Question: We have 5 dichotomous items based on the same reading text. What would happen if we replace the dichotomies with a rating scale item scored 0-5?
Answer: This is an attractive option if the target dichotomies exhibit more local dependence than the other items on the test. Let us first suppose that the 5 items fit the Rasch model about as well as the other items on the test. Then here is a standard polytomous rating scale model:
and here is an example of one of the five items:
Then, for these to be equivalent,
and similarly for Di4, Dii5. The overall difficulty of a polytomous item is given by
giving Fik = Di - Dik as the Rasch-Andrich thresholds.
In general, the overall difficulty of the polytomous item will not match the average difficulty of the equivalent dichotomous items. If the local origin of the measurement scale is set at the average difficulty of the items, it is seen that the local origin will also change location, so that the person measures will change value. But all person measures will change by about the same amount.
Example: Dichotomous item difficulties are {1,2,3,4,5} logits. Then the Dix are {.55, 1.87, 3.0, 4.13, 5.45} logits.
So what will happen if your 5 items are more locally dependent than the other items on the test? If the local dependence increases their fit to the Rasch model, then they will be more discriminating than 5 independent Rasch dichotomous items. Consequently the {Dik} will be less dispersed than the values given above.
But if the local dependence is on a secondary dimension, so that it decreases their fit to the Rasch model, then the {Dik} will be more dispersed than the values given above.
John Michael Linacre
Rating Scale Equivalence to Dichotomies, Linacre J.M. Rasch Measurement Transactions, 2006, 20:1 p. 1052
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