Germán Rodríguez
Generalized Linear Models Princeton University

### 3.3 The Comparison of Two Groups

We start our applications of logit regression with the simplest possible example: a two by two table. We study a binary outcome in two groups, and introduce the odds ratio and the logit analogue of the two-sample $$t$$ test.

#### 3.3.1 A 2-by-2 Table

We will use the contraceptive use data classified by desire for more children, as summarized in Table 3.2

Table 3.2. Contraceptive Use by Desire for More Children

 Desires Using Not Using All $$i$$ $$y_i$$ $$n_i-y_i$$ $$n_i$$ Yes 219 753 972 No 288 347 635 All 507 1100 1607

We treat the counts of users $$y_i$$ as realizations of independent random variables $$Y_i$$ having binomial distributions $$B(n_i,\pi_i)$$ for $$i=1,2$$, and consider models for the logits of the probabilities.

#### 3.3.2 Testing Homogeneity

There are only two possible models we can entertain for these data. The first one is the null model. This model assumes homogeneity, so the two groups have the same probability and therefore the same logit

$\mbox{logit}(\pi_i) = \eta.$

The m.l.e. of the common logit is $$-0.775$$, which happens to be the logit of the sample proportion $$507/1607=0.316$$. The standard error of the estimate is 0.054. This value can be used to obtain an approximate 95% confidence limit for the logit with boundaries $$(-0.880,-0.669)$$. Calculating the antilogit of these values, we obtain a 95% confidence interval for the overall probability of using contraception of $$(0.293,0.339)$$.

The deviance for the null model happens to be 91.7 on one d.f. (two groups minus one parameter). This value is highly significant, indicating that this model does not fit the data, i.e. the two groups classified by desire for more children do not have the same probability of using contraception.

The value of the deviance is easily verified by hand. The estimated probability of 0.316, applied to the sample sizes in Table 3.2, leads us to expect 306.7 and 200.3 users of contraception in the two groups, and therefore 665.3 and 434.7 non-users . Comparing the observed and expected numbers of users and non-users in the two groups using Equation 3.13 gives 91.7.

You can also compare the observed and expected frequencies using Pearson’s chi-squared statistic from Equation 3.14. The result is 92.6 on one d.f., and provides an alternative test of the goodness of fit of the null model.

#### 3.3.3 The Odds Ratio

The other model that we can entertain for the two-by-two table is the one-factor model, where we write

$\mbox{logit}(\pi_i) = \eta + \alpha_i,$

where $$\eta$$ is an overall logit and $$\alpha_i$$ is the effect of group $$i$$ on the logit. Just as in the one-way anova model, we need to introduce a restriction to identify this model. We use the reference cell method, and set $$\alpha_1=0$$. The model can then be written

$\mbox{logit}(\pi_i) = \left\{ \begin{array}{ll}\eta&i=1\ \eta+\alpha_2&i=2 \end{array} \right.$

so that $$\eta$$ becomes the logit of the reference cell, and $$\alpha_2$$ is the effect of level two of the factor compared to level one, or more simply the difference in logits between level two and the reference cell. Table 3.3 shows parameter estimates and standard errors for this model.

Table 3.3. Parameter Estimates for Logit Model of
Contraceptive Use by Desire for More Children

 Parameter Symbol Estimate Std. Error $$z$$-ratio Constant $$\eta$$ $$-$$1.235 0.077 $$-$$16.09 Desire $$\alpha_2$$ 1.049 0.111 9.48

The estimate of $$\eta$$ is, as you might expect, the logit of the observed proportion using contraception among women who desire more children, $$\mbox{logit}(219/972)= -1.235$$. The estimate of $$\alpha_2$$ is the difference between the logits of the two groups, $$\mbox{logit}(288/635)-\mbox{logit}(219/972)=1.049$$.

Exponentiating the additive logit model we obtain a multiplicative model for the odds:

$\frac{\pi_i}{1-\pi_i} = \left\{ \begin{array}{ll} e^\eta& i=1\ e^\eta e^{\alpha_2}& i=2 \end{array} \right.$

so that $$e^\eta$$ becomes the odds for the reference cell and $$e^{\alpha_2}$$ is the ratio of the odds for level 2 of the factor to the odds for the reference cell. Not surprisingly, $$e^{\alpha_2}$$ is called the odds ratio.

In our example, the effect of 1.049 in the logit scale translates into an odds ratio of 2.85. Thus, the odds of using contraception among women who want no more children are nearly three times as high as the odds for women who desire more children.

From the estimated logit effect of 1.049 and its standard error we can calculate a 95% confidence interval with boundaries $$(0.831, 1.267)$$. Exponentiating these boundaries we obtain a 95% confidence interval for the odds ratio of $$(2.30, 3.55)$$. Thus, we conclude with 95% confidence that the odds of using contraception among women who want no more children are between two and three-and-a-half times the corresponding odds for women who want more children.

The estimate of the odds ratio can be calculated directly as the cross-product of the frequencies in the two-by-two table. If we let $$f_{ij}$$ denote the frequency in cell $$i,j$$ then the estimated odds ratio is

$\frac{ f_{11} f_{22} } { f_{12} f_{21} }.$

The deviance of this model is zero, because the model is saturated: it has two parameters to represent two groups, so it has to do a perfect job. The reduction in deviance of 91.7 from the null model down to zero can be interpreted as a test of

$H_0: \alpha_2=0,$

the significance of the effect of desire for more children.

An alternative test of this effect is obtained from the m.l.e of 1.049 and its standard error of 0.111, and gives a $$z$$-ratio of 9.47. Squaring this value we obtain a chi-squared of 89.8 on one d.f. Note that the Wald test is similar, but not identical, to the likelihood ratio test. Recall that in linear models the two tests were identical. In logit models they are only asymptotically equivalent.

The logit of the observed proportion $$p_i=y_i/n_i$$ has large-sample variance

$\mbox{var}(\mbox{logit}(p_i)) = \frac{1}{\mu_i} + \frac{1}{n_i-\mu_i},$

which can be estimated using $$y_i$$ to estimate $$\mu_i$$ for $$i=1,2$$. Since the two groups are independent samples, the variance of the difference in logits is the sum of the individual variances. You may use these results to verify the Wald test given above.

#### 3.3.4 The Conventional Analysis

It might be instructive to compare the results obtained here with the conventional analysis of this type of data, which focuses on the sample proportions and their difference. In our example, the proportions using contraception are 0.225 among women who want another child and 0.453 among those who do not. The difference of 0.228 has a standard error of 0.024 (calculated using the pooled estimate of the proportion). The corresponding $$z$$-ratio is 9.62 and is equivalent to a chi-squared of 92.6 on one d.f.

Note that the result coincides with the Pearson chi-squared statistic testing the goodness of fit of the null model. In fact, Pearson’s chi-squared and the conventional test for equality of two proportions are one and the same.

In the case of two samples it is debatable whether the group effect is best measured in terms of a difference in probabilities, the odds-ratio, or even some other measures such as the relative difference proposed by Sheps (1961). For arguments on all sides of this issue see Fleiss (1973).

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