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Easyiest way to separate one group from another?

Hello,

I am a student from Germany and I am using your sawtooth software to conduct a CBC.
My problem is that I've got three different studies for my thesis, but I would like to deal with them in one Survey. So what I am looking for is a possibility to separate the resondents of my survey equally and randomly in three groups. As I don't know how many respondents are going to participate I'd like to implement a quota question which randomly chooses one of the three surveys for each of the respondent.
Unfortunately I don't know how to deal with that problem, it would be an enormous help if sb. could be so kind to tell me how to do it.

Otherwise I have to use three different links, which is kind of impractical

Unfortunately I am running out of time, I'd like to run my survey tomorrow or the day after...

I am looking forward to your answers.

Please help a student in trouble. ;-)
asked Dec 11, 2013 by Moritz

1 Answer

+1 vote
One (and maybe the easiest) way to do this is to set up a quota question at the start. Create 3 cells - one for each respondent group. Make each cell "Always qualify" in the settings. Set an upper limit for each cell, maybe one third the total for each cell.

Then via the "Advanced" tab select "Randomize all" for "Check cell membership".

When a respondent gets to this question the survey will randomly allocate them to one of the cells that have not yet reached their upper limit.

You can then use this quota question in your skip logic to direct respondents to the correct part of the survey.

Make sense?

Cheers
Russell
answered Dec 11, 2013 by rapizel Bronze (1,420 points)
Indeed it does,
thank you Russell. Still, If I create that question according to my understanding it means that the respondent only gets to another question, if the first question has been answered by the first quota? Or am I wrong? I would prefer a solution where user 1 is linked to my first survey, user two to my second and the third to my third survey.
Is that possible too?
Nevertheless, thank you very much for your immediate answer.

cheers!
Moritz
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