2
$\begingroup$

I conducted an experiment in which participants listened to sentences while looking at pictures about the sentences on a computer screen. Whether at a given time point a participant looked at the left half of a picture or the right half of a picture was recorded via an eye-tracker.

The study had a 2 by 2 within subjects and within items design. If time was not an issue here, I could use the model below (using R code). (Here, I assume varying random intercepts and random slopes for both subject and item.)

lmer(look ~ iv1 * iv2 + (1 + iv1*iv2 | subject) + (1 + iv1*iv2 | item), 
     family = "binomial", data=data)

However, given that participants listened to sentences each of which lasted a few seconds, where they looked at (left vs. right) would vary across time. So one way to model time might be to include time as a covariate (it may even be necessary to natural polynomials):

lmer(look ~ iv1 * iv2 + time + (1 + iv1*iv2 | subject) + (1 + iv1*iv2 | item), 
     family = "binomial", data=data)

But things get complicated since (1) it is possible that time interacts v1 and v2 and (2) I probably need to somehow model time in the random effect terms. An additional complication is that eye-tracking data is fairly large. The particular set of data that I am working with currently has 5 million rows, so running even the simplest multilevel logistic regression can be fairly time consuming.

So my question is, given my design, what would be a good way to model time.

$\endgroup$
1
  • $\begingroup$ What exactly is iv1 and iv2 ? Maybe I am looking it in a wrong way, but are you sure you need this random effect structure? (aside the "keeping it maximal" rationale) You allow for correlation between the intercept deviations of item (or subject) and the iv1*iv2 effect deviations within the levels of item. Those are a lot of parameters... Maybe just (time| item) (and subject obviously) are enough to test your assumptions? (time and 1 + time are equivalent specifications by the way) Check your model residuals. $\endgroup$
    – usεr11852
    Commented Nov 4, 2013 at 23:56

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.