Very uneven group assignment

[nadine]

Hi Jeremy,

In the advanced tutorial in the documentation it says:

If a table contains a column named group or row, PennController will automatically alternate which group of items is run! In this case, there are two groups, so every odd-numbered participant will only see the 4 items from the A group, and every even-numbered participant will only see the 4 items from the B group (or vice versa).

I took this to mean that, also with more than two groups, participants will be distributed evenly over these groups. I just ran an experiment with seven groups, and these are the numbers of participants per group:

A     5
B    11
C     4
D     1
E    13
F     2
G     6

Could you explain what went wrong and how to fix it?

Thank you!

-Nadine

• This topic was modified 1 year, 6 months ago by nadine.

[Jeremy]

Hi Nadine,

Page 10 in the advanced tutorial also says:

PennController uses an internal counter to counterbalance group assignment. By default, the counter increases by one at the end of an experiment, which rotates which experimental item group is run.

11 and 13 of your participants opened the link to your experiment when the counter had a value whose modulo 7 was 1 or 4, respectively

You could use SetCounter to insert a trial at the very beginning of your experiment to increase the counter at the very beginning instead, but that won’t ensure that your participants start with a different counter value (eg. if multiple people click the link at the same time) or that participants with a different counter value will complete your experiment (anyone can decide to leave before finishing)

The most certain way to collect an even number of participants is to collect your data in batches, circulating a link with ?withsquare=0, one with ?withsquare=1, one with ?withsquare=2, etc. and run more participants for each link as needed after you check your results file

Jeremy

[nadine]

Thank you for explaining! This makes sense. I read the whole tutorial, but didn’t really stop to think what the consequences are of increasing the counter at the end only. I’ll use the URL-based strategy next time.

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