Ron Heck, Fall 2011
EDEP 768E: Seminar in Multilevel Modeling
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Week 4: Notes
Investigating a Random Slope
We can begin with Model 3 and add a Random slope parameter. If you remember, we have the
slope model for low SES as a fixed parameter. Where there is a random slope, we often refer to
the model as a “slopes-as-outcome” model.
When we have fixed level-1 predictors at level 2 we show them like this:
 2 j   20 .
If we wish to examine the lowSES slope (  2 j ) as randomly varying, we have to add the
variance parameter to the model ( u2 j ). So we will now have the following:
 2 j   20  u2 j .
The new model will consist of the level-1 model:
Yij  0 j  1 j femaleij  2 j lowsesij   ij ,
the level-2 intercept model:
0 j   00   01zenroll _ y   02 schcontext   03schqualcomp   04teacheff  u0 j ,
and the level-2 slope model:
 2 j   20  u2 j .
When we substitute them all into one combined equation we will have the complete model with
10 parameters to estimate (7 fixed effects, 2 random effects, and the residual).
Yij   00   01 zenroll _ y j   02 schcontext j   03schqualcomp j   04teacheff j 
 10 femaleij   20lowsesij  u2 jlowsesij  u0 j   ij
(EDEP) Week 4: Investigating a Random Slope
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The output (Table 1) confirms 10 parameters to estimate (7 fixed effects, 2 random level-2
effects, and the level-1 residual).
Table 1
Next we can look at the fixed effects (Table 2). Our primary interest initially, however, is in the
variance components.
Table 2
The variance components table (Table 3) provides evidence that the student SES-math
achievement slope varies across schools (Wald Z = 49.31, p < .003). Keep in mind that because
variances cannot be below 0 (i.e., negative), we can use a one-tailed test for their significance
(i.e., this means dividing the significance level by 2). Note that the default covariance matrix of
random effects at level 2 has a variance for the intercept and a variance for the slope, but there is
no covariance between them.
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This is referred to as a diagonal covariance matrix, since there is no off-diagonal covariance.
You can see that the off-diagonal element (the covariance) is fixed at 0.
2
 Intercept
0 


2
 Slope
 0

In the variance components table (Table 3) you can see that the level-2 variances for the intercept
and slope are 62.23 and 49.31, respectively. Both are significant. The results, therefore, suggest
we might build a model to explain variance in the slope parameter across schools.
Table 3
The fit of Model 3 with a random slope has the following fit indices (Table 4). We can use the
AIC to determine whether the model with random slope fits better than Model 3. For Model 3 (9
parameters) the -2 LL is 176,082.640. For the current model it is 176,058.224. The difference is
distributed as chi square (24.416, 1 df). The required chi-square at p = .05 is 3.84. Therefore, we
can conclude that the model with random low SES slope represents an improved fit to the data
over the model with the slope fixed.
Table 4
(EDEP) Week 4: Investigating a Random Slope
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Building a Model to Explain the Random Slope
We would like to build a parallel model to the intercept and then only if variables are not
significant, can we drop them out of the final model.
We will build the interactions through opening the “Fixed Effect” box and clicking on “Build
Nested Terms.” After clicking on that, we will put “lowses” in the “Build Term” box using By*
(lowses*Zenroll_y). Do the same for each of the 4 school predictors in model 3. Click continue.
Now we have added four effects to the model. We can confirm 14 effects to estimate. Below is
the -2LL and other indices (Table 5).
Table 5
We can see that the model with 14 parameters (-2LL = 176022.629) estimated fits better than the
model with 10 parameters (-2LL = 176058.224). We can construct a chi-square test with 4
degrees of freedom (176058.224 - 176022.629 = 35.395) with 4 df). The required chi-square
with 4 df is 9.49 at p = .05. Hence, the current model represents an improved fit.
The fixed effects (Table 6) suggest there are two promising effects (lowSES*schcontext) and
lowSES*schqualcomp. The meaning is that in schools 1-SD above the grand mean in terms of
more challenging student composition (i.e., higher levels of students on free/reduced lunch,
receiving languages services), the impact of individual lowSES on math outcomes is diminished
(4.73, p < .01). So the combined effect in those schools is -14.56 + 4.73 = -9.83. Hence there is
less of a gap in achievement for low SES students in these schools. This suggests the effect of
student composition “diminishes” the negative effect of low SES on achievement. As we might
suspect, in schools where there are high numbers of low SES students, the impact of individual
low SES status on achievement is lessened considerably. On the other hand, in schools where
very few students require these types of special academic programs, the impact of being lowSES
on outcomes is exaggerated (-14.56 + -4.73 = -19.29). Regarding school quality, the results
suggest that the gap between lowSES and achievement is actually larger (-1.71), p < .08). This
suggests that in higher quality schools in terms of academic focus, being low SES is actually
more consequential in terms of achievement (-14.56 + (-1.71). For example, achievement
expectations may be higher in such schools.
(EDEP) Week 4: Investigating a Random Slope
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Table 6
We could decide to remove the variables in the slope model that are not significant (or we can
leave them in the model.
The variance table (Table 7) there is still considerable variability in school intercepts and the
random slope to be explained between schools.
Table 7
The last thing we might want to do is examine whether there is a relationship between the
intercept and the slope. We can obtain an unstructured covariance matrix (UN). It looks like the
following:
2
 Intercept
 IS 

.
2
 Slope
  IS

This change will add one random parameter to be estimated in the model. We can see whether
the model’s fit is improved by adding the random covariance. We can examine the change in 2LL for this model with one additional parameter estimated.
(EDEP) Week 4: Investigating a Random Slope
For the model with 15 parameters the fit indices are as follows (Table 8):
Table 8
The likelihood ratio test (i.e., the difference in model -2LL) is as follows 176,022.629 –
176,000.884 = 21.745 (1df). The required chi-square coefficient with 1 df at p = .05 is 3.84. We
can conclude that the model with unstructured covariance matrix fits the data better than the
diagonal covariance matrix.
Notice that with the random low SES slope, the two variables from the previous model (i.e.,
lowses*schcontext, lowses*schqualcomp) are both significant at p < .05 (Table 9).
Table 9
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(EDEP) Week 4: Investigating a Random Slope
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The covariance results are the following (Table 10):
Table 10
The table suggests that the covariance between the slope and intercept is negative and significant
(Wald Z = -3.513, p < .01).
We can also obtain the correlation by specifying (UNR). The fit indices and parameter estimates
are the same, but the covariance between the intercept and slope is expressed as a correlation
(instead of a covariance). We can see (Table 11) that school achievement is moderately and
negatively related (r = -.68, p < .01) to the slope describing the impact of low student SES on
achievement.
Table 11
Examine the variance accounted for by the intercepts at each level.