In a test evaluation scenario, latent variable analyses are used to represent the ability of the test-taker and the difficulty of the question. A 1PL model uses a latent, random log-odds threshold, $\alpha$ to gauge the likelihood of correct response by a test taker. This is called a 1PL model.

In this 1PL dichotomous model, where $\alpha = 1$, the zero ability people are the people who only have chance accuracy, or the average ability people (who could have very high accuracy)?


closed as unclear what you're asking by jbowman, kjetil b halvorsen, mdewey, Carl, Michael Chernick Nov 9 '18 at 15:48

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    $\begingroup$ Please edit your question to make it more clear. You use some technical terms like "1PL" or "alpha" that may be interpreted differently by different people. $\endgroup$ – Tim Nov 8 '18 at 14:06
  • $\begingroup$ @Tim - it's always necessary to include some terminology that is specific to the area (IMHO) in a question. If someone doesn't know what 1PL is, they won't know the answer, and people searching for answers about 1PL models will typically search for that term. (Searching 1PL irt on Google takes you to the Wikipedia page for IRT, which explains it.) I edited the question a little to expand it. $\endgroup$ – Jeremy Miles Nov 8 '18 at 17:26
  • $\begingroup$ $\alpha = 1$ means the log odds of correct response from a 0 ability (average performer) is `plogis(1) = 0.73. 0 ability is average performance. Average performance is always at least as good as chance accuracy unless questions are set up to solicit the wrong response more often $\endgroup$ – AdamO Nov 8 '18 at 17:30
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    $\begingroup$ @JeremyMiles I know the terminology because I worked with IRT models and had a paper on them. I'm just asking to use wording that makes it accessible for broader audience. Moreover the Greek letters have no objective meanings, so "alpha" could mean different things in some specific context. $\endgroup$ – Tim Nov 8 '18 at 17:32
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    $\begingroup$ @JeremyMiles Acronym usage for journal articles are uniform in that one always spells out what the acronym stands for in words, unless those acronyms have become words in the dictionary sense, e.g., laser, radar. On this site, although journal rules do not apply per se, readership reaction to elliptical material lacking in links to explain otherwise simple material tend to be treated as stylistically unclear. Very wide accessability of the written word and elegant prose on this site are oft rewarded, and the contrary oft penalized. Overtly casual text is oft down voted. $\endgroup$ – Carl Nov 9 '18 at 1:00

The ability parameter zero corresponds to the 50% probability of a correct answer for the average difficult item. This is the standard notation of a one parameter logistic model (1PL) in the item response theory (IRT) framework:

(1) $P(x_{ij} = 1|\theta_i, \beta_j) = \frac{exp(\theta_i − \beta_j)}{1 + exp(\theta_i−\beta_j)}$

The probability of answering an item is the combination of two independent forces, the subject ability ($\theta$) and item difficulty ($\beta$). The inclusion of a different discrimination parameter per item ($\alpha_i$) leads to a two parameter logistic model (2PL):

(2) $P(x_{ij} = 1|\theta_i, \beta_j) = \frac{exp[\alpha_i(\theta_i − \beta_j)]}{1 + exp[\alpha_i(\theta_i − \beta_j)]}$

The 1PL is a 2PL with all the discrimination parameters are set to 1 ($\alpha_i$ = 1).

If $\theta$ and $\beta$ are zero in equation 1, their difference is zero, thus the probability of getting item right is 0.5. This happens when the location of the Item Characteristic Curve (ICC) is zero and responder has ability equals to zero, which is the situation of item q2 in the following figure (from STATA ITEM Response THEORY REFERENCE Manual RELEASE 15).

enter image description here

  • $\begingroup$ Can you define what the "average difficult item" is? I don't think this accurately reflects the question $\endgroup$ – philchalmers Nov 8 '18 at 14:54
  • $\begingroup$ Thanks for the update, but I still don't see why $\beta = 0$ is the 'average difficulty'. Are you suggesting that the difficulty parameters must be constrained to sum to 0 by construction? $\endgroup$ – philchalmers Nov 8 '18 at 16:27
  • $\begingroup$ Average on a logit scale. Would you like to suggest edits on my answer? $\endgroup$ – paoloeusebi Nov 8 '18 at 17:14
  • $\begingroup$ See my answer above. It's not that $\beta$ must equal zero to have in infection point, its just that the $\theta$ must match the difficulty parameter for this to occur. Whether the $\beta$ is the average across the items or not is irrelevant. $\endgroup$ – philchalmers Nov 8 '18 at 19:45

Assuming that your 1PL definition is

$$P(x = 1 | \theta, \alpha) = \frac{1}{1 + \exp{[-1\cdot (\theta - \alpha)}]}$$

then no, when $\theta = 0$ and $\alpha = 1$, $P(x = 1 | \theta, \alpha) \ne 0.5$.

The form of $P(x = 1 | \theta, \alpha) = 0.5$, commonly referred to as the inflection point, occurs only when $\theta = \alpha$; in other words, when the difficulty of the item matches the ability of the participant. The is generally why the $\alpha$ parameters are referred to as 'difficulty parameters', because larger $\alpha$ values clearly require higher ability values before the probability of positive endorsement becomes close to 1.


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