this but a possible solution may rely strongly on the data and context of the problem.

Suppose two models are under consideration, where one model is a special case or "reduced" form of the other obtained by setting \(k\) of the regression coefficients (parameters) equal to zero. The larger model is considered the "full" model, and the hypotheses would be

\(H_0\): reduced model versus \(H_A\): full model

Equivalently, the null hypothesis can be stated as the \(k\) predictor terms associated with the omitted coefficients have no relationship with the response, given the remaining predictor terms are already in the model. If we fit both models, we can compute the likelihood-ratio test (LRT) statistic:

where \(L_0\) and \(L_1\) are the max likelihood values for the reduced and full models, respectively. The degrees of freedom would be \(k\), the number of coefficients in question. The p-value is the area under the \(\chi^2_k\) curve to the right of \( G^2)\).

To perform the test in SAS, we can look at the "Model Fit Statistics" section and examine the value of "−2 Log L" for "Intercept and Covariates." Here, the reduced model is the "intercept-only" model (i.e., no predictors), and "intercept and covariates" is the full model. For our running example, this would be equivalent to testing "intercept-only" model vs. full (saturated) model (since we have only one predictor).

Larger differences in the "-2 Log L" values lead to smaller p-values more evidence against the reduced model in favor of the full model. For our example, \( G^2 = 5176.510 − 5147.390 = 29.1207\) with \(2 − 1 = 1\) degree of freedom. Notice that this matches the deviance we got in the earlier text above.

Also, notice that the \(G^2\) we calculated for this example is equal to 29.1207 with 1df and p-value <.0001 from "Testing Global Hypothesis: BETA=0" section (the next part of the output, see below).

Testing the null hypothesis that the set of coefficients is simultaneously zero. For example, consider the full model

\(\log\left(\dfrac{\pi}{1-\pi}\right)=\beta_0+\beta_1 x_1+\cdots+\beta_k x_k\)

and the null hypothesis \(H_0\colon \beta_1=\beta_2=\cdots=\beta_k=0\) versus the alternative that at least one of the coefficients is not zero. This is like the overall F−test in linear regression. In other words, this is testing the null hypothesis of the intercept-only model:

\(\log\left(\dfrac{\pi}{1-\pi}\right)=\beta_0\)

versus the alternative that the current (full) model is correct. This corresponds to the test in our example because we have only a single predictor term, and the reduced model that removes the coefficient for that predictor is the intercept-only model.

Coefficients:
             Estimate Std. Error z value Pr(>|z|)    
(Intercept)  -1.82661    0.07858 -23.244  < 2e-16 ***
parentsmoke1  0.45918    0.08782   5.228 1.71e-07 ***
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

An alternative statistic for measuring overall goodness-of-fit is the Hosmer-Lemeshow statistic.

This is a Pearson-like chi-square statistic that is computed after the data are grouped by having similar predicted probabilities. It is more useful when there is more than one predictor and/or continuous predictors in the model too. We will see more on this later.

To calculate this statistic:

1. Group the observations according to model-predicted probabilities ( \(\hat{\pi}_i\))
2. The number of groups is typically determined such that there is roughly an equal number of observations per group
3. The Hosmer-Lemeshow (HL) statistic, a Pearson-like chi-square statistic, is computed on the grouped data but does NOT have a limiting chi-square distribution because the observations in groups are not from identical trials. Simulations have shown that this statistic can be approximated by a chi-squared distribution with \(g − 2\) degrees of freedom, where \(g\) is the number of groups.

Warning about the Hosmer-Lemeshow goodness-of-fit test:

• It is a conservative statistic, i.e., its value is smaller than what it should be, and therefore the rejection probability of the null hypothesis is smaller.
• It has low power in predicting certain types of lack of fit such as nonlinearity in explanatory variables.
• It is highly dependent on how the observations are grouped.
• If too few groups are used (e.g., 5 or less), it almost always fails to reject the current model fit. This means that it's usually not a good measure if only one or two categorical predictor variables are involved, and it's best used for continuous predictors.

