1
$\begingroup$

I am performing a Wilcoxon signed rank test in R, for two paired samples, where I have used the following:

wilcox.test(abs_error_gics, abs_error_sbp, alternative = "two.sided", mu=0, conf.int=T, conf.level = 0.99, paired = TRUE)

wherein I get the following output:

data:  abs_error_gics and abs_error_sbp
V = 48485000, p-value = 0.00000002249
alternative hypothesis: true location shift is not equal to 0
99 percent confidence interval:
 0.00768364 0.02082407
sample estimates:
(pseudo)median 
    0.01426058

Obviously, I can reject the null hypothesis and say that the difference in medians is not zero. However, from the following table:

enter image description here

what I want to report in my result table, is how much larger the median on average is expected to be for GICS, compared to SBP, in the pairwise difference row. However, I am under the impression that this pairwise difference median, CANNOT exceeed the simple difference of medians? i.e. the simple difference from my table is 0.9%. From the R code I posted, I used paired = TRUE, since both GICS and SBP comes from the same underlying data. Doing this, yielded a pseudo-median larger than the simple difference, which should not be possible in my opinion? However, running it again with paired = FALSE, I get a pseudo-median of 0.89% (i.e. smaller than the simple difference). Can someone explain if my thinking is correct, or?

My data can be found here:

Link to dataset

$\endgroup$

bumped to the homepage by Community yesterday

This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.

  • 3
    $\begingroup$ "I can reject the null hypothesis and say that the difference in medians is not zero" This is an error made in many books. The statistic in question is not the difference in medians but the median of pairwise averages of the pair-differences (including each pair-difference with itself). It's possible to construct examples where the sample medians are identical but the test rejects the null. $\endgroup$ – Glen_b Apr 25 at 11:31
  • $\begingroup$ I am sorry, but I lost it after ... "averages of the pair-differences". My thinking was, that the median of pairwise averages is simple; I simply create a vector which takes the average between x and y, and take the median of this vector. But what comes after that, I cannot understand. Can you help? $\endgroup$ – Philip Apr 25 at 16:35
  • $\begingroup$ You have the order of operations reversed ("The A of B of C" means do C then B then A). $\,$ Step1. Take pair-differences, creating a new set of data. $\,$ Step 2. Take values from this new set two at a time (for $i\leq j$), to create all possible pair-averages (including $i=j$) $\,$ Step 3. Calculate the median of those pair average. That's the Hodges-Lehmann one-sample estimator, applied to the pair-differences. There's a corresponding population quantity to this in the population of pair-differences. $\endgroup$ – Glen_b Apr 25 at 23:38
0
$\begingroup$

From ?wilcox.test:

Optionally (if argument conf.int is true), a nonparametric confidence interval and an estimator for the pseudomedian (one-sample case) or for the difference of the location parameters x-y is computed. (The pseudomedian of a distribution F is the median of the distribution of (u+v)/2, where u and v are independent, each with distribution F. If F is symmetric, then the pseudomedian and median coincide. See Hollander & Wolfe (1973), page 34.) Note that in the two-sample case the estimator for the difference in location parameters does not estimate the difference in medians (a common misconception) but rather the median of the difference between a sample from x and a sample from y.

$\endgroup$
  • $\begingroup$ Thank you for providing the input from the manual. I have read it previously, but am still having a hard time figuring out if it's the statistic I want to report. I want a statistic that says the median will, on average, be 1.4 percentage points (since data is percentages) larger for x compared to y. Is that what the pseudo median gives me? $\endgroup$ – Philip Apr 25 at 11:37
  • $\begingroup$ Look at posts that @Glen_b linked to in his comment $\endgroup$ – Łukasz Deryło Apr 25 at 12:11
  • $\begingroup$ I have read all the posts, but to no avail. I updated my post with additional info, where I concretized my question. $\endgroup$ – Philip Apr 25 at 17:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.