# Using Wilcoxon Ranksum text with Equal sample medians [duplicate]

We have two independent samples, with skewed distribution from two different populations x, and y. When we compute the summary for these numeric vectors, we get following output,

summary(x)
Min. 1st Qu.  Median    Mean 3rd Qu.    Max.
1.000   1.000   1.000   2.219   3.000 116.000
length(x)=25312
summary(y)
Min. 1st Qu.  Median    Mean 3rd Qu.    Max.
1.000   1.000   1.000   1.129   1.000  19.000
length(y)=49832


As seen, they have equal sample medians with this data. But the intuition is that median for population1(from which sample x is taken) should be greater than median for population2(from which sample y is taken). Though sample medians are same, will it be worthwhile running Wilcoxon Ranksum test for this example? And, can I also get help with the syntax for one-sided Wilcoxon(in R). To be more precise, when we write

wilcox.test(x,y,alternative="greater",paired = FALSE)


Does this mean that median of x is greater than median of y or the other way around? Help is appreciated.

• It is great that you posted summary statistics. Unfortunately, R neglects to report the sizes of the datasets: could you tell us what they are? – whuber Dec 30 '19 at 20:10
• Please see my edits.Size is added for two samples.thanks – jayant Dec 30 '19 at 20:14
• Those sample sizes are so large you do not need a formal test. You ought to progress immediately to a more advanced stage of analysis where you characterize the differences between the distributions. Start with a QQ plot of the two datasets. – whuber Dec 30 '19 at 20:17
• The Wilcoxon test doesn't add any information that that isn't already obvious from the summary statistics (and will be made abundantly clear with a QQ plot). Although you could apply it (using a suitable adjustment for the huge numbers of ties), why bother? – whuber Dec 30 '19 at 20:27
• A qqplot comparing the 2 distributions directly might be more informative than two separate qqnorm plots. See the last "Usage" example on the R qqnorm help page. – EdM Dec 30 '19 at 21:44

• There is a controversy around the statement that Wann-Whitney's Null hypothesis is the 2 populations [distributions] are equal. This null is more apt for Kolmogorov-Smirnov. Two perfectly identical shape - say, normal - populations with the same centre but different variances won't be distinguished by MW. MW is all about stochastic dominance vs stochastic balance. Or, in other words, it is about the (in)equality of the "location of gravity". – ttnphns Dec 30 '19 at 22:12