In the model statement, the option lackfit tells SAS to compute the HL statistic and print the partitioning. For our example, because we have a small number of groups (i.e., 2), this statistic gives a perfect fit (HL = 0, p-value = 1). Instead of deriving the diagnostics, we will look at them from a purely applied viewpoint. Recall the definitions and introductions to the regression residuals and Pearson and Deviance residuals.

The Pearson residuals are defined as

\(r_i=\dfrac{y_i-\hat{\mu}_i}{\sqrt{\hat{V}(\hat{\mu}_i)}}=\dfrac{y_i-n_i\hat{\pi}_i}{\sqrt{n_i\hat{\pi}_i(1-\hat{\pi}_i)}}\)

The contribution of the \(i\)th row to the Pearson statistic is

\(\dfrac{(y_i-\hat{\mu}_i)^2}{\hat{\mu}_i}+\dfrac{((n_i-y_i)-(n_i-\hat{\mu}_i))^2}{n_i-\hat{\mu}_i}=r^2_i\)

and the Pearson goodness-of fit statistic is

\(X^2=\sum\limits_{i=1}^N r^2_i\)

which we would compare to a \(\chi^2_{N-p}\) distribution. The deviance test statistic is

\(G^2=2\sum\limits_{i=1}^N \left\{ y_i\text{log}\left(\dfrac{y_i}{\hat{\mu}_i}\right)+(n_i-y_i)\text{log}\left(\dfrac{n_i-y_i}{n_i-\hat{\mu}_i}\right)\right\}\)

which we would again compare to \(\chi^2_{N-p}\), and the contribution of the \(i\)th row to the deviance is

\(2\left\{ y_i\log\left(\dfrac{y_i}{\hat{\mu}_i}\right)+(n_i-y_i)\log\left(\dfrac{n_i-y_i}{n_i-\hat{\mu}_i}\right)\right\}\)

We will note how these quantities are derived through appropriate software and how they provide useful information to understand and interpret the models.

The concepts discussed with binary predictors extend to predictors with multiple levels. In this lesson we consider \(Y_i\) a binary response, \(x_i\) a discrete explanatory variable (with \(k = 3\) levels, and make connections to the analysis of \(2\times3\) tables. But the basic ideas extend to any \(2\times J\) table.

We begin by replicating the analysis of the original \(3\times2\) table with logistic regression.

First, we re-express the data in terms of \(Y_i=\) number of smoking students, and \(n_i=\) number of students for the three groups based on parents behavior (we can think of \(i\) as an index for the rows in the table).

Then we decide on a baseline level for the explanatory variable \(x\) and create \(k − 1\)indicators if \(x\) is a categorical variable with \(k\) levels. For our example, we set "Neither" as the baseline for the parent smoking predictor and define a pair of indicators that take one of two values, specifically,

\(x_{1i}=1\) if parent smoking = "One" ,
\(x_{1i}=0\) otherwise

\(x_{2i}=1\) if parent smoking = "Both",
\(x_{2i}=0\) otherwise.

If these two indicator variables were added as columns in the table above, \(x_{1i}\) would have values \((0,1,0)\), and \(x_{2i}\) would have values \((1,0,0)\). Next, we let \(\pi_i\) denote the probability of student smoking for the \(i\)th row group so that finally, the model is

\(\log\left(\dfrac{\pi_i}{1-\pi_i}\right)=\beta_0+\beta_1 x_{1i}+\beta_2 x_{2i}\)

for \(i=1,\ldots,3\). Thus, the log-odds of a student smoking is \(\beta_0\) for students whose parents don't smoke ("neither" group), \(\beta_0+\beta_1\) for students with one smoking parent ("one" group), and \(\beta_0+\beta_2\) for students with both smoking parents. In particular, \(\beta_1\)  is the difference in log odds or, equivalently, the log odds ratio for smoking when comparing students with one smoking parent against students with neither smoking parent. Similarly, \(\beta_2\) represents that when comparing students with two smoking parents against students with neither smoking parent.

Based on the tabulated data, where we saw \(1.42=416 ( 1168)/(188(  1823))\) and \(1.80=400(1168)/(188(1380))\), we're not surprised to see \(\hat{\beta}_1=\log(1.42)=0.351\) and \(\hat{\beta}_2=\log(1.80)=0.588\). The estimated intercept should be \(\hat{\beta}_0=\log(188/1168)=-1.826\).

