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I have some count data on woodland species which appears to be fairly normally distributed (See histogram below). I am fitting it as the response variable in a glm with multiple explanatory variables. As the distribution appears normal, in this instance is it better to use a Gaussian family for the glm despite it being count data? I have looked at similar questions but none are really asking the same thing as this.

Count data

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    $\begingroup$ George Box famously said that "all models are wrong, but some are useful". In this case, the question really is whether a Poisson GLM or OLS is more useful. Strictly speaking, OLS is inappropriate because you have count data. Then again, it is more flexible since it models an additional parameter, the variance. An alternative would be negative binomial distribution. Do you have any particular reason for wanting to avoid a Poisson or negbin GLM? $\endgroup$ Commented Jul 7, 2022 at 10:52
  • $\begingroup$ Hi, thanks - no there is no reason to avoid Poisson or neg Binomial - but the significance of some of my x variables changes depending on whether I use normal or poisson (I haven't tried neg binomial yet). As the results are so different I want to be sure to use the most appropriate distribution for the data. I am very new to stats so unsure of the best approach. $\endgroup$
    – user197410
    Commented Jul 7, 2022 at 12:15
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    $\begingroup$ More fundamentally: your choice of a model should not depend (usually) on the univariate distribution of the response. Given these are low counts, it would be advisable to use a Poisson model as your first choice and then study how well it explains the data. The Normal-looking envelope of your bar chart could easily be explained by Normal-looking variation among the explanatory variables. $\endgroup$
    – whuber
    Commented Jul 7, 2022 at 14:05
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    $\begingroup$ See the simialr Q at stats.stackexchange.com/questions/142338/… $\endgroup$ Commented Jul 7, 2022 at 19:59
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    $\begingroup$ A subtle point is that this is a faulty histogram due to the strange way in which the bins were determined (by software default). Visually, two bins have been combined at the left and half the remaining bins are empty (which leads to a very deceptive visualization in many cases; in this case, it makes the data look a little more symmetrically distributed than they are). You can make it into a correct histogram by specifying bins explicitly (put the breakpoints at $-1/2,1/2,3/2,\ldots,$ etc) or more simply by tabulating the counts and drawing them as a barplot. $\endgroup$
    – whuber
    Commented Jul 7, 2022 at 21:10

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You appear to be looking at the marginal distribution of the counts. Only the conditional distribution matters (although addressing the converse situation, see my answer to: What if residuals are normally distributed, but y is not?). To assess this properly, fit a model and look at the residuals.

There are several issues with fitting an incorrect type of model to data (e.g., an OLS regression for count data):

  1. The predicted values can go outside of the possible range (e.g., $\hat{y}<0$). It should be easy to check this. Using spline functions for continuous explanatory variables may allow the model to fit well enough within the range of your covariates (but extrapolation should be considered verboten).
  2. The residual distribution will have non-constant variance. This should also be easy to check, and you could always use a sandwich estimator for testing.
  3. The data will not be normal (i.e., they are discrete). This is not really a big deal for testing parameters, as you seem to have a lot of data. It would be very sketchy if you want to make prediction intervals.
  4. It may well not be the right way to think about your situation. This is a toughie, but only you can ultimately say.

All in all, it isn't clear you should use OLS based on what you've presented here. It may be acceptable, or may be possible to make it acceptable, but you'll have to check and think carefully about the results, and your situation and goals.

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  • $\begingroup$ This is really helpful, thanks. I will update when I have had a bit more experimentation $\endgroup$
    – user197410
    Commented Jul 14, 2022 at 17:13

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