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I want to predict a variable that has a likert format (satisfaction 1 to 4). However, the independent variables available to me are a series of count variables (e.g. counting frequency up to 6 incidents) and categorial (e.g. race or sex). I started to fit a multinomial logistic regression model (using multinom() in R), but then realised most of my explanatory variables may not be not suitable. Could I treat count data as continuous? I don't think so, but I would appreciate some advice from somebody with experience in this type of situation. Or is there a better statistical approach I could take? Thank you

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    $\begingroup$ You may also try datascience.stackexchange.com $\endgroup$ – bstrain Dec 3 '19 at 13:09
  • $\begingroup$ Could you please put your previous efforts here? This is a platform for asking technical questions, if you want to ask more about statistics you could visit stat.stackexchange.com. $\endgroup$ – Farhood ET Dec 3 '19 at 13:10
  • $\begingroup$ Thank you, Esther and bstrain. $\endgroup$ – Eli Dec 3 '19 at 13:17
  • $\begingroup$ Hi Farhood, I tried to fit a multinomial logistic regression model, but then I became unsure because of what described above. I'll visit stat.stackexchange.com Sorry, I automatically reach for stackoverflow when I'm in doubt. $\endgroup$ – Eli Dec 3 '19 at 13:19
  • $\begingroup$ Hi Eli, you can do a ordinal regression using the package ordinal in R. Multinominal is used to predict probabilities of multiple classes, and this is not your case for likert score. It can take both categorical and continuous categories. Fit it first, see where your count predictors makes sense. If it doesn't, you can try logging it. $\endgroup$ – StupidWolf Dec 3 '19 at 13:50
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Maybe you should transform your categorical variables in a series fo dummies.

if you have a variable that is color= (blue, red, green) you can create: a dummy that is equal to 1 if blue and 0 if not. a dummy that us equal to 1 if green and 0 if not.

similarly you can treat count data with dummies, based on a threshold (i.e. more than X is 1, less than X is 0)

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  • $\begingroup$ That's a very good suggestion, thank you! $\endgroup$ – Eli Dec 3 '19 at 13:16

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