How Much is Enough?

Problem 1: When I define boundaries of grade levels, what confidence interval should I define around these boundaries? The person measures have standard errors of 0.25 - 0.3 logits, which are equivalent to about 3 to 5 marks on an 80 mark test. A 95% confidence interval around the boundary contains ±2 S.E.. This is about 1 to 1.2 logits wide, or 15 to 20 marks! This seems very high, and to make a nonsense of the grade boundaries, which are themselves about 1 to 1.5 logits apart. Have I got the wrong end of the stick?
Julian Williams, University of Manchester

Answer: First, let's check that your results are reasonable. The highest precision possible with an 80 item dichotomous test is to have all items targeted exactly at a boundary. Then the S.E. at that boundary is 1/(80*0.5*0.5) 0.22 logits. Your test is only slightly less precise than this. When the S.E. is 0.3 logits, then 1 logit 1/SE² 11 score points. So that 0.3 logits 3½ points, confirming your results. An 80 item test may be too short for the precision you want. Your two-sided 95% confidence interval about a grade boundary is indeed about 1 logit wide. But is this the interval you want for decision making? More likely all you want is a high confidence, say 75%, that the child has exceeded the grade's lower boundary. Then the child is assigned to the grade if the child's measure is .68 S.E. = .68*0.3 = 0.2 logits 3 score points above the grade's lower boundary.

Problem 2: We want to anchor the scale against last year's test results. In particular we want to define certain "grade boundaries" on the scale which will be equivalent to last year's grade boundaries. I propose to include a set of anchor items whose difficulty parameters are near the grade boundaries defined for last year's test. How many anchor items will I need?

Answer: Let's assume that last year's test was successfully calibrated, and so that you have "good" item difficulties for the anchor items and exact logit measures for the grade boundaries. Each ability measure on the 80 item test has an S.E. of about 0.3 logits on this year's test. But we want to translate these measures onto last year's frame of reference. This will introduce extra imprecision due to the equating. If the equating worked perfectly with an average p-value of .8 (80% success by sample on common items), the S.E. of the equating constant would be 1/(sample size * common items * .8 * (1-.8)). If the sample size is 1,000, and we want the S.E. of the equating constant to be negligible, e.g., 0.03 logits, i.e., 1/10th of the S.E. of the person measure, then you want 1/(.03*.03*1000*.8*.2) = 7 common items. In practice, anchor items are not perfectly efficient. Some may have become "new" items, and so will need to be unanchored. 10 anchor items permits a margin of error. Choose items spread fairly uniformly across the measurement segment of interest.

Problem 3: The goodness of fit usually accepted for the infit mean square is from 0.8 to 1.2. This a "rule of thumb": is there a scientific basis to this?

Answer: The values 0.8 to 1.2 are based on the analysis of well-behaved data from multiple-choice tests. Exact standard errors for fit statistics can be computed (see Wright & Masters, 1982, p. 100ff.), but they are of limited use. The measurement challenge is to extract as much useful information as possible from inevitably messy data. The focal question is not "do the data fit the model (perfectly)", but "do the data fit the model well enough to construct useful measures". This is easy to check. Peal off bad data in layers and compare the resulting measures with scatter plots. When the plots shows no meaningful change, the remaining data are good enough. Alternatively, start with a core of "good" data, and layer on successively more doubtful data until the measures start to degrade. A few analyses like this will give you better information for the crucial fit range for your data than can be provided by any statistical theory or expert pontification.

John M. Linacre

How much is enough? Linacre J.M., Williams J. … Rasch Measurement Transactions, 1998, 12:3 p. 653.


Please help with Standard Dataset 4: Andrich Rating Scale Model



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