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Please don't close my question, it really is not a duplicate, no other answer on this forum is relevant to my case. Also, I have been advised that if I submit the spss output of my non-normally distributed model, someone will have a look at it and advise me. I have been waiting for some help for weeks. It's very important, please.

My data show deviation from normality of the residuals (as evidenced by the Shapiro results) but they say that anova is strong enough to 'survive' that. My dependent variable (emotion mean) is a mean score done from two ordinal variables. I wonder if this is what causes the normality problem. I have two IVs (gender and relationship type).

Could someone have a peek at my results and advise me if I can ignore it (the non-normal distribution) and still do my two-way anova, please?

The following are the results of the normality testing for residuals:

enter image description here enter image description here

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  • $\begingroup$ See this answer stats.stackexchange.com/questions/5680/… $\endgroup$
    – DevD
    Aug 1 at 18:53
  • $\begingroup$ @DevD I did read that. It is a very different anova. I need help with the residuals of my particular DV and anova. I was told to post the output. $\endgroup$
    – lisaarthur
    Aug 1 at 19:02
  • $\begingroup$ What is the model you have fit? Is there just one predictor? $\endgroup$
    – mkt
    Aug 1 at 19:11
  • $\begingroup$ @mkt That's not a predictor. That's normality results for the residuals for my DV. My previous question had the normality distribution of the DV for both of my IVs but it got closed. And it was ignored before that and people asked me for the results for the residuals. So I posted the residuals this time. They said that is what's important. But if you would like to see it, it is here stats.stackexchange.com/questions/582159/… $\endgroup$
    – lisaarthur
    Aug 1 at 19:35

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The QQ plot looks "reasonably" Normal given that there are implicit constraints on the values the residuals can take because:

  • The independent variables are gender (two levels) and relationship type (also two levels).
  • The dependent variable is meanEmotion defined as the average of two Likert-type ordinal variables.

That's why there are 107 observations (the degrees of freedom of the Shapiro Wilk test) but we see only about 20 distinct residuals in the QQ plot: there is overplotting (points plotted on top of each other).

Since you know — even before doing the analysis — that the Normal distribution is only an approximation in your case, you shouldn't over-interpret the result of the Normality tests. Your data would be better modeled with an ordinal logistic regression instead of the ANOVA, which assumes (among other things) that the response is continuous. @EdM points to the UCLA Stats tutorials which have a section about ordinal logistic regression.

This answer by @ChristianHennig discusses some pitfalls of checking model assumptions: Testing Model Assumptions in R.

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  • $\begingroup$ Thank you for your kind response. Both IVs have 2 levels. Here is full information on my model and the normality outputs for each of the IVs on the DV: stats.stackexchange.com/questions/582159/… $\endgroup$
    – lisaarthur
    Aug 1 at 20:08
  • $\begingroup$ "There are 107 observations but only about 20 distinct residual values." - Is that bad? Did I make a mistake somewhere? $\endgroup$
    – lisaarthur
    Aug 1 at 20:09
  • $\begingroup$ It's more of a misunderstanding rather than an error: given the nature of your data, the Normal distribution can only be an approximation. But then you perform not one but two tests for Normality. $\endgroup$
    – dipetkov
    Aug 1 at 20:19
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    $\begingroup$ A Normal random variable can take any value of the real line. Your response can take only a limited set of distinct values. If the original variables have levels 1, 2, 3, 4, 5, the average of two of those can be 1, 1.5, 2, 2.5, 3, 3.5, 4, 4.5, 5. And there are 4 possible combinations of the IVs. Hence the limited possibilities for the residual values. This looks suspicious to the Kolmogorov Smirnov test, hence the low p-value. But it's to be expected given the nature of your data! That's why I say not to read too much into that p-value. $\endgroup$
    – dipetkov
    Aug 1 at 20:31
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    $\begingroup$ +1 in particular for the suggestion to try ordinal logistic regression. The OP might want to look at this UCLA OARC web page for links to examples of how to do it. $\endgroup$
    – EdM
    Aug 1 at 20:51

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