Question: "One thing that I struggle with is why one doesn't add up the item difficulties for the items endorsed/answered correctly in order to estimate a person's ability level?"
Response: Yes, this is perplexing. Let's look at the simplest case. Suppose that we have 100 equally difficult items. which are at the local origin of the scale, and so are of 0 logit difficulty. A high performer would get most of them, say 90, correct. What ability level does that indicate? No matter how we add up, subtract or multiply the item difficulties, the high performer would be reported with an ability of 0 logits, exactly the same as everyone else. So simply adding up difficulties is not enough. The ability estimate is based not only on the difficulty of all the items encountered, but also the degree of success on those items. In this simplest case, the logit ability of the high performer is the average item difficulty, 0, plus the log-success-to-failure-ratio, loge(90/10) = 2.2 logits.
Rasch ability estimated from adding item difficulties, Rasch Measurement Transactions, 2004, 18:2 p. 979
|Rasch Measurement Transactions (free, online)||Rasch Measurement research papers (free, online)||Probabilistic Models for Some Intelligence and Attainment Tests, Georg Rasch||Applying the Rasch Model 3rd. Ed., Bond & Fox||Best Test Design, Wright & Stone|
|Rating Scale Analysis, Wright & Masters||Introduction to Rasch Measurement, E. Smith & R. Smith||Introduction to Many-Facet Rasch Measurement, Thomas Eckes||Invariant Measurement: Using Rasch Models in the Social, Behavioral, and Health Sciences, George Engelhard, Jr.||Statistical Analyses for Language Testers, Rita Green|
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