Abstract
A common problem is comparing two independent treatment groups in terms of some random variable Y when there is some covariate X. Typically the comparison is made in terms of E(Y|X). However, highly skewed distributions occur in psychology (Micceri, 1989), and so a better method might be to compare the two groups in terms of the median of Y given X, say M(Y|X). For the jth group, assume that M(Y|X) = βjX + αj. The βs and αs can be estimated with the Brown‐Mood procedure, but there are no results on how one might test H0:β1 = β2 or H0:α1 = α2. Let and be the estimates of α and β, respectively. One of the more obvious approaches is to use a jackknife estimate of the variance of and then assume that the resulting test statistics have a standard normal distribution. This approach was found to be unsatisfactory. Some alternative procedures were considered, some of which gave good results when testing H0:α1 = α2, but they were too conservative, in terms of Type I errors, when testing H0:β1 = β2. Still another procedure was considered and found to be substantially better than all others. The new procedure is based on a modification of the Brown‐Mood procedure and a bootstrap estimate of the standard errors. Some limitations of the new method are noted.
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More From: British Journal of Mathematical and Statistical Psychology
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