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) = beta-(j)X + alpha-j. The beta-s and alpha-s can be estimated with the Brown-Mood procedure, but there are no results on how one might test H0:beta-1 = beta-2 or H0:alpha-1 = alpha-2. Let alpha-triple-overdot and beta-triple-overdot be the estimates of alpha and beta, respectively. One of the more obvious approaches is to use a jackknife estimate of the variance of alpha-triple-overdot and beta-triple-overdot, 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:alpha-1 = alpha-2, but they were too conservative, in terms of Type I errors, when testing H0:beta-1 = beta-2. 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.