# [Rasch] Rasch S.E at the cut score...reciprocal?

Imogene Rothnie imogene.rothnie at sydney.edu.au
Fri Nov 1 10:39:28 EST 2013

```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?

Imogene:

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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