I am conducting a vote-counting meta-analysis where the only information I have from each study is the direction of the treatment effect (positive or negative) and the sample size. My plan is to fit a binary logistic regression with treatment effect direction as the response variable and several study-level characteristics as the predictors. I would like to weight the studies by their sample size (i.e., give more weight to larger studies), but I'm not sure how to do this in a binary logistic regression.

With a continuous response variable, it would be possible to add weights like this in R using glm:

glm(y ~ x, weights = sample_size, family = gaussian)

However, for a binary response (family = binomial), the weights are treated as the number of trials, which does not make sense in my situation (each study is a single Bernoulli trial).

Is there a reasonable way to assign weights to observations in a binary logistic regression?

  • $\begingroup$ "the weights are treated as the number of trials" This is not necessarily true. The weights correspond to how much each observation is weighed in the likelihood. A weight of 5 is like having the same observation occur 5 times (and contribute 5 times to the likelihood). $\endgroup$
    – Noah
    Dec 19, 2023 at 17:28
  • $\begingroup$ A bigger issue is whether an analyses on such amazingly coursened data can be reliable. Think of the case where there is no true effect. The observed sign of an effect will be a random Bernoulli observation with probability 0.5, and unless you have hundreds of studies the proportion of positive effects could be > 0.5 easily by chance. $\endgroup$ Dec 20, 2023 at 9:28

1 Answer 1


In short I believe the answer is no, you cannot just use the sample size of the different trials (and ignore everything else about the trials).(As you rightly assume, weighting a positive trial by the sample size is inappropriate, since you didn't have n successes.

I think you need to add a lot more detail (and it's not a logistic regression question). assume that your data from each trial was normally distributed. the relevant variable is

$$\frac{(x-\mu)}{\sigma /\sqrt{n}}$$

where n is the sample size. A trial with low sample size but low standard deviation may be more 'significant' than a trial with high sample size but also higher sample size.

Without additional information, I believe ignoring the sample size would be just as valid, on the assumption each trial was not underpowered.


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