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Paired groups randomization

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6 years 9 months ago #155059 by Damzee
Paired groups randomization was created by Damzee
Hi everyone!

I got a question about randomization that involves question group pairs, that stay together despite the randomization.
If only we got another "layer" of grouping this should be easy, but without that, i can't find a solution and im stuck.


A few sentences about the questionnaire and the problem:

The questionnaire is set to appear by question groups. The question groups each contain one picture

and two related questions. The question groups are paired. Group A contains an obscured picture

while group B contains the original one unfold.

Basically, the questionnaire first shows the obscured picture with the two related questions (group

A), then the non-obscured one with the its two related questions ( group B ).

So the proper order [format] of the question groups goes like this: 1A-1B / 2A-2B / 3A-3B and so on.

However, in some cases the pictures are strongly related or depict the same thing. In these cases we

would like to show all the obscured pictures with the questions (group A) first, then to show the non-

obscured ones with the questions ( group B ).

In this case the question groups’ proper order [format] goes like this: 1A-2A / 1B-2B

Objective:

Randomize the question groups in a way that they stay in pairs, except for the above mentioned

cases.

Example:

Question groups in pairs:

1A–1B / 2A–2B / 3A*–3B* / 4A*–4B* / 5A–5B

The group pairs marked with an asterisk change according to this:

As mentioned above, the 3rd and 4th group are strongly related to each other and they follow the

A-A / B-B format.

Bearing this in mind the new order will be like this:

1A-1B / 2A-2B / 3A*-4A* / 3B*-4B* / 5A-5B

If we want to successfully randomize this, we should aim for the random appearance of the question

pairings, while the unique cases keep their above mentioned format.

Example:

2A-2B / 5A-5B / 3A-4A / 3B-4B / 1A-1B

We should not just keep in mind that the A-B question groups –despite the randomization– stay in

pairs, but that the unique cases should be A-A pairing followed by their respective B-B pairing.

The A-A / B-B pairing may contain more element than 2. So it could be 3 ( A-A-A / B-B-B ) or more.


Anybody got any idea how can i pull this off? Is it even possible?


Thanks in advance!
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