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st: help with R-square of xtreg,fe and areg


From   JAYESH KUMAR <[email protected]>
To   <[email protected]>
Subject   st: help with R-square of xtreg,fe and areg
Date   Fri, 17 Oct 2003 23:20:49 +0530 (IST)

Dear Users,

I have a basic question regarding the difference in xtreg,fe and areg.
Which R-square should I report in my results? R-sq: within, between or
overall obtained from xtreg,fe or shall I report R-squared or Adjusted
R-squared obtained from the areg.

I am inclosing the output from the two commands. As you can clearly see
that both results are same in terms of coeff, p-value, F-stats, etc. The
only difference is with different R-squares. I am bit puzzled here, which
R-sq should I report in my final tables. I am using Stata 7.

Any suggestion would be of help. Is there any reference, in which I can
find the differences in R-squares, in terms of interpretation, rather than
derivation?

TIA,

-Jayesh Kumar
************************************************************************
JAYESH KUMAR,
Research Scholar,
Indira Gandhi Institute Of Development Research (IGIDR),
Gen. Arun Kumar Vaidya Marg,
Santosh Nagar, Goregaon (East), Mumbai-400065, INDIA.
Tel # + 91 (22) 2840 0919/0920/0921  Extn. 591(Office) 263(Residence)
Fax # + 91 (22) 2840 2752/2026
visit: www.igidr.ac.in/~jayesh
SSRN Papers on the web at:
http://papers.ssrn.com/sol3/cf_dev/AbsByAuth.cfm?per_id=333715
************************************************************************
When I don't know what I'm doing I'm doing Research!




OUT PUT:
================================================================================


. xtreg inc age pq_a,fe

Fixed-effects (within) regression               Number of obs      =      5132
Group variable (i) : ind                        Number of groups   =      2524

R-sq:  within  = 0.0485                         Obs per group: min =         1
       between = 0.0045                                        avg =       2.0
       overall = 0.0054                                        max =         7

                                                F(2,2606)          =     66.35
corr(u_i, Xb)  = -0.8909                        Prob > F           =    0.0000

------------------------------------------------------------------------------
         inc |      Coef.   Std. Err.      t    P>|t|     [95% Conf. Interval]
-------------+----------------------------------------------------------------
         age |  -.0107904   .0009367   -11.52   0.000    -.0126272   -.0089536
        pq_a |   9.67e-06   .0000694     0.14   0.889    -.0001265    .0001458
       _cons |   .3441254   .0208627    16.49   0.000     .3032161    .3850346
-------------+----------------------------------------------------------------
     sigma_u |  .26712239
     sigma_e |  .07998241
         rho |  .91772273   (fraction of variance due to u_i)
------------------------------------------------------------------------------
F test that all u_i=0:     F(2523, 2606) =     4.40          Prob > F = 0.0000




. areg inc age pq_a,absorb(ind)

                                                       Number of obs =    5132
                                                       F(  2,  2606) =   66.35
                                                       Prob > F      =  0.0000
                                                       R-squared     =  0.8109
                                                       Adj R-squared =  0.6277
                                                       Root MSE      =  .07998

------------------------------------------------------------------------------
         inc |      Coef.   Std. Err.      t    P>|t|     [95% Conf. Interval]
-------------+----------------------------------------------------------------
         age |  -.0107904   .0009367   -11.52   0.000    -.0126272   -.0089536
        pq_a |   9.67e-06   .0000694     0.14   0.889    -.0001265    .0001458
       _cons |   .3441254   .0208627    16.49   0.000     .3032161    .3850346
-------------+----------------------------------------------------------------
        ind  |     F(2523, 2606) =      4.399   0.000        (2524 categories)




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