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Re: st: One-sided rank sum test
From
Nick Cox <[email protected]>
To
"[email protected]" <[email protected]>
Subject
Re: st: One-sided rank sum test
Date
Wed, 15 Jan 2014 12:57:32 +0000
The help for -ranksum- explains the result of -porder()- as the
probability that the variable for first group is larger than the
variable for the second group. It's not a P-value of any flavour at
all; it is just a probability as so defined.
Many people like to focus on it as (without paradox) a natural
parameter for this so-called non-parametric problem.
Roger Newson's spectacular -somersd- (SSC, SJ, etc.) can be thought of
as an extended virtuoso riff on that theme. Classically, this grows
out of Kendall's tau and other related work.
Nick
[email protected]
On 15 January 2014 12:48, Corten, R. (Rense) <[email protected]> wrote:
> Dear list,
>
> I'd like to perform a one-sided rank sum test. By default, the ranksum command returns a two-sided p-value. There is the option 'porder', but I'm not sure how to interpret the result; in any case, the additional probability returned by porder is generally *larger* than the two-sided version.
>
> For instance:
> sysuse auto
> ranksum price, by(foreign) porder
>
> Ho: price(foreign==Domestic) = price(foreign==Foreign)
> z = -1.041
> Prob > |z| = 0.2980
>
> P{price(foreign==Domestic) > price(foreign==Foreign)} = 0.423
>
> So, two questions:
> 1. How to interpret the probability generated by 'porder' (0.423, in this case)?
> 2. How to obtain the p-value for a two-sided test? Can I just divide the two-sided p-value (0.2980 in this case) by two?
>
> Best, Rense
>
>
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