# Generalized estimating equations output in SPSS

I am hoping to confirm that I have a suitable way to analyse the different proportions of people who are categorized as left lateralised on the one hand, or bilateral/right lateralised on the other in two different tasks.

I cannot use an ordinary logistic regression (or chi square test) as the conditions are repeated measures.

I have used the Generalized Estimating Equations option in SPSS to allow for the within subjects individual intercepts to vary as for repeated measures, but am wondering how best to interpret the output to show that the proportion of those in each category differs between the two tasks.

Is it correct that I need to re-estimate the model but remove condition from my fixed effects and then compare measures of model fit in each estimation for a significant difference (e.g.using the AIC)? Or should I just stick to looking for a significant Wald statistic and leave it at that?

My final question is how to interpret the pairwise comparisons from this type of analysis. This shows a significant difference between my two conditions, but as I understand it to be a log odds value I'm not quite sure where to go with it.

I do apologise if I haven't made my problems clear, I am quite new to this.

• Your first sentence makes me wonder exactly what outcome you're modeling and where exactly the repeated measurements are coming from. It sounds like your repeated measurements are just the left and right hand binary outcomes for each person. Or, are the repeated measurements the two different tasks, with the left/right measurements collapsed to a single binary outcome? If its the latter, then is there a reason you're analyzing both tasks in a single model? Is that justified? It would simplify the analysis to do two separate models and may actually be the better thing to do. – Macro Jul 18 '12 at 23:52
• (cont) It seems that if you're putting both tasks into the same model, you're constraining any potential covariates to have the same effect for each task, which may not be sensible. If there are no covariates and your goal is "show that the proportion of those in each category differs between the two tasks", this is a simple test of two proportions and can be done without the model. – Macro Jul 18 '12 at 23:56
• By the way, you certainly cannot use AIC in generalized estimating equations models, as the estimates produced are not maximum likelihood estimators. – Macro Jul 19 '12 at 12:44
• The outcome variable is whether an individual is left lateralized or not (based originally on a continuous measure, but it is standard in the literature to classify laterality categorically in this way). My first predictor is task (which is a repeated measure). So I understand from your comment that I could just use a test of proportions to see if the numbers of people who were left lateralised were comparable. However, I also have a covariate of number of words produced, which we can indeed assume to have the same effect in both tasks (two verbal fluency tasks). – Heather Jul 20 '12 at 14:48
• cont. The reason I did not use a normal binary logistic regression was because I thought I would be violating the assumption of independence of errors due to repeated measures. When I tried to run the GLM mixed procedure on SPSS I a) was given error messages about the Hessian matrix (!?) and b) couldn’t ascertain from the ouptput any estimates of differences in odds ratios for the groups or any way of comparing models. So I hoped the GEE might be suitable instead. – Heather Jul 20 '12 at 14:49

It is true that betas are log odds, however, you can exponentiate them and then interpret the result as an odds ratio. If odds ratios aren't sufficiently intuitive (in my experience, people aren't born with the ability to think in odds ratios, but you can learn to use them), you can solve for two cases that have covariate values that seem typical, or are of interest to you, and that are identical except that in the first case CONDITION=0 and the other CONDITION=1. Exponentiating both will yield the two odds; computing $odds/(odds+1)$ in each case will yield two probabilities. Remember that these probabilities and their difference hold only for that exact combination of covariate values. Thus, if you want to know about what happens with a different set of covariate values, you have to go through the process again.