Kruskal-Wallis, assumption of non-homogeneity

Hello there,

I am trying to do a Kruskal-Wallis test, but I am finding that some of my data sets do not satisfy the assumption of homogeneity of variances. What am I supposed to do in this situation?

Let me provide some background on my analysis. I am looking at changes in shipping traffic over time. My data consists of daily distance values (distance traveled by all ships on a given day), for three different time periods. These data are not normally distributed (due to many 0 values, i.e., days with no travel), so that is why I am using non-parametric tests.

So, my data consist of 2 columns: Distance (scale variable) and Period (nominal variable). The three periods (years 2009,2010, and 2011) are coded as 1, 2, and 3 and serve as my groups.

I also am doing this 6 times over - because there are 6 shipping lanes that I am interested in looking at.

I followed these steps to test the assumption that the distributions are roughly the same: Basically, I used SPSS to rank my data, then I got the means for the respective groups, then I got the absolute differences between those, and then performed a one-way ANOVA on these absolute differences. This is supposed to show no statistical significance so that you fail to reject the null hypothesis that the variances are the same - in other words, your data satisfy the assumption of homogeneity. Then you can continue on with the Kruskal-Wallis test.

So what is one to do when this assumption is not met? Also, if anyone has any better ideas for a more appropriate statisical analysis - by all means let me know! I really appreciate it. I've been advised to use a chi square, but I am not sure I can do that since my observations (distance) are not nominal (they are numerical). Am I right to think that? Or can you use numerical data in a chi square? The advantage of using Kruskal-Wallis was that I would be able to tell exactly where the significant difference was coming from, instead of only knowing that there was a significant difference somewhere.

Ok, I think that covers it. Thank you!