# Interpretation of a Mann Whitney Wilcoxon test

In my assignment, I have a table showing chemical analyses of water in two separate years. I am trying to test if there is a significant difference in concentrations of major ions between the two years.

For example, to keep it sample. I want to compare if sodium concentrations are different in year 2016, compared to year 2017. The sample sizes are not the same (n > 30 for each) and they are not normally distributed data. Therefore, I used Mann Whitney Wilcoxon to test for differences in R.

I get a p-value, and I understand what the two possible outcomes are. Let's say that in my specific case, I get a p-value of 0.0003 at alpha = 0.05, indicating that there is a significant change. But I don't really know how to report this in-text (in my interpretation). This is what I have right now.

"Statistical analyses showed that sodium concentrations in 2016 were significantly different from those in 2017 (P < 0.05) " Is this a correct interpretation? Any help is appreciated! Thanks.

The null hypothesis of the rank sum test is:

$$H_{0}: P(X_{A} > X_{B}) = 0.5$$

In plain language that is: the probability of a random observation of $$X$$ in group $$A$$ being greater than a random observation in group $$B$$ is one half. If this is true then you are just as likely to see $$X_{A}>X_{B}$$ as to see $$X_{A}.

The alternative hypothesis is:

$$H_{A}: P(X_{A} > X_{B}) \ne 0.5$$

In plain language that is: the probability of a random observation of $$X$$ in group $$A$$ being greater than a random observation in group $$B$$ is not one half.

The implication for $$H_{A}$$ is that your are more likely to observe a greater value in either group $$A$$ or $$B$$.

If you can make two additional assumptions:

1. The distributions of $$X$$ in $$A$$ and $$B$$ both have the same shape; and
2. The variances of $$X$$ in $$A$$ and $$B$$ are equal

then you can interpret $$H_{A}$$ as finding evidence of median difference, or mean difference (evidence of location shift).

• Thanks. About point 2. Since this data is not normally distributed, I can evaluate the variance using Levene test? – Henry Mar 14 '18 at 23:01
• @Henry Yes... probably using the Brown-Forsyth variant based on the median. – Alexis Mar 14 '18 at 23:18
• @Henry You can express your thanks by clicking the up-arrow and/or the check mark next to my answer. :) – Alexis Mar 14 '18 at 23:29