# The Multilevel Model for Change (Ch 3 of ALDA) – revisited

In my previous post I talked about how to replicate the example in Chapter 3 of ALDA. It was mostly about finding the dataset and modifying the R code on the UCLA web site so it works. In this post I want to talk a little more about the statistics.

Recall the example involves two groups of infants: one assigned to a program to enhance cognitive functioning and the other acting as a control. Cognitive measurements were taken at three different time periods for both groups. Did the group assigned to the program perform differently than the control group? To answer this question the authors postulate a linear model where cognitive test results are explained by time, as follows:

$Y = \beta_{0} + \beta_{1}*time$

But the intercept and slope coefficients in that model are modeled as follows:

$\beta_{0} = \gamma_{00} + \gamma_{01}*program$
$\beta_{1} = \gamma_{10} + \gamma_{11}*program$

So we have two levels of modeling happening at the same time. The first level concerns within-person change while the second level concerns between-person differences in change. We can consolidate the two levels into one formula, like this:

$Y = \gamma_{00} + \gamma_{10}*time + \gamma_{01}*program + \gamma_{11}*time*program$

So we have an intercept and three coefficients. When we fit the model to the data, we get:

$Y = 107.84 - 21.13*time + 6.85*program + 5.27*program*time$

All of which are significant coefficients. When program = 0, our linear model is $Y = 107.84 - 21.13*time$. When program = 1, our linear model is $Y = 114.69 - 15.86*time$. The intercept in these models is interpreted as the cognitive score at the first measurement (when the infants were 12 months old). We can see that infants in the program had a higher performance at the first measurement than those not in the program: 114.69 – 107.84 = 6.85. The slope tells us the rate of decline of cognitive performance (decline?). We see the infants in the program had a slower rate of decline over time: -15.86 versus -21.13. Or put another way: -21.13 – -15.86 = -5.27, which is the coefficient of the interaction. That is, the difference in slopes between the two groups is -5.27.

Now it’s interesting to note that the model does not appear to make use of the fact that each subject contributed three waves of data. Our dataset has 309 records. But those 309 records are in 103 groups. Those 103 groups are the 103 infants. Each infant contributed three cognitive test scores over time. The dataset has a variable, ID, to indicate those groups. But ID is not in our model. Shouldn’t we make use of that information? Well, in fact we did. Have a look at the R code:

lme(cog ~ time*program, data=ch3, random = ~ time | id, method="ML",
control=list(opt = "optim"))


Notice how “id” indicates the grouping structure in the “random” argument. Time is specified as the random effect and “| id” indicates it is grouped by “id” (i.e., the 103 infants). In so many words, this allows us to capture the variability in each infant’s own change trajectory. We can think of plotting the cognitive test score for one infant over time and fitting a line to those three points. There will be some error in that line. But not as much error than if we fit a line to all infants in a group over the three times. In this latter scenario we’re not accounting for the grouping of the measurements by infant. We can actually see what happens if we don’t account for this grouping by doing an Analysis of Covariance (ANCOVA).

With ANCOVA, we’re basically doing regression with continuous and categorical variables. The usual approach to ANCOVA is to think of doing a regular ANOVA analysis but blocking on a continuous variable. For example, comparing cholesterol levels (y) between a treated group and a reference group adjusted for age (x, in years). We’re interested in the treatment effect but we want to account for the effect of age.

We can naively do ANCOVA with the Chapter 3 example from ALDA as follows:

lm(cog ~ program + time + time:program, data=ch3)


Look at the results:

Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept)   107.841      1.773  60.822  < 2e-16 ***
program         6.855      2.363   2.901  0.00399 **
time          -21.133      2.747  -7.694 1.99e-13 ***
program:time    5.271      3.660   1.440  0.15087


Now compare those to the multilevel modelling results, obtained from the call to lme() above:

                 Value Std.Error  DF   t-value p-value
(Intercept)  107.84074  2.048799 204  52.63608  0.0000
time         -21.13333  1.903664 204 -11.10140  0.0000
program        6.85466  2.730259 101   2.51063  0.0136
time:program   5.27126  2.536850 204   2.07788  0.0390


Notice the similarities? That's right, both return the same model coefficients! But compare the difference in standard errors. The most dramatic is the interaction between time and program. In the ANCOVA analysis the interaction appears to be insignificant (SE = 3.7; p = 0.15). But in the multilevel model it's significant at the 5% level (SE = 2.5; p = 0.03). We see that the ANCOVA model does not take into account the change trajectories at the individual level, and is thus not sensitive enough to detect the significant difference in rates of cognitive decline between the infants in the program and those in the control group.

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