Home Health Quantitative Methods for HIV/AIDS Research

LR Tests for Uniformity

One-Sample Problem

Let

The full log-likelihood contribution for the ith subject is XjLi '/(Po' b, y), where

Because we are assuming a Poisson process for testing events, independent increments imply that we may treat observations within a subject as independent from one another.

Now' letting log Ц? 0' p' y) = 1 X=1 'i/(P0'b' c)'

asymptotically under H0 (McCullagh and Nelder 1989). To accommodate intervals of differing widths, this statistic may be modified to

where y° = (у0,yK)' and y0 = log(tj-tj-1) -log(ti), j = 2,...^.

Two-Sample Problem

Now suppose we would like to compare the regularity of testing between two groups. There are two possible questions of interest now:

• 1. Do the two groups differ from one another in their regularity of testing?
• 2. Does either group differ from uniformity in its testing patterns?

The first question does not make reference to any baseline definition of uniformity, and will fail to reject the null hypothesis even if both groups are substantially different from uniform in their testing patterns as long as they deviate from uniformity in roughly the same way. The second question is more restrictive, but is perhaps more important from a practical standpoint, as it will reject the null hypothesis if there is any deviation from uniformity in either or both of the groups with respect to testing patterns.

To formulate this in terms of model parameters, let y2, ..., yk be, as before, the parameters corresponding to the dummy variables for interval in the first group, with sample size m1; and let б2, ., 5k be the same parameters for the second group, with sample size m2. To address our first question, the null hypothesis is

while for the second question, the null hypothesis is

Now let Xi be the treatment indicator, that is, the covariate taking value 1 if the subject is in the treated group and 0 if control. Let

and let zi now include all covariates as well as all treatment-by-covariate interactions, with в likewise appropriately augmented; this means that в now has dimension 2p. The parameter aQ - P0 is the main effect of treatment, so that for a subject in the treatment group, aQ = a0 + log F1, where a0 is analogous to p0 in (5.2), and Fj is analogous to Gj in Equation 5.3.

To test the null hypothesis (5.12), we have

Likewise, to test (5.13), we have

The test in (5.14) is equivalent to a test of the time-by-treatment interaction in a longitudinal analysis (Fitzmaurice et al. 2004). The test in (5.15), by contrast, is a joint test of the hypothesis of no time effect in either treatment group. Note that in neither test are we concerned with the main effect for treatment. We thereby allow for the baseline rate of the process to differ between the two treatment groups, as we are only interested in the pattern of testing over time. The fact that the probability mass falling into first interval does not appear separately in the set of parameters (i.e., y2, ..., yk,

• 62,..6k) used for the test is not relevant, because both null hypotheses imply their equivalence between the two treatment groups: if, as in null hypothesis
• (5.12) , уj = 6j for all j > 2, then logGj = logFj, which implies that log^ = logG

for all j > 2 as well. Therefore, if Yj = 6j for all j > 2, Gi = F1 also. The same argument applies for null hypothesis (5.13), since this is a special case of

(5.12) with Yj = 6j = 0 for all j > 2, so G1 = F1 = 1/k.

Found a mistake? Please highlight the word and press Shift + Enter
 Related topics