A common problem is comparing two independent treatment groups in terms of some random variableYwhen there is some covariateX. Typically the comparison is made in terms ofE(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 ofYgivenX, sayM(Y|X). For thejth group, assume thatM(Y|X) = βjX +αj. Theβsand αs can be estimated with the Brown‐Mood procedure, but there are no results on how one might testH0:β1= β2orH0: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 testingH0:α1= α2, but they were too conservative, in terms of Type I errors, when testingH0: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