Statalist


[Date Prev][Date Next][Thread Prev][Thread Next][Date index][Thread index]

st: RE: benchmarking + determine the peers


From   "Martin Weiss" <martin.weiss1@gmx.de>
To   <statalist@hsphsun2.harvard.edu>
Subject   st: RE: benchmarking + determine the peers
Date   Tue, 30 Sep 2008 16:36:39 +0200

See this FAQ by NJC:
http://www.stata.com/support/faqs/data/members.html


HTH
Martin


-----Original Message-----
From: owner-statalist@hsphsun2.harvard.edu
[mailto:owner-statalist@hsphsun2.harvard.edu] On Behalf Of Bastian Steingros
Sent: Tuesday, September 30, 2008 4:33 PM
To: statalist@hsphsun2.harvard.edu
Subject: st: benchmarking + determine the peers

Dear Stata users,

I am searching a function in stata to program peer groups.
In my benchmarking study about American industries I want to compare
companies by its sales, earnings etc. 
I use the three digit industry code to classify the industries (sic-codes).
The peer group must be within the same industry.
Here, the purpose is to compute the mean or median of the sales and to see
if the respective firm's performance  is below or above the mean or the
median.

But, the peer group of firm X contains all the companies with the same digit
code but *not* firm X (otherwise I would compare X with itself)

To conclude:

year   sic-code firm  sales mean median
1990   333         X      10     ?         ?
1990   333         Y      20     ?         ?
1990   333         Z      30     ?         ?
---------------------------------------------
1990   334         A      21     ?         ?
1990   334         B       5      ?        ?
---------------------------------------------
1991   etc. ...

so, in 1990
the peer group mean/median for X must contain (20+30)  but not 10
the peer group mean/median for Y must contain (10+30)  but not 20
the peer group mean/median for Z must contain (10+20)  but not 30
and so on...

is there a way to program is stata something like:

by year sic-code, sort: egen median/mean *without the sales of the firm X /
firm Y etc.*

(Please note, it is basically more important for me to obtain the *median*
than the mean peer group due to the outlier problem of the mean function.)

So, please help me find a solution to the median peer group first.

Many thanks in advance!

Bastian





-- 
Psssst! Schon vom neuen GMX MultiMessenger gehört? Der kann`s mit allen:
http://www.gmx.net/de/go/multimessenger
*
*   For searches and help try:
*   http://www.stata.com/help.cgi?search
*   http://www.stata.com/support/statalist/faq
*   http://www.ats.ucla.edu/stat/stata/


*
*   For searches and help try:
*   http://www.stata.com/help.cgi?search
*   http://www.stata.com/support/statalist/faq
*   http://www.ats.ucla.edu/stat/stata/



© Copyright 1996–2014 StataCorp LP   |   Terms of use   |   Privacy   |   Contact us   |   What's new   |   Site index