# Why would the median be used to summarize ordinal data instead of the mode?

I am working with ordinal data (produced from Likert-types scales). I am of the understanding that results should be presented as a mode, which makes sense to me. However, when working in SPSS and utilising the Kruskal Wallis test, results are presented as a median (or mean rank).

I am therefore a little confused as to how best to present this data.

• "Present" in what sense? Do you just want to provide descriptive / summary statistics in a paper? Something else? – gung - Reinstate Monica May 9 '17 at 17:03
• Perhaps present was the wrong word. I suppose I am looking more to understand why SPSS would present these as medians when ordinal data is more usually presented as a mode. – RStev May 9 '17 at 17:55
• So is your question really just, 'why would the median be used to summarize ordinal data instead of the mode'? – gung - Reinstate Monica May 9 '17 at 19:23
• The median can be defined without too much difficulty for ordinal data because ... they can be put in order. That's the easy bit. In practice ties can reduce its utility. The more contentious bit is to argue that often the mean works well in practice for ordinal data, whatever the measurement theorists say. More in stats.stackexchange.com/questions/67551/… and in @David Lane's answer. – Nick Cox May 10 '17 at 0:48
• "How best to present the data" isn't only determined by its classification in Stevens' typology.-- your own needs, those of your audience, the particular things you're trying to convey and so on are all part of choosing. The median can be calculated on ordinal or higher on Stevens' scale, the mode on any of them. ... Why not present both? – Glen_b -Reinstate Monica May 10 '17 at 9:24

• Nonparametric statistics can certainly provide inference about population parameters (in the case of the Mann-Whitney, for example, the inference is about the Hodges-Lehmann difference -- $\text{median}(X_i-Y_j)$ for $i$, $j$ randomly selected from their populations) . If a difference in means was a suitable thing for the inference in this case, one could perform nonparametric inference about the difference in means quite easily, such as via a permutation test. What makes nonparametric tests nonparametric is that the distributional model is not finite-parametric. – Glen_b -Reinstate Monica May 10 '17 at 9:27