This is difficult to read.
You should try -findit- using keywords like
trimmed mean, Hodges-Lehmann, etc.
Phil Ender has a program called -robmean-.
The trimean is computable after -summarize,
detail- as
0.25 * (r(p25) + r(p75)) + 0.5 * r(p50)
In general, some of these problems are
approachable as weighted means, so you
need to prepare a weight variable before
using -summarize- with weights.
Nick
[email protected]
> -----Original Message-----
> From: [email protected]
> [mailto:[email protected]]On Behalf Of Andreas
> Aschbacher
> Sent: 16 March 2004 14:34
> To: [email protected]
> Subject: Re: st: L-estimator
>
>
> sorry I have seen now that on e-mail-editor there is no such
> abridgement
> as I wrote before ! I hope it is possible for you to read and
> understand
> anyhow. andreas
>
> Dear fellows !
> >
> > I have 300 values in radioactive-measurement in one column :
> > now I want to compute :
> >
> > 1) - Tn = sum 1 to 300 [a(i).x(i)]
> 0 < alpha <
> > 0.5
> > a(i) = 0
>
> >
> > for i <= n1 or i >= n- n1 + 1
> > a(i) = (n1 + 1 -alpha*n)/((1
> -2*alpha)*n)
> > for i = n1 + 1 or i = n - n1
> > a(i) = 1/((1-2*alpha)*n)
>
> >
> > for n1 + 1 < i < n - n1
> >
>
> >
> > with : n1 = [alpha*n],n = 300
> > / in <a
> href="http://www.ntsearch.com/search.php?q=europe&v=55";>europe
> </a> we say L - estimator type I/
> >
> > 2) - Tn = sum 1 to 300 [a(i).x(i)]
> 0 < alpha <
> > 0.5
> > a(i) = (1/n)*[x(n1 +1)/x(i)]
>
> >
> > for i <= n1
> > a(i) = 1/n
>
> >
> > for n1 + 1 <= i <= n - n1
> > a(i) = (1/n)*[x(n -
> n1)/x(i)]
> >
> > for i >= n - n1 +1
> >
>
> >
> > with : n1 = [alpha*n],n = 300
> > / in <a
> href="http://www.ntsearch.com/search.php?q=europe&v=55";>europe</a>
> we say L - estimator type II/
> >
> > 3) - Tn = 0.25 *[x(n1 + 1) + x(n - n1)] +
> 0.5*median-value
> >
> > 0 < alpha < 0.5
> >
>
> >
> > with : n1 = [alpha*n],n = 300
> >
> > / in <a
> href="http://www.ntsearch.com/search.php?q=europe&v=55";>europe</a>
> we say trimean /
> >
> > 4) - Tn = 0.3 *[x(n1 + 1) + x(n - n1)] +
> 0.4*median-value
> >
> > 0 < alpha < 0.5
> >
> > / in <a
> href="http://www.ntsearch.com/search.php?q=europe&v=55";>europe</a>
> we say L - estimator type III/
> >
> > 5)if I create (x(i) + x(j))/2 that I get 300^2 values,I
> want to compute
> > median of this -
> > we say wilcoxon-appliance,I hope you will understand my
> terminologie
> > / in <a
> href="http://www.ntsearch.com/search.php?q=europe&v=55";>europe</a>
> we say L - estimator type IIII/
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