Concerning the paper: Permutation Tests for Comparing Two Populations, Journal of Mathematical Sciences & Mathematical Education, Vol. 3 No.3, pg 25, Ferry Butar Butar, Ph.D., Jae-Wan Park

There are two kinds of product inspectors: A (not-experienced) and B (experienced) and after each had inspected 200 items, their work was checked and the errors counted:

A: 30, 35, 26, 40, 36, 20, 45, 31, 33, 29, 21, 48
B: 31, 15, 25, 19, 28, 17, 19, 18, 24, 10, 20, 21

the means are mA=32.83, mB=20.58 , d0=12.25.

In order to get a more vivid picture I ordered the data

A: 20, 21, 26, 29, 30, 31, 33, 35, 36, 40, 45, 48
B: 10, 15, 17, 18, 19, 19, 20, 21, 24, 25, 28, 31

The Authors use a permutation test of the variability (based on the median of samples) and claim that "there was not difference relative to the number of mistakes", but I do not agree.

a) Based on 40´000 permutations for each sample I found that [18.90, 22.26] contains 95% of the Population differences corresponding to the 0.025, 0.975 quantiles. Therefore, d0=12.25 is outside

b) I performed www.socscisstatistics.com/tests/mannwhitney/default2.aspx U-value = 14.5 under critical 5% signifiance, two-tails, = 37 Z-score,p-value= 0.001.

Am I wrong?


1 Answer 1


This is a test of variability and the measure the paper uses is the median not the mean. The key phrase in the paper is:

Since means of the two treatments are unknown, we use the sample medians which are 32, and 19.5, respectively. Calculate the absolute value of the observed minus the median for each treatment.

They then go on to calculate The ratio of median deviance for the original observations etc. I don't think you've been calculating the proper things.

(Your comments are hard to understand and don't add to the original question, so I'd suggest deleting them or incorporating them into the original question.)


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