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Messages - A.Kirsch

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Questions and Answers / Re: Differential Item Functioning
« on: December 14, 2020, 07:30:13 AM »
Hi Dan,

yes, you are right, I wanted to apply the ETS categories here too. I did the MH with other grouping criteria as well. But for the grouping criterion I applied here (high/low testscore), a MH was not possible, because testscore/ability is already the matching variable of the MH (?!)

Thanks a lot. Now it is clear. I simply disregarded the mean shift in my considerations,
but in principle I have understood the procedure.

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Questions and Answers / Re: Differential Item Functioning
« on: December 11, 2020, 10:36:24 AM »
Hi Dan,

Perhaps my mistake is that I tried to think of the two group-specific parameters as calibrated separately?

Tables and calculation are attached.

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Questions and Answers / Re: Differential Item Functioning
« on: December 09, 2020, 01:23:53 PM »
Hi Dan,

I got confused when trying to merge the significance test of DIF [Chi Square = (EST/SE)^2] described in the tutorial with the item specific Wald-Test (e.g. Draba, 1977, Strobl, 2012)*, which is defined as W= DIF^2 / SE(1)^2+SE(2)^2.
If I put the estimated parameters and standard errors for both groups into this formula, the result is twice (EST/SE)^2.

But why?

*https://www.rasch.org/memo25.htm

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Questions and Answers / Re: Differential Item Functioning
« on: December 08, 2020, 09:10:07 AM »
Hi Eveline, hi guys,

I know that the topic is a bit older, but I read it a few days ago and I'm a bit unsure now.

Based on the DIF tutorial by Adams & Wu (2010) and many others (e.g. Le, 2009), I have so far assumed that DIF is the absolute difference between the two estimated difficulties of the two groups. So it should be (abs.) sum of the both group-specific values given in term 3 for item i (deviation from the average difficulty of the item i), shouldn't it?

Thus, EST/SE would only be the significance of the deviation of one group from the average difficulty of the items, while the significance of DIF in relation to the item-specific Wald-Test could be determined by Chi square = [Est(1)/SE(1)]^2 + [Est(2)/SE(2)]^2 = 2 (Est/SE)^2, df=2-1.  Where EST corresponds to values in term 3, 1=Group 1 and 2=Group 2, or?

Thanks
Alex

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