[R] R-help Digest, Vol 217, Issue 20

[Bert Gunter]

... or a George Box (I believe) said: The crucial "Declaration of
Independence."


Bert Gunter

"The trouble with having an open mind is that people keep coming along and
sticking things into it."
-- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )


On Sat, Mar 20, 2021 at 4:25 PM John Maindonald <john.maindonald using anu.edu.au>
wrote:

> No, it is not distribution free.  Independent random sampling is assumed.
> That is a non-trivial assumption, and one that is very often not true or
> not
> strictly true.
>
>
> John Maindonald             email: john.maindonald using anu.edu.au<mailto:
> john.maindonald using anu.edu.au>
>
>
> On 21/03/2021, at 00:00, r-help-request using r-project.org<mailto:
> r-help-request using r-project.org> wrote:
>
> From: Jiefei Wang <szwjf08 using gmail.com<mailto:szwjf08 using gmail.com>>
> Subject: Re: [R] about a p-value < 2.2e-16
> Date: 20 March 2021 at 04:41:33 NZDT
> To: Spencer Graves <spencer.graves using effectivedefense.org<mailto:
> spencer.graves using effectivedefense.org>>
> Cc: Bogdan Tanasa <tanasa using gmail.com<mailto:tanasa using gmail.com>>, Vivek Das <
> vd4mmind using gmail.com<mailto:vd4mmind using gmail.com>>, r-help <
> r-help using r-project.org<mailto:r-help using r-project.org>>
>
>
> Hi Spencer,
>
> Thanks for your test results, I do not know the answer as I haven't
> used wilcox.test for many years. I do not know if it is possible to compute
> the exact distribution of the Wilcoxon rank sum statistic, but I think it
> is very likely, as the document of `Wilcoxon` says:
>
> This distribution is obtained as follows. Let x and y be two random,
> independent samples of size m and n. Then the Wilcoxon rank sum statistic
> is the number of all pairs (x[i], y[j]) for which y[j] is not greater than
> x[i]. This statistic takes values between 0 and m * n, and its mean and
> variance are m * n / 2 and m * n * (m + n + 1) / 12, respectively.
>
> As a nice feature of the non-parametric statistic, it is usually
> distribution-free so you can pick any distribution you like to compute the
> same statistic. I wonder if this is the case, but I might be wrong.
>
> Cheers,
> Jiefei
>
>
>
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>
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