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]]>Multiplying by 1/2 keeps the final number between 0-1. However, this is not the only way to calculate the item discrimination. Another way is to use the percentage correct rather than the number of correct items.

In terms of the distribution, this is more art than science. It is at the discretion of the teacher to determine what is most appropriate given the content and the particular context.

For reference, consider Brown (2003), “language assessment principles and classroom practices”

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]]>I have a couple of questions

I do not fully understand this formula: Item discrimination = # items correct of strong group – # items correct of weak group

1/2(total of two groups).

Why the .5*total… ? could you explain that a bit? or have I got caught in semantics or layout?

Secondly what is a “reasonable” distribution of students picking the answers. Don’t you mean that an even distribution between the three distractors? We do expect the students to pick the right answer and hopefully students know the right answer. Therefore I should mean that the interesting thing is the plausibility of the distractors. What is the scientific research that you use for your opinion in this matter?

Thirdly which references do you rely on in this article?

Looking forward to your responses, have a good day

Cita Nørgård, teaching and learning consultant at University of Southern Denmark.

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]]>Auntie Teri

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]]>Very interesting…!

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