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Re: Antw: Re: st: xtfrontier


From   "Alexander Kalb" <kalb@zew.de>
To   <statalist@hsphsun2.harvard.edu>
Subject   Re: Antw: Re: st: xtfrontier
Date   Mon, 02 Jul 2007 12:43:01 +0200

Thank you Ahmed and Nicola, these explanations were very helpful!

Alex

>>> <nicola.baldini2@unibo.it> 29.06.2007 11:36 >>>
You should interpret the results like this: the most efficient units
operates at an efficiency level which is about 1/1.44=70 of its
theorerical maximum. While the numbers are economically surprising (you
can't believe that a firm survives working at one seventh of its
efficiency level --- but some niche car builders as Ferrari or
Lamborghini may not be driven to efficiency by market forces as car
builders in more competitive market segments - as Wolkswagen or Ford),
the concept is not. Firms usually have some inefficiencies (they are
called X-inefficiency). Those who appear to you as inefficiencies are
slack resources that firms usually need to face environmental
changes/external shocks/unexpected events.
If you really want to rescale, you need to divide each TE by 1.439743,
as suggested by Ahmed, not to substract .439743 (remember that TE are
exponentiated coefficients)! Then you can interpret the resulting
coefficients as units producing their output below the *most* efficient
unit.

Nicola

At 02.33 28/06/2007 -0400, "Alexander Kalb" wrote:
>When I estimated my stochastic cost frontier I thought exactly the
same.
>But when I predicted the efficiency scores (with the command predict
>efficiency, te), I got estimates lying between 1.439743 (most
efficient)
>and 6.81706 (most inefficient). The question now is, how one can
>interpret this results, since the most efficient unit does not have
the
>number one (as expected) but the number 1.439743. You simply could
>rescale the efficiency scores by substracting .439743 from all
numbers
>to get one unit with an efficiency score with 1. Then a unit with,
say
>an efficiency score of 1.20, can be interpreted as producing its
output
>with 20% above the efficient level. But I don't know if this is the
>right way.
>
>Alex 
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