Help with a random slope interpretation PreviousNext
Mplus Discussion > Multilevel Data/Complex Sample >
 Daniel posted on Tuesday, April 05, 2005 - 8:25 am
I'm not sure how to interpret these findings. I wanted to test wether beliefs about the risk of smoking had different effects on depression for different levels of daily smoking (binary 1=yes, 0=no). Here are a couple of pieces (select pieces) of code and results.

Model: slope | dep11 on risk1-risk3 risk6 risk7;
slope on daily11 gender;


Estimates S.E. Est./S.E.

DAILY11 0.275 0.126 2.188
GENDER 0.233 0.070 3.327

DEP11 17.576 4.158 4.227
SLOPE -0.667 0.280 -2.383

Residual Variances
DEP11 95.347 25.037 3.808
SLOPE 0.022 0.092 0.243

Do these results mean that the the effect of perceived risk on depression symptoms is negative, but being a daily smoker increases the slope by .275? This to me would imply that there is a different effect between daily smokers and non-daily smokers on the rate of change for a unit increase in risk perceitons (so to speak).
 BMuthen posted on Wednesday, April 06, 2005 - 3:11 am
Model: slope | dep11 on risk1-risk3 risk6 risk7;

The above statement specifies that the slopes for all of the covariates are equal. I don't think this is what you mean. Also, there is a simpler way to do this. You can use the DEFINE command to create the product of risk of smoking and levels of daily smoking and then use this interaction term as a covariate.
 Rinus Voeten posted on Monday, May 09, 2005 - 12:32 pm
The Mplus user's guide (version 3) says in connection with Example 9.1: "The residual covariance between s and y is fixed at zero as the default. This default can be overridden." (p. 212).
I wonder why this covariance is fixed; is that related with the estimation method (MLR)? How can the default be overridden? Will the estimated intercept-slope covariance be in the output when the default is overridden?
 Linda K. Muthen posted on Monday, May 09, 2005 - 1:05 pm
Defaults are chosen to reflect what is most commonly seen in practice or to facilitate model estimation. The default can be overridden by adding s WITH y; to the MODEL command. If you add this statement, that parameter will be in the results.
 Rinus Voeten posted on Tuesday, May 10, 2005 - 1:46 am
Thanks for your quick response. It worked.

One may quicker run into estimation problems when the intercept-slope covariance is left free to be estimated. It has happened to me a few times. In the multilevel regression literature, however, it is advised to never fix the intercept-slope covariance, because the size of this covariance depends on the position of the scale origin of the predictor. In the social sciences, the scale origin is for most variables arbitrary. Snijders and Bosker (1999, p. 69-70) make this argument. Do you have another opinion on this issue?
 bmuthen posted on Tuesday, May 10, 2005 - 5:44 am
I am aware of this argument and don't disagree. The default is that all random effects are uncorrelated in order to not complicate the models, but we should probably modify that to correlate with the intercepts.
 Sylvana Robbers posted on Thursday, October 01, 2009 - 10:08 am
Dear dr. Muthen,

Quick question. Do cases with fewer than 3 datapoints contribute differently to the estimation of the slope parameter than cases that have at least 3 datapoints?

 Linda K. Muthen posted on Friday, October 02, 2009 - 8:54 am
Yes because they provide less information.
 Steven John posted on Thursday, December 02, 2010 - 9:25 am

I have a question about how to interpret the estimates in a random slope model. It is not std regressions coefficients as I understand it, but unstandardized? How can I interpret the strength of the positive slope 0.635?

Model: slope | Y ON X;

Between Level

Estimate S.E. Est./S.E. P-Value


0.635 0.225 2.819 0.005
 Linda K. Muthen posted on Friday, December 03, 2010 - 8:40 am
If the results you show are from the regular output, the coefficient is the raw coefficient. The random slope in the regression of y on x varies over clusters. It has a mean and variance. When you regress the random slope on z you see that z has a significant positive relationship with s. If you ask for STANDARDIZED in the OUTPUT command, you will obtain an R-square for s which tells the percent variance is s explained by z.
 Steven John posted on Sunday, December 05, 2010 - 1:44 am

STANDARDIZED (STD, STDY, STDYX) options are not available for TYPE=RANDOM.
Request for STANDARDIZED (STD, STDY, STDYX) is ignored.

I'm currently running Mplus version 5 and is this why I don't get any STDXY output? I will get it starting with version 6?

