Testing Composite Hypotheses about Fixed Effects (Ch 4 of ALDA)

I’m still on my ALDA kick here, this time posting about section 4.7 of Chapter 4. In my last post I talked deviance-based hypothesis tests in the context of model building. Recall you have to be aware of which method of estimation you used when you deploy the deviance statistic. Namely, if you used Restricted Maximum Likelihood (RML) to estimate model parameters, you can only use the deviance statistic to test hypotheses about variance components. This is an important point as many programs default to RML, such as the lme4 package in R and SAS PROC MIXED. But the deviance statistic is not the only tool for testing composite hypotheses.

That leads us to section 4.7 and the Wald statistic. The Wald statistic allows you to “test composite hypotheses about multiple effects regardless of the method of estimation used. This means that if you use restricted methods of estimation, which prevent you from using deviance-based tests to compare models with different fixed effects, you still have a means of testing composite hypotheses about sets of fixed effects.” (p. 122) Sounds good to me!

The authors give two examples, one of which I want to review in this post. As usual they don’t show you how to do the test using statistical software. Unfortunately either does the UCLA stats consulting page for ALDA. So I had to figure it out on my own.

Let’s reset the example study motivating the work in this chapter. 82 adolescents were surveyed on alcohol use. Some of the variables collected included:

  • alcuse, a rating-scale measure of alcohol use (the response)
  • age_14, age of participant centered about 14 (14 = 0, 15 = 1, 16 = 2)
  • coa, an indicator whether or not participant is a child of an alcoholic (1 = yes, 0 = no)
  • cpeer, a rating-scale measure of alcohol use among peers centered on its sample mean of 1.018
  • id, an arbitrary level to group persons

These variables are part of Model F, the model of interest in section 4.7, which aims to explain the variability in alcuse. (Models A through E precede Model F in the earlier model-building portion of the chapter.) Here’s Model F in its multilevel form:

level 1
Y_{ij} = \pi_{0i} + \pi_{1i}*age14_{ij} + \epsilon_{ij}

level 2
\pi_{0i} = \gamma_{00} + \gamma_{01}*coa + \gamma_{02}*cpeer + \zeta_{0i}
\pi_{1i} = \gamma_{10} + \gamma_{12}*cpeer + \zeta_{1i}

So this model posits that individuals have a liner trajectory over time (level 1), and that the parameters themselves of that linear trajectory differ between individuals based on coa and cpeer (level 2).

We can combine the two levels into one scary-looking composite representation of the model:
Y_{ij} = \gamma_{00} + \gamma_{01}*coa + \gamma_{02}*peer + \gamma_{10}*age14 + \gamma_{12}*peer*age14 + \zeta_{0i} + \zeta_{1i}*age14 + \epsilon_{ij}

Then we can estimate the parameters of that model in R with the following code:

alcohol1 <- read.table("http://www.ats.ucla.edu/stat/r/examples/alda/data/alcohol1_pp.txt", 
                       header=T, sep=",")
attach(alcohol1)
library(lme)
model.f1 <- lmer(alcuse ~ coa + cpeer*age_14 + (age_14 | id), alcohol1, REML = FALSE)
summary(model.f1)

And now we are ready to test composite hypotheses about this model. The first example in the book asks whether the average child of non-alcoholic parents - with an average value of peer - drinks no alcohol at age 14 (intercept = 0) and remains abstinent over time (slope = 0). So we set coa and cpeer both equal to 0 in our composite model and we're left with this:

Y_{ij} = \gamma_{00} + \gamma_{10}*age14 + \zeta_{0i} + \zeta_{1i}*age14 + \epsilon_{ij}

Thus our question is essentially asking if the slope and intercept in this model are 0. Or to state it formally, our composite null hypothesis is as follows:

H_{0}: \gamma_{00} = 0 \: and \: \gamma_{10} = 0

Now to carry out this test, we need to express this hypothesis as a general linear hypothesis in matrix notation. First let's restate the hypothesis using our fixed effects:

1\gamma_{00} + 0\gamma_{01} + 0\gamma_{02} + 0\gamma_{10} + 0\gamma_{12} = 0
0\gamma_{00} + 0\gamma_{01} + 0\gamma_{02} + 1\gamma_{10} + 0\gamma_{12} = 0

We have weighted our coefficients so that the only two we're interested in are viable. Now we create a matrix of the weights. This is easier and faster to show in R than trying to use LaTeX, which I'm not even sure I can pull off with the Word Press plugin I'm using.

C <- matrix(c(1,0,0,0,0,0,0,0,1,0), nrow=2, byrow=TRUE)
C
     [,1] [,2] [,3] [,4] [,5]
[1,]    1    0    0    0    0
[2,]    0    0    0    1    0

Now using the matrix we just created we can conduct a linear hypothesis test using the linearHypothesis function available in the car package, like so:

library(car)
linearHypothesis(model.f1, C)

This returns a Wald statistic of 51.03 on 2 degrees of freedom, almost matching the book which reports 51.01. The p-value is practically 0, which means we reject this composite hypothesis.

Now it's nice to know there's an R function that will calculate the Wald statistic, but what is it? How can we find out? The following code reveals the source of linearHypothesis:

print(getAnywhere(linearHypothesis.lme))

In it we see the following calculation:

SSH <- as.vector(t(L %*% b - rhs) %*% solve(L %*% V %*% t(L)) %*% 
        (L %*% b - rhs))

That's it. So we have some matrix multiplication happening. In this calculation L is our hypothesis matrix, b is our fixed effects, rhs means "right hand side" (which in our example is 0) and V is the variance-covariance matrix of the model parameters.

If we wanted to calculate the Wald statistic by hand, we could do the following:

# extract the model coefficients
b <- matrix(summary(model.f1)@coefs[,1],nrow=5)

# create the "right-hand side"
q <- matrix(c(0,0),nrow=2)

# extract the variance-covariance matrix
V <- vcov(model.f1)

# calculate the Wald statistic
W <- t(C%*%b - q) %*% solve(C%*% V %*%t(C)) %*% (C%*%b - q)

To calculate the p-value, use the pchisq function:

pchisq(as.numeric(W),2,lower.tail=FALSE)

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