[Rasch] Rasch S.E at the cut score...reciprocal?
thomas.salzberger at gmail.com
Fri Nov 1 10:58:43 EST 2013
The S.E. is 1 over the square root of Information.
So, 0.05 = 1 over the square root of 0.25 times x (= the number of items
0.25 is the maximum information a dichotomous item can provide (at
So, the 1600 items would all need to be located at the cut off.
I can't right follow your argument about your number of candidates. Aren't
we interested in the S.E. of the person estimate?
All the other candidates do not really help in this regard (apart from the
fact that the item location estimates are more precise, of course).
Why not just take the S.E. from the person(raw score) closest to the
cut-off? And then add/substract the S.E. and see where that gets you (in
terms of logit and raw score metric)?
2013/11/1 Imogene Rothnie <imogene.rothnie at sydney.edu.au>
> Thanks very much Mike ****
> ** **
> I don’t know where the 0.25 in your below calculation comes from? ****
> ** **
> ..the test in question only has 50 items but has over 2000 candidates..so
> we would still expect small item logit S.E.s but the current conversion is
> giving us a borderzone of most of the test!****
> ** **
> *From:* rasch-bounces at acer.edu.au [mailto:rasch-bounces at acer.edu.au] *On
> Behalf Of *Mike Linacre
> *Sent:* Friday, November 01, 2013 9:38 AM
> *To:* rasch at acer.edu.au
> *Subject:* Re: [Rasch] Rasch S.E at the cut score...reciprocal?****
> ** **
> Your logic about the S.E. of raw scores is correct: smallest at the
> extremes, biggest in the center.
> See http://www.rasch.org/rmt/rmt204f.htm
> To have an S.E. as small as 0.05 logits, the test must have at least
> 1/(0.05*0.05*0.25) = 1600 dichotomous items, so that 40 is 2.5% of the raw
> score range.
> Mike L.
> On 11/1/2013 8:23 AM, Imogene Rothnie wrote:****
> I have recently read some Rasch digests about error at the cut point of
> the test. These suggest that to convert the Rasch logit S.E. at the cut
> point of the test to raw score units (or approximately) one should take
> the reciprocal of the S.E. of the nominated logit value. ****
> HOWEVER, as the S.E. gets smaller (good) the reciprocal gets bigger! And
> so the raw score range within the boderzone becomes larger. I must be
> missing something? ****
> E.g we have a test with very small S.E. at cut logit measure of 0.05…
> 1/0.05 is 20- a borderzone of 40marks…whereas an S.E. of 0.15, hence
> 1/0.15 would give you a raw score range of 14…****
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Thomas.Salzberger at gmail.com
Thomas.Salzberger at wu.ac.at
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