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Hello everybody,

I have a question regarding a labled choice experiment. I know binary DCEs, where the decision is to accept the combination of attributes or not. Levels vary only between choice sets. But my setting is a bit different. I have three alternatives. The levels of all attributes are the same within a choice set and cary only between different choice sets. Only the labels comprise a potential difference for respondents. Is such a design possible?

Best regards
Micael

Yes that is definitely possible. Attributes that are constant across alternatives within a choice set are called scenario variables. In the utility function you would add constants but also the scenario variables in all alternatives except one, where the coefficients can be different across alternatives.

You can see an example in this paper: https://journals.plos.org/plosone/article?id=10.1371/journal.pone.0142507

Michiel

Michiel, thank you very much for your response and the helpful example!

Using your example, I have a question regarding the data structure in the panel data set. So if seek shelter ist the first, escape is the second and staying ist the third and base alternative, I will estimate two coefficients for seeking shelter and escape. And for these two alternatives coeffients for all attributes. So how do I code the level for the base alternative and its attributes? Is it like in the following table?

Choice Set Seeking shelter Escape Staying Poeple leaving
1 1 0 0 20
1 0 1 0 20
1 0 0 1 20
2 1 0 0 80
2 0 1 0 80
2 0 0 1 80
…

Best regards
Michael

How you code it depends on the software that you use for model estimation. Nlogit has a different format from Apollo or Biogeme, so I refer to the manuals of those software estimation tools.

As stated in that paper, the utility of the base alternative is simply set to zero, and all coefficients and impacts are relative to this base alternative. If a dummy variable has 3 levels, then you would only include 2 levels in the utility function. This would likely result in something like this:

1 1 0 20
1 0 1 20
1 0 0 0
2 1 0 80
2 0 1 80
2 0 0 0

Michiel