# Choosing right test to compare two populations

I have two populations whose size is 36 and whose values I want to compare.

Example

pop1={-1.6,-1.55,-0.819,-4.86,-5.15,-21,...};
pop2={-3.5,-5.29,-2.11,-6.01,-2.91,-14,...};


I want to understand whether or not the two populations are statistically different and, if so, to know which one has larger values. Here's what I was thinking about doing:

1. I was thinking about simply using a MannWhitney test to compare the medians, but I'm not sure there is something else that I could use.

2. I was also thinking about doing a ranked comparison

Because the two populations' values are somewhat related, I was thinking about attributing an index to each comparison as follows

      value             index
val1=-1.6,-3.5            1
val2=-1.55,-5.29          1
val3=-0.819,-2.11         1
val4=-4.86,-6.01          1
val5=-5.15,-2.91          2
val6=-21,-14              2
...


The vector v_index would then be compared with another vector v_control only with 1s or 2. I would then determine whether the median of v_index-v_control is statistically different than 0.

What do you guys think?

• Use t-Test for paired two sample for means in the MS Excel tools. – Jskakfdf May 5 '15 at 19:23
• Please tell us what you mean by "statistically different". It's not quite the same thing to compare medians as it is to compare means, for instance, and neither is remotely the same as comparing the underlying distributions. – whuber May 5 '15 at 19:52
• Well, the issue here is that I'm testing the likelyhood of my original vector (with 1 and 2) is to have all 1s, or assessing whether all values in pop1 are larger than pop2. Whether they have similar distributions or not is not very relevant for this case, hence why I was testing whether the medians were different between the original vectors. – Sosi May 5 '15 at 22:45
• Note that without some additional assumption(s), the Wilcoxon-Mann-Whitney is not strictly a comparison of medians (on the other hand, if you make the relevant assumption(s) it as easily becomes a comparison of means) – Glen_b May 6 '15 at 2:19
• The test is actually of the population median of pairwise differences between random values from the two distributions is 0. There are some discussions of it on site. To make that the same as the difference of population means (when it exists) or the difference in medians, one way that's usually quite relevant is to assume the distributions are the same shape apart from a possible location shift. Then not only is it testing that the means or medians are equal under the null, but the two-sample Hodges-Lehmann estimate of the shift also estimates the difference of population means (/medians etc) – Glen_b May 6 '15 at 10:53