What is the acceptable range for discrimination parameters with the two-parameter IRT model? I have information from Baker (2001) but would love additional information. I have heard discrimination parameters must be above 0.5 to be acceptable but am having trouble finding a citation for that value.

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    $\begingroup$ Seems rather arbitrary, though it probably had to do with the author's justification for the information criteria. Slopes that are too small don't discriminate individuals well, and might as well be dropped from the test. It really depends on the context anyway, and what matters more is the contribution of all the items for test response functions rather than individual item contributions. $\endgroup$ Commented Oct 12, 2015 at 11:06

2 Answers 2


I think that the cut-off of 0.5 is related to the historical factor analysis tradition of requiring a standardized loading of |0.3| (where loadings can range from -1 to 1). To see this, first transform the discrimination value into the standardized loading form with the formula

$$f = \frac{a}{\sqrt{(1 + a^2)}} $$

where $a = \alpha / 1.702$ because of the scaling of the logistic metric compare to the normal ogive. In this case, $\alpha = .5$, therefore the standardized loading is $f = 0.2818$, or approximately 0.3. This matches very closely to the standardized loading cut-off, and is probably the reason why Baker adopted it.

Note that in the same spirit of traditional factor analysis cut-off, this criteria is rather arbitrary and should generally be discouraged. You should be more interested in the meaning of the item to understand if this value is reasonable and contributes some amount of information to your test (indeed, some low discriminating items even have some use at the beginning stages of a computerized adaptive test), and also be mindful of the sampling variability via estimates such as the standard error.

  • $\begingroup$ Useful answer. Question: if I use your mirt package and get IRT parameters using coef() with IRTpars = TRUE, does this use the same 1.702 scaling factor? In other words, if that gives me an a = .50, does that correspond to an f = .28 as in your example? $\endgroup$
    – Jon
    Commented Sep 20, 2017 at 19:01
  • $\begingroup$ @psychometriko the IRTpars = TRUE just performs a transformation into a classical parameterization which is still in the logistic metric. So you'd need to still divide by 1.702 when converting to standardized slopes (see the output from summary(model) to check) $\endgroup$ Commented Sep 20, 2017 at 19:59
  • $\begingroup$ @philchalmers,what's F,h2 in the summary(model)?what's the relationship with a? $\endgroup$
    – kittygirl
    Commented Jun 5, 2019 at 3:32

In the 2PLM, the discrimination parameter is on the logit scale and thus theoretically could take on any value from - to + infinity. I would not necessarily agree that discrimination parameters have to be above .5 to be acceptable; recall also that they can certainly be negative as well. Size of discrimination parameters will also depend on sample size, response patterns in the indicators, and the corresponding bivariate contingency tables.


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