The goodness-of-fit statistics \(X^2\) and \(G^2\) from this model are both zeroes because the model is saturated. Suppose that we fit the intercept-only model as before by removing the predictors from the model statement:

proc logistic data=smoke descending;
class s1 s2 (ref=first)/ param=ref;
model y/n = / scale=none lackfit;
output out=predict pred=prob reschi=pearson resdev=deviance;
title 'Logistic regression for intercept-only model';
run;

    Null deviance:  3.8366e+01  on 2  degrees of freedom
Residual deviance: -3.7215e-13  on 0  degrees of freedom
AIC: 28.165

> ## hosmerlem() takes the vector of successes, predicted vector of success and g=# of groups as input
> ## produce the vector of predicted success "yhat"
> yhat=rowSums(response)*predict(smoke.logistic, type="response")
> yhat
1 2 3
400 416 188

> hosmerlem(response[,1], yhat, g=3) ## here run 3 groups
X^2 Df P(>Chi)
"-1.00593633284243e-24" "1" "."

We have shown that analyzing a \(2\times 3\) table for associations is equivalent to a binary logistic regression with two dummy variables as predictors. For \(2\times J\) tables, we would fit a binary logistic regression with \(J − 1\) indicator variables.

Starting with the (full) model

\(\log\left(\dfrac{\pi}{1-\pi}\right)=\beta_0+\beta_1 x_1+\beta_2 x_2\)

the null hypothesis \(H_0\colon\beta_1=\beta_2=0\) specifies the intercept-only (reduced) model:

\(\log\left(\dfrac{\pi}{1-\pi}\right)=\beta_0\)

In general, this test has degrees of freedom equal to the number of slope parameters, which is 2 in this case. Large chi-square statistics lead to small p-values and provide evidence to reject the intercept-only model in favor of the full model.

The likelihood-ratio statistic is

\(G^2 = −2 (\log \mbox{ likelihood from reduced model}−(−2 \log \mbox{ likelihood from full model}))\)

and the degrees of freedom is \(k=\) the number of parameters differentiating the two models. The p-value is \(P(\chi^2_k \geq G^2)\).

For our example, \(G^2 = 5176.510 − 5138.144 = 38.3658\) with \(3 − 1 = 2\) degrees of freedom. Notice that this matches

Likelihood Ratio 38.3658 2 <.0001

from the "Testing Global Hypothesis: BETA=0" section. Here is the model we just looked at in SAS.

data smoke;
input s1 s2 $ y n ;
cards;
0 1 400 1780
1 0 416 2239
0 0 188 1356
;
proc logistic data=smoke descending;
class s1 s2 (ref=first)/ param=ref;
model y/n = s1 s2 / scale=none lackfit;
output out=predict pred=prob reschi=pearson resdev=deviance;
run;

proc logistic data=smoke descending;
class s1 s2 (ref=first)/ param=ref;
model y/n = s1 / scale=none lackfit;
output out=predict pred=prob reschi=pearson resdev=deviance;
run;

Compare this to

model y/n = s1 /scale=none lackfit;

For the test of the significance of a single variable \(x_j\)

\(H_0\colon\beta_j=0\) versus \(H_A\colon\beta_j\ne0\)

we can use the (Wald) test statistic and p-value. A large value of \(z\) (relative to standard normal) or \(z^2\) (relative to chi-square with 1 degree of freedom) indicates that we can reject the null hypothesis and conclude that \(\beta_j\) is not 0.

