I have skewness and kurtosis values of 0.447 and -0.861 respectively. There are 540 observations and the standard deviation is 1.662. I have seen posts and suggestions saying between -1,1 and 2,2 should be fine but I haven't been able to find a consensus. If it is not strictly normally distributed, is it close enough that running tests that are robust to slight deviations will still produce good enough results?

enter image description here

  • $\begingroup$ Why are you interested in normality of this sample? $\endgroup$
    – Tim
    Mar 22, 2020 at 16:22
  • $\begingroup$ I wanted to do ANOVA and normality is required $\endgroup$
    – JW_98
    Mar 22, 2020 at 16:49
  • 3
    $\begingroup$ ANOVA does not assume that marginal distributions of the variables are normal. $\endgroup$
    – Tim
    Mar 22, 2020 at 17:26
  • $\begingroup$ Sorry could you expand further? The figure shows the number of people who took x days of the week off $\endgroup$
    – JW_98
    Mar 22, 2020 at 17:38
  • $\begingroup$ ANOVA does not require that the variables be normal, but it does require that the DV be continuous and this one is not (see my answer and OP's comment on my answer and my edit). $\endgroup$
    – Peter Flom
    Mar 22, 2020 at 18:21

1 Answer 1


Your variable is "days" and ranges from 0 to 7 (so, it seems like it is days per week).

It can't be normal. IF the histogram was close to normal then, for some purposes, you might get away with treating it as normal (although I probably wouldn't do this) but yours is not close to normal. It's not even unimodal.

To get better advice on what to do, please tell us the goals of your work, what your research questions and hypotheses are and so on.

EDIT in response to comment by the OP

Then this is a count variable and you should do a count regression model. The usual starting point is Poisson regression, but I, for one, have never had a model that met the assumptions of Poisson. So, I'd start with a negative binomial regression.

Is this a dependent variable? An independent variable? Something else?

  • $\begingroup$ I want to see if a policy had an effect on the number of sick days taken by people in a chosen occupation. I am using ANOVA to see if there are significant differences between the chosen occupation and 3 similar occupations before and after the policy was implemented. This would be the dependent vairable and the occupaitons would be the independent variable $\endgroup$
    – JW_98
    Mar 22, 2020 at 16:48
  • $\begingroup$ If 7 is an upper limit, negative binomial makes about as much sense as Poisson (i.e. not much); its only advantage could be an ability to (perhaps) get closer to the variance, but predictions won't respect a looming upper boundary like that. Binomial might make some sense (the bimodality could be because it's a mixture over distributions because we're looking at a marginal, not a conditional), but I bet it's not really binomial either. $\endgroup$
    – Glen_b
    Mar 23, 2020 at 6:42
  • $\begingroup$ ... Nevertheless binomial would be a reasonable starting point for thinking about what sort of model to use, just from consideration that it's a count of days in a week that something happened. One likely issue would be dependence over time (within the reference week), that would change the distribution away from binomial. $\endgroup$
    – Glen_b
    Mar 23, 2020 at 6:48

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