I have a logistic HLM model with one level-1 predictor and without level-2 predictors. Random variance components are significant for intercepts, but far from significant (p>.5) for slopes. In my understanding this means that there is some variation in the level-1 intercepts but no or very little variation in the level-1 slopes. Hence, if I add a level-2 predictor to the model, this predictor could predict level-1 intercepts but not level-1 slopes (because there is no variation to predict, right?).

However, if I add a level-2 predictor, I find that both the intercepts and the slopes are significantly predicted. The cross-level interaction looks very nice and is very much in line with my hypotheses. But how can there be a significant cross-level interaction with no varation in level-1 slopes? How can I interpret this result?

In this related thread (which was very satisfyingly answered) I cannot find an answer to my question: Can I probe cross-level interactions without random slope in a hierarchical linear model?

  • $\begingroup$ I would find your question easier to understand if you were more specific about the data and perhaps supplied the model formula that you used. $\endgroup$
    – Placidia
    Jan 30, 2015 at 12:48
  • 1
    $\begingroup$ Ok: The data are occasions (level 1) nested within persons (level 2). At each occasion, the person reports the strength nicotine craving in the last hour on a scale from 1 to 6 (level-1 predictor), and whether or not a cigarette was smoked (binary outcome). The level-2 predictor is a personality test score. As indicated by a nonsignificant random variance component for the slope, the effect of craving on smoking is about the same across persons. However, as indicated by a significant cross-level interaction, the effect of craving on smoking is stronger in persons with a high personality score. $\endgroup$
    – MWolff
    Jan 30, 2015 at 13:36

1 Answer 1


The cross-level interaction is possible when the variance of random slopes is (close to) zero. In fact it never is exactly zero, and one decides whether it is a zero choosing some criteria, depending on which you may come to different conclusions. Wald test, likelihood ratio test and bootstrapped confidence intervals are most used criteria to decide whether variance of random slopes is actually zero. They may come out with different results.

Random slopes in logistic HLM are assumed to be continuous, they play as a dependent variable in cross-level interactions. Therefore, your question may be reduced to a more simple one: is it possible to get significant effect when the variance of the dependent variable (slopes in your case) is very small. It is possible, because significance of the effect is higher if the sample size is larger, variance of independent variable is larger and residual is smaller.

Consider this example. The variance of y is 0.00000895 which some tests would identify as zero. At the same time, the effect of x is highly significant (mostly because the residual variance is also really small).

> x <- runif(100, 0, 1)
> e <- rnorm(n=100, mean=1, sd=0.001)
> y <- 0.01*x + e
> var(y)
[1] 0.00000895
> model<- lm(y ~ x)
> summary(model)

lm(formula = y ~ x)

        Min          1Q      Median          3Q         Max 
-0.00268828 -0.00068731 -0.00004948  0.00072524  0.00187309 

             Estimate Std. Error t value            Pr(>|t|)    
(Intercept) 1.0001458  0.0002009 4978.37 <0.0000000000000002 ***
x           0.0097976  0.0003340   29.34 <0.0000000000000002 ***

Going back to your original question, the point is that this kind of significant cross-level interaction has no or little meaning as it explains a negligibly small proportion of variance in your dependent variable.

  • $\begingroup$ The OP was (a) doing logistic regression (b) using a mixed effects or multi-level model. Your answer has neither of these features. Can you edit it to be an answer to the OP's question. $\endgroup$
    – mdewey
    Oct 11, 2016 at 13:33
  • 1
    $\begingroup$ @mdewey, his contention is that the random slopes are continuous & serve as a dependent variable. Whether or not this is correct, it does address the OP's situation & is an attempt to answer the Q. $\endgroup$ Oct 11, 2016 at 14:05
  • $\begingroup$ @gung, fair enough I have re-read it several times now and see what he was driving at. $\endgroup$
    – mdewey
    Oct 11, 2016 at 14:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.