Coefficients:
             Estimate Std. Error z value Pr(>|z|)    
(Intercept)  -1.82661    0.07858 -23.244  < 2e-16 ***
parentsmoke1  0.34905    0.09554   3.654 0.000259 ***
parentsmoke2  0.58823    0.09695   6.067  1.3e-09 ***
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Confidence Intervals: An approximate \((1 − \alpha)100\)% confidence interval for \(\beta_j\) is given by

\(\hat{\beta}_j \pm z_{(1-\alpha/2)} \times SE(\hat{\beta}_j)\)

For example, a 95% CI for \(\beta-1\) is

\(0.3491 \pm 1.96 (0.0955) = (0.16192, 0.5368)\)

Then, the 95% CI for the odds-ratio of a student smoking, if one parent is smoking in comparison to neither smoking, is

\((\exp(0.16192), \exp(0.5368)) = (1.176, 1.710)\)

Since this interval does not include the value 1, we can conclude that student and parents' smoking behaviors are associated. Furthermore,

• The estimated conditional odds ratio of a student smoking between one parent smoking and neither smoking is \(\exp(\beta_1) = \exp(0.3491) = 1.418\).
• The estimated conditional odds ratio of a student smoking between both parents smoking and neither smoking is \(\exp(\beta_2) = \exp(0.5882) = 1.801\).
• The estimated conditional odds ratio of a student smoking between both parents smoking and one smoking is \(\exp(\beta_2)/\exp(\beta_1) = 1.801/1.418 = 1.27 = \exp(\beta_2-\beta_1) = \exp(0.5882 − 0.3491) = 1.27\). That is, compared with a student who has only one parent smoking, a student who has both parents smoking has an odds of smoking 1.27 times as high.

Recall the \(3\times2\times2\) table that we examined in Lesson 5 that classifies 800 boys according to S = socioeconomic status, B = whether a boy is a scout, and D = juvenile delinquency status.

Because the outcome variable D is binary, we can express many models of interest using binary logistic regression. Before handling the full three-way table, let us consider the \(2\times2\) marginal table for B and D as we did in Lesson 5. We concluded that the boy scout status (B) and the delinquent status (D) are associated and that (from the table below)

the estimated log-odds ratio is

\(\log\left(\dfrac{33( 360)}{64 (343)}\right)=-0.6140\)

with a standard error of \(\sqrt{\dfrac{1}{33}+\dfrac{1}{343}+\dfrac{1}{64}+\dfrac{1}{360}}=0.2272\). That is, we estimate that being a boy scout is associated a lower log-odds of delinquency by 0.614; the odds-ratio is 0.541. Now let’s fit the logistic regression model

\(\log\left(\dfrac{\pi}{1-\pi}\right)=\beta_0+\beta_1 x\)

where \(x\) is a dummy variable

\(x\) = 0 if non-scout,
\(x\) = 1 if scout.

options nocenter nodate nonumber linesize=72;

data scout1;
   input S $ y n;
   cards;
scout     33 376
nonscout  64 424
;

proc logist data=scout1;
  class S / param=ref ref=first;
  model y/n = S / scale=none;
  run;

model y/n = / scale=none;

#### corresponds to the data scout1 in scout.SAS
#### Fits a Logistic regression with S="scout" or "nonscout"

S=factor(c("scout","nonscout"))
Sscout=(S=="scout")
Snonscout=(S=="nonscout")
y=c(33,64)
n=c(376,424)
count=cbind(y,n-y)
result=glm(count~Sscout+Snonscout,family=binomial("logit"))
summary(result)

Coefficients: (1 not defined because of singularities)
              Estimate Std. Error z value Pr(>|z|)    
(Intercept)    -1.7272     0.1357 -12.732  < 2e-16 ***
SscoutTRUE     -0.6140     0.2272  -2.702  0.00688 ** 
SnonscoutTRUE       NA         NA      NA       NA    
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

   Null deviance: 7.6126e+00  on 1  degrees of freedom
Residual deviance: 4.0190e-14  on 0  degrees of freedom
AIC: 15.083

Now let us do a similar analysis for the \(3\times2\) table that classifies subjects by S and D:

Two odds ratios of interest are

\((53(236)) / (34( 212)) = 1.735\)
\((53( 255)) / (10 (212)) = 6.375\)

We estimate that the odds of delinquency for the S = low group are 1.735 times as high as for the S = medium group, and 6.375 times as high as for the S = high group. The estimated log odds ratios are

\(\log 1.735 = .5512\) and \(\log 6.375 = 1.852\)

and the standard errors are

\(\sqrt{\dfrac{1}{53}+\dfrac{1}{212}+\dfrac{1}{34}+\dfrac{1}{236}}=0.2392\)

\(\sqrt{\dfrac{1}{53}+\dfrac{1}{212}+\dfrac{1}{10}+\dfrac{1}{255}}=0.3571\)

Now let us replicate this analysis using logistic regression. First, we re-express the data in terms of \y_i\) = number of delinquents and \(n_i\) = number of boys for the \(i\)th row (SES group):

Then we define a pair of dummy indicators,

\(x_1=1\) if S=medium,
\(x_1=0\) otherwise,

\(x_2=1\) if S=high,
\(x_2=0\) otherwise.