 Linda K. Muthen posted on Sunday, December 05, 2010 - 11:09 am
With TYPE=RANDOM it is not clear how to standardize parameters given that the variance of y varies with the values of x.
 Andre Plamondon posted on Monday, March 26, 2012 - 11:35 am
I was wondering how to interpret the effect of a covariate on a random slope. Does a positive association mean that the level-1 slope is more positive at high levels of the level-2 covariate? Or does is mean that the level-1 slope is more pronounced at high levels of the level-2 covariate?
 Bengt O. Muthen posted on Monday, March 26, 2012 - 6:32 pm
It means that the random slope value increases when the level-2 covariate increases. If the random slope is positive or not depends on what its mean is and what the level-2 covariate value you're considering is.
 Andre Plamondon posted on Monday, March 26, 2012 - 7:10 pm
So if the intercept of the slope is negative and the level-2 covariates has a positive effect on the slope, that means that the level-1 relationship is negative (which makes sense in my case) and at higher levels of the level-2 covariate the slope is more positive (i.e., makes it closer to zero). Is that correct?

I guess computing a cross-level interaction term would be much simpler but I still want to be able to explain the results.
 Bengt O. Muthen posted on Monday, March 26, 2012 - 8:18 pm
This depends on the mean of the level-2 covariate. Say that it has mean zero (e.g. by centering). At (conditional on) covariate value 0 the mean of the slope is the same as its intercept and is negative given your negative intercept estimate for the random slope. But conditioning on a level-2 covariate value above is zero mean (positive value), the mean of the slope may be positive depending on whether the covariate effect is larger than the intercept. So you want to evaluate the slope value at some interesting level-2 covariate values.
 Andre Plamondon posted on Tuesday, March 27, 2012 - 9:08 am
So I can just substract the unstandardized effect from the intercept of the slope to see if it gets past zero?

Thanks for the explanation, it's much clearer now.
 Bengt O. Muthen posted on Tuesday, March 27, 2012 - 9:25 am
Yes, if by unstandardized effect you mean b*w, where w is the level-2 covariate and b its slope.
 Andre Plamondon posted on Tuesday, March 27, 2012 - 4:20 pm
Ok, so if the slope intercept is -1.2, the centered covariate's range goes from -4 to 5, and its effect on the slope is -.20, then the level-1 unstandardized slope varies from -1.2+(-4*-.20) to -1.2+(5*-.20). Is that it?
 Bengt O. Muthen posted on Tuesday, March 27, 2012 - 5:09 pm
 Jinni Su posted on Wednesday, August 22, 2012 - 10:56 am
Hi, I have multiple within-level covariates, x1, x2, and x3, and I would like to specify them as random slopes and see how they vary as a function of between-level factors B1,B2.

is the following syntax correct?


slope | Y on x1 x2 x3;
Y slope on b1 b2;
y with slope;

or should I do as follow:


slope1| y on x1;

slope2| y on x2;

slope3| y on x3;
y slope1 slope2 slope3 on b1 b2;
y with slope1 slope2 slope3;

Another question, is creating cross-level interaction terms as covariates analytically equivalent to specifying random slopes? If yes, under the model command, shall I enter the interaction terms as within or between?

 Bengt O. Muthen posted on Wednesday, August 22, 2012 - 3:14 pm
You need to take the second approach.

They are close but not equivalent because the Between level specification says that the DVs have residuals. See the u1 residual on slide 45 in our Topic 7 course handout on our web site.
 anonymous posted on Tuesday, May 14, 2013 - 11:26 am
I am attempting to conduct a complex twolevel random analysis. There are two cluster variables: TID (class) and TUTORG (tutor group). We're examining the effects of two active tutoring treatments (TRT) on performance (ponavgline). We also are controlling for pretesting on performance (pravgline). I am interested in examining the random effects of treatment (tutoring) clusters. First, is the model specified correctly below? Second, where do I find the random effects of tutoring clusters in the output?

NAMES ARE STID TID TUTORG TRT prnavgline ponavgline PRcalc POcalc prnptot ponptot
USEVARIABLES ARE TID TUTORG TRT prnavgline ponavgline;
WITHIN = prnavgline;
cluster = TID TUTORG;
missing are blank;
s1 | ponavgline ON prnavgline;
ponavgline s1 ON TRT
 Bengt O. Muthen posted on Tuesday, May 14, 2013 - 6:26 pm
It looks like you specified this correctly. You can request factor scores for both s1 and ponavgline on Between.
 anonymous posted on Thursday, May 16, 2013 - 6:21 am
Am I correct in assuming that the factor score of s1 and ponavgline will be the random effects of tutoring clusters? If not, where do I find this in the output?
 Linda K. Muthen posted on Thursday, May 16, 2013 - 8:20 am
To obtain factor scores, use the FSCORES option of the SAVEDATA command.
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