# Is it possible to calculate a log-odds ratio with repeated trials?

Background and Problem

This is a continuation of a question posted earlier (here) in which I describe a meta-analysis combining between- and within-subject designs using log-odds ratios (OR) as the metric of interest. In short, I indicated that I would have access to the raw data from a series of experiments on the same topic using either design. I was asked to aggregate them, and also to test for design effects. The formula necessary to calculate a within-subject OR comparable to typical between-subject OR calculations was helpfully provided in one of the answers.

I have since received the data and discovered that the structure is not quite as anticipated. In my earlier post, I said that the within-subject condition would involve two responses (i.e., a binary response for each of two conditions). This remains true. However, I had anticipated that the between-subject condition would involve only a single, binary response (in one of the two conditions), but have discovered it involves two binary responses (each in the same condition). This means that whereas the within-subject OR must be based on only a single response for each condition, the between-subject OR based on two responses in a single condition for each subject.

My Question:

With this in mind, I would like to ask:

1. Is it possible to calculate a log-odds ratio from binary data involving multiple responses from a given subject comparable to the within-subject log-odds ratio described here? Or is this comparison no longer possible using an OR?
2. While I have reason to prefer a standard meta-analytic approach, if this is no longer possible, could my question instead be resolved using a logistic multi-level modelling approach with participant as a random effect nested within experiment?
• So, just to clarify, they were allocated to a condition and then each gave exactly two binary responses which have equal status? So you could create a 0, 1, 2 variable if you chose? Sep 26, 2016 at 16:00
• @mdewey That is correct. For each participant in the between-subject experiments there were two trials from the same condition (randomly assigned) with binary responses, so I could count the number of occurrences of the event in question for a given subject, resulting in a count (0, 1, 2) or a proportion (0, .5, 1). Sep 26, 2016 at 20:08
• @mdewey Just to follow-up on your comment – it sounded as though you might know the answer but simply never got around to posting it. I am still trying to decide upon the best course of action, in case you have any advice. Nov 13, 2016 at 17:40
• I am afraid I cannot see how to do this but that may just be my ignorance. If it is still a live issue for you then why not edit the extra information into your question which will bring the question back up onto the front page and hope someone more knowledgeable than me sees it. Nov 14, 2016 at 13:23