Let \(\pi\) = probability of delinquency. Then, the model

\(\log\left(\dfrac{\pi}{1-\pi}\right)=\beta_0+\beta_1 x_1+\beta_2 x_2\)

says that the log-odds of delinquency are \(\beta_0\) for S = low, \(\beta_0+\beta_1\) for S = medium, and \(\beta_0+\beta_2\) for S = high.

options nocenter nodate nonumber linesize="72";
data new;
	input S $ y n;
    cardsl
low		53 265
medium		34 270
hight		10 265
;

proc logist data=new;
class S / order data param=ref ref=first;
model y/n = S scale=none;
run;

S=factor(c("low","medium","high"))
y=c(53,34,10)
n=c(265,270,265)
count=cbind(y,n-y)
Smedium=(S=="medium")
Shigh=(S=="high")
result=glm(count~Smedium+Shigh,family=binomial("logit"))
summary(result)

Call:
glm(formula = count ~ Smedium + Shigh, family = binomial("logit"))

Deviance Residuals: 
[1]  0  0  0

Coefficients:
            Estimate Std. Error z value Pr(>|z|)    
(Intercept)  -1.3863     0.1536  -9.027  < 2e-16 ***
SmediumTRUE  -0.5512     0.2392  -2.304   0.0212 *  
ShighTRUE    -1.8524     0.3571  -5.188 2.13e-07 ***
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

(Dispersion parameter for binomial family taken to be 1)

    Null deviance: 3.6252e+01  on 2  degrees of freedom
Residual deviance: 6.8390e-14  on 0  degrees of freedom
AIC: 20.942

Number of Fisher Scoring iterations: 3

Notice that we did not say anything here about the scout status. We have "ignored" that information because we collapsed over that variable. Next, we will see how this plays out with logistic regression.

In this example, we could have also arranged the input data like this:

A SAS program for fitting the same model is shown below.

data new;
   input S $ B $ y n;
   cards;
low      scout      11   54
low      nonscout   42  211
medium   scout      14  118
medium   nonscout   20  152
high     scout       8  204
high     nonscout    2   61
;

proc logist data=new;
  class S / order=data param=ref ref=first;
  model y/n = S / scale=none;
  run;

The parameter estimates from this new program are exactly the same as before:

But the overall fit statistics are different! Before, we had \(X^2=0\) and \(G^2=0\) because the model was saturated (there were three parameters and \(N = 3\) lines of data). But now, the fit statistics are:

The model appears to fit very well, but it is no longer saturated. What happened? Recall that \(X^2\) and \(G^2\) are testing the null hypothesis that the current model is correct, versus the alternative of a saturated model. When we disaggregated the data by levels of B, using six input lines rather than three, the current model did not change but the saturated model did; the saturated model was enlarged to six parameters.

Another way to interpret the overall \(X^2\) and \(G^2\) goodness-of-fit tests is that they are testing the significance of all omitted covariates. If we collapse the data over B and use only three lines of data, then SAS is unaware of the existence of B. But if we disaggregate the data by levels of B and do not include it in the model, then SAS has the opportunity to test the fit of the current model—in which the probability of delinquency varies by S alone—against the saturated alternative in which the probability of delinquency varies by each combination of the levels of S and B. When the data are disaggregated, the goodness-of-fit tests are actually testing the hypothesis that D is unrelated to B once S has been taken into account—i.e., that D and B are conditionally independent given S.

Here’s another way to think about it. The current model has three parameters:

• an intercept, and
• two indicators for S.

