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I'm trying to use KS test to determine whether one group of data is scholastically dominates another. So I'm studying dataset regarding performance of companies, which are divided into 2 groups. Instead of comparing mean values for this two groups, I follow [1] and want to compare distributions using KS test (Table 3). They do two tests: one sided (A less then B) and two sided (equality). For that I use STATA's ksmirnov command, the problem is how to interpret the output. It return D and p but what one can conclude from these values is not clear for me. For instance, for my groups it returns:

. ksmirnov performance, by(myGroup)
Two-sample Kolmogorov-Smirnov test for equality of distribution functions

 Smaller group       D       P-value  Corrected
 ----------------------------------------------
 0:                  0.0047    0.972
 1:                 -0.1635    0.000
 Combined K-S:       0.1635    0.000      0.000

The 0 is checking hypothesis that group0 has smaller values then group1. The 1 for hypothesis that group0 has larger values then group1. But I do not understand how to interpret D and p. What is the unit of D and is it big enough to accept hypothesis (for instance, for the confidence 0.05)?

[1] http://www.etsg.org/ETSG2012/Programme/Papers/329.pdf

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    $\begingroup$ $D$ is a difference of cdfs (cdfs give probabilities). The p-value is also a (conditional) probability. Are you unfamiliar with how p-values relate to hypothesis testing? $\endgroup$ – Glen_b -Reinstate Monica Oct 12 '14 at 17:41
  • $\begingroup$ So how is D useful? I mean what kind of interpretation one can give to, for instance, output above? p-values as far as I understood are responsible for acceptance of the H. $\endgroup$ – Kirill Lykov Oct 13 '14 at 12:38
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    $\begingroup$ D is the measure of how discrepant the two ECDFs are. See the diagram here. The p-value is the probability of obtaining a test statistic result at least as extreme the one observed if the null were true. $\endgroup$ – Glen_b -Reinstate Monica Oct 13 '14 at 12:41
  • $\begingroup$ So having the table above, I conclude that there are many values in group0 which are smaller values then group1 (p-value = 0.972) yet the difference is small. And another hypothesis (group0 has larger values) might be rejected yet there are some extreme cases when the difference is huge (0.16)? $\endgroup$ – Kirill Lykov Oct 13 '14 at 13:18
  • $\begingroup$ The p-value you quote doesn't support the conclusion in your first sentence. You don't reject alternative hypotheses (as in your second sentence). $\endgroup$ – Glen_b -Reinstate Monica Oct 13 '14 at 14:39
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The Ds are the test statistics and they derive from the differences between the empirical cumulative distribution functions of the two groups. Therefore, they are the differences of probabilities. The p-values have their normal interpretation: if $pval \leq \alpha$, reject the null hypothesis; where $\alpha$ is a predetermined significance level.

Stata also gives an additional p-value for the non-directional hypothesis (Combined K-S), corrected for small samples.

Examples and details of what Stata does are in [R] ksmirnov, including the math in the Methods and formulas section.

An example of a "manual" computation of the Ds is:

clear
set more off

*------ example data -----

use http://www.stata-press.com/data/r12/ksxmpl

*----- manual computation -----

bysort group: cumul x, gen(cumd)

sort x

gen cumd1 = cumd 
replace cumd1 = cumd1[_n-1] if group != 1

gen cumd2 = cumd 
replace cumd2 = cumd2[_n-1] if group != 2
replace cumd2 = cumd2[2] in 1

gen diff = cumd1 - cumd2

summarize diff, meanonly
display  "" _n ///
         "Results are:" _n ///
         "This is D+ : `r(max)'" _n ///
         "This is D- : `r(min)'"

line cumd1 cumd2 x, sort // graph the cdfs

*----- direct computation -----

ksmirnov x, by(group)

The "manual" approach is from [1], which Stata cites in its manual.

[1] Riffenburgh, R. H. 2005. Statistics in Medicine. 2nd ed. New York: Elsevier.

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  • $\begingroup$ Am I right that having the STATA output I have in the post I can conclude that H0 is accepted because p=0.972 and H0 is rejected since p=0? $\endgroup$ – Kirill Lykov Oct 13 '14 at 12:36

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