Testing equal medians between two variables?

[River Huang]

Dear All, I have the following on two variables

* Example generated by -dataex-. To install: ssc install dataex
clear
input float(id odiff odiff1)
 1 0  0
 2 0  0
 3 0  0
 4 0  0
 5 1  0
 6 0  0
 7 0  0
 8 0  0
 9 0  0
10 0  0
11 0  0
12 0  0
13 0  0
14 2  1
15 2  3
16 0  0
17 0  0
18 0 -1
19 0 -1
20 0  0
21 0  0
22 1 -1
23 0  0
24 1  2
25 1  0
26 3  2
27 0  0
28 2  2
29 2  1
30 1  0
end

I think I can use

ttest odiff = odiff1

to test whether the means of odiff and odiff1 are equal. However, Is there a way to test whether their medians are equal or not? Thanks.

[Carlo Lazzaro]

River:
perhaps this Stata thread can be helpful: https://www.statalist.org/forums/for...clinical-trial

[Nick Cox]

The Wilcoxon-Mann-Whitney test is often cited as doing this, but in general that's not what it does. It's better regarded (slightly non-standard view ahead) as estimating P(X > Y) with a test as a side-effect.

https://www.stata-journal.com/articl...article=st0253 is worth a look.

Somers' d is also worth a look. More generally, check out Roger Newson's papers and programs.

[Joseph Coveney]

I can use

ttest odiff = odiff1

to test whether the means of odiff and odiff1 are equal. However, Is there a way to test whether their medians are equal or not? Thanks.

I don't understand what it is you're after.

I can see the paired t-test: you're testing whether the mean difference is zero, that is, whether the mean of individual odiff - odiff1 values is zero.

But what does it mean "to test whether their medians are equal" when you have paired data?

From the way that you've stated your question, I assume that you're not interested in the paired t-test's direct analogue: whether the median difference is zero.

Or are you? If so, then

help signtest

because it specifically does that.

[River Huang]

＠Carlo Lazzaro, @Nick Cox, and @Joseph Coveney: Thank you for all your suggestions. I asked this question for my friend. I will pass the suggestions to her, and see what is exactly in her mind.