But the alternative has six parameters. We can think of these six parameters as an intercept and five dummies to distinguish among the six rows of data, but then it's not easy to see how the current model becomes a special case of it. So, we think of the six parameters as

• an intercept,
• two dummies for S,
• one dummy for B, and
• two interaction terms for SB.

Now it has become clear that the current model is a special case of this model in which the coefficients for B and the SB interactions are zero. The overall \(X^2\) and \(G^2\) statistics for the disaggregated data are testing the joint significance of the B dummy and the SB interactions.

So, should we aggregate, or should we not? If the current model is true, then it doesn’t matter; we get exactly the same estimated coefficients and standard errors either way. But dis-aggregating gives us the opportunity to test the significance of the omitted terms for B and SB.

Therefore, it often makes sense to dis-aggregate your dataset by variables that are not included in the model, because it gives you the opportunity to test the overall fit of your model. But that strategy has limits. If you dis-aggregate the data too much, the \(n_i\)s may become too small to reliably test the fit by \(X^2\) and \(G^2\).

In this part of the lesson we will consider different binary logistic regression models for three-way tables and their link to log-linear models. Let us return to the \(3\times2\times2\) table:

As we discussed in Lesson 5, there are many different models that we could fit to this table. But if we think of D as a response and B and S as potential predictors, we can focus on a subset of models that make sense.

Let \(\pi\) be the probability of delinquency. The simplest model to consider is the null or intercept-only model,

\(\log\left(\dfrac{\pi}{1-\pi}\right)=\beta_0\) (1)

in which D is unrelated to B or S. If we were to fit this model in PROC LOGISTIC using the disaggregated data (all six lines), we would find that the\(X^2\) and \(G^2\) statistics are identical to those we obtained in Lesson 5 from testing the null hypothesis "S and B are independent of D". That is, testing the overall fit of model (1), i.e., intercept only model, is equivalent to testing the fit of the log-linear model (D, SB), because (1) says that D is unrelated to S and B but makes no assumptions about whether S and B are related.

After (1), we may want to fit the logit model that has a main effect for B,

\(\log\left(\dfrac{\pi}{1-\pi}\right)=\beta_0+\beta_1 x_1\) (2)

where

\(x_1=1\) if B=scout,
\(x_1=0\) otherwise.

If the data are disaggregated into six lines, the goodness-of-fit tests for model (2) will be equivalent to the test for (DB, SB) log-linear model, which says that D and S are conditionally independent given B. This makes sense, because (2) says that S has no effect on D once B has been taken into account.

The model that has main effects for S,

\(\log\left(\dfrac{\pi}{1-\pi}\right)=\beta_0+\beta_2 x_2+\beta_3 x_3\) (3)

where

\(x_2=1\) if S = medium,
\(x_2=0\) otherwise,

\(x_3=1\) if S = high,
\(x_2=0\) otherwise,

says that B has no effect on D once S has been taken into account. The goodness-of-fit tests for (3) are equivalent to testing the null hypothesis that (DS, BS) fits, i.e. that D and B are conditionally independent given S.

The logit model

\(\log\left(\dfrac{\pi}{1-\pi}\right)=\beta_0+\beta_1 x_1+\beta_2 x_2+\beta_3 x_3\) (4)

has main effects for both B and S, corresponds to the model of homogeneous association which we discussed in Lesson 5. We could not fit the model at that time, because the ML estimates have no closed-form solution, but we calculated CMH statistic and Breslow-Day statistic. But with logistic regression software, fitting this model is no more difficult than for any other model.

This model says that the effect of B on D, when expressed in terms of odds ratios, is identical across the levels of S. Equivalently, it says that the odds ratios describing the relationship between S and D are identical across the levels of B. If this model does not fit, we have evidence that the effect of B on D varies across the levels of S, or that the effect of S on D varies across the levels of B.

Finally, the saturated model can be written as

\(\log\left(\dfrac{\pi}{1-\pi}\right)=\beta_0+\beta_1 x_1+\beta_2 x_2+\beta_3x_3+\beta_4x_1x_2+\beta_5 x_1x_3\)

which has the main effects for B and S and their interactions. This model has \(X^2=G^2=0\) with zero degrees of freedom (see scout.sas).