Hello Kate,

For both concerns you mention, OLS (ordinary least squares) regression is not a good candidate.

Your dilemma about independence could be solved by using a fixed definition of the error count, such as "the number of errors made on the first attempt" on the task of interest. If, however, you're tracking the same participants over time, then the model has to shift to allow for correlated measures (as in a repeated measures design).

Concerning alternatives to poisson models for count data, here is a site that lists and briefly explains the most common ones: https://stats.idre.ucla.edu/stata/seminars/regression-models-with-count-data/

Good luck with your work.

Hello Kate,

For both concerns you mention, OLS (ordinary least squares) regression is not a good candidate.

Your dilemma about independence could be solved by using a fixed definition of the error count, such as "the number of errors made on the first attempt" on the task of interest. If, however, you're tracking the same participants over time, then the model has to shift to allow for correlated measures (as in a repeated measures design).

Concerning alternatives to poisson models for count data, here is a site that lists and briefly explains the most common ones: https://stats.idre.ucla.edu/stata/seminars/regression-models-with-count-data/

Good luck with your work.

There are several options here. The two that spring to mind first are generated estimating equations and multilevel (aka linear mixed) models.

I'm more familiar with multilevel models - which are generally more flexible. If you are using R you can fit these with the lme4 package (amongst others) or Bayesian software such as the brms package.

For count data there are Poisson, quasipoisson and negative binomial options to handle likely over dispersion and you can specify random intercepts to handle dependency between observations associated with a single individual/case.

e.g.,

library(lme4)

fit1

What is the empirical distribution of your count variable?

Hi , It is a normal distribution. Daniel Wright

The Dependent variable is to count how many mistakes participants made in a test including 5 questions. In fact, the 5 questions are similar to each other. Participants may learn from the first mistake and avoid making mistakes in the remaining questions. In Poisson regression, it assumes that every event should be independent. In addition, the dependent variable is in a normal distribution. I am not sure whether I still can apply Poisson regression.

Kate Xx , how can a count variable up to five be Normally distributed. It is discrete. Further, what you appear to have is what Mosteller and Tukey (1977) called a counted fraction (out of 5) or a proportion, so you probably should be thinking more of a binomial model (e.g., logistic regression and a quasi-binomial may be useful), but please provide the empirical distribution. Given this is a number out of 5 just give the frequencies for 0,1,2,3,4,5.

Sorry, I just find it is not normally distributed. It seems I should consider the negative binomial regression model.

I do not recommend you doing a negative binomial regression. Given you had been saying it was normally distributed, it is likely better that you work with a person in your institution on the statistics. That will likely be the best way to get the most out of these data.

Hello Kate,

From the histogram, I can see that your data might follows a zero-inflated Poisson distribution. In this case, you can consider the zero-inflated Poisson regression.

See here:

http://bitly.ws/99wh

For the problem of independence, I agree with David Morse .

Good luck with your work!

I am not sure about the dependence. When I see the graph, I would assume that every participant delivered exactly 1 value: the amount of errors made during the task. Therefore, there should be no dependence between the data. But maybe we need more information. On the other hand, I agree with Daniel, this can be quite challenging and you should consult a statistician in your department.

Rainer Duesing , I think she was meaning dependence within each person, so if you thought it would be Poisson without this dependence, you might expect extra Poisson variation. Of course this isn't Poisson, it is a number out of 5, so is treated as a single value per person it would be a proportion, and a quasi-binomial term might be appropriate. But given that the person presumably did much research and contemplation before writing a question, and believed the distribution was Normal, ...

Given that the response are counts "out of 5" (so the domain is {0,1,2,3,4,5}): the number of mistakes made in a series of 5 trials. This calles for a binomial model. However, it might in fact be too strict to assume that the probability of making a mistake is the same across all subjects, so that a beta-binomial model seems to be appropriate.

However, since the counts are pretty concentrated close to 0 and the frequencies of higher counts are (more or less) exponentially decreading, I suppose that a negative binomial model would work similarily well (estimating a mistake-rate rather then a mistake-probability).

Hi Kate Xx ,

Based on your information “Participants may learn from the first mistake and then try to avoid mistakes in the latter”, it is interesting that after the opportunity to learn from mistakes, there are still mistakes.

Two consideration: (1) human behaviour 😊 ; (2) there are dependence among measures (depending on the type of “human behaviour”)

If your outcome is measured as before-after, it is a kind of repeated measure analysis. Therefore, you have a panel data. In addition, since your outcome is count and zero skewed, your outcome should be Poisson (of course zero inflated).

For a panel dataset you can perform panel regression or generalized estimated equation (GEE) instead. Since your condition is not suitable for population averaged analysis, I prefer panel regression with random effect option.

However, you are using SPSS. I could not find panel regression for Poisson outcome in SPSS. Instead, you can perform GEE with Poisson outcome.

Analyze --> generalized linear models --> GEE --> dialog box --> choose Poisson log linear in the second step of the dialog box.

Why it does not follow Poisson distribution?!

Anyway, you can choose negative binominal in the mentioned dialog box instead of Poisson loglinear.