Wald Test to find out if a coefficient is significantly more negative (with xtlogit)

[Antonio Garzon]

Hi:

I have been asked by some reviewers to use a Wald Test to check if the coefficient of one of the variables in my model is significantly more negative than another one in the same model. My understanding of the Wald test is that it allows you to evaluate the difference between nested models. I have looked at different documents on Wald test and the "test" document in the Stata manual and haven't really found anything that could help me with the request from the reviewers. I wonder if anybody could help me with this?
Thanks
Antonio

[Clyde Schechter]

Use -lincom- to test the difference between the two coefficients.

Such a test is a test of nested models. When you do -lincom coeff_a - coeff_b- you are, in effect, testing the free model with the constrained model whereby coeff_a is equal to coeff_b. Such models are, through linear algebra, equivalent to testing a model that contains a variable, call it c, which is equal to a+b, along with either a or b, and the nested submodel which contains c alone.

[Antonio Garzon]

Thank you. That resolved my issue.

[Antonio Garzon]

Hi Clyde:
Just a quick follow up question on the interpretation of lincom. Following the below, what would be the most accurate interpretation? You mentioned that lincom in fact tests, using my example below, a model that contains a variable which is equal to SEa2 + FE3 along with SEa2, and the submodel with SEa2 + FE3 alone. So using this interpretation of lincom what does the coefficient -1.990835 indicate? Thanks

. xtlogit ceased RnDALLIANCE YEAR ROLE PERCFAIL PRIORTH1a RARE PROCOMP PROCOMP2 RnDm SEa2 
>  FE3, fe
note: multiple positive outcomes within groups encountered.
note: 769 groups (1605 obs) dropped because of all positive or
      all negative outcomes.

Iteration 0:   log likelihood = -364.64294  
Iteration 1:   log likelihood = -327.67305  
Iteration 2:   log likelihood =  -326.3469  
Iteration 3:   log likelihood =  -326.3451  
Iteration 4:   log likelihood =  -326.3451  

Conditional fixed-effects logistic regression   Number of obs      =      1145
Group variable: fid                             Number of groups   =        79

                                                Obs per group: min =         2
                                                               avg =      14.5
                                                               max =       110

                                                LR chi2(11)        =    232.62
Log likelihood  =  -326.3451                    Prob > chi2        =    0.0000

------------------------------------------------------------------------------
      ceased |      Coef.   Std. Err.      z    P>|z|     [95% Conf. Interval]
-------------+----------------------------------------------------------------
 RnDALLIANCE |  -.3522932   .2703902    -1.30   0.193    -.8822483    .1776618
        YEAR |   .2189495   .0387531     5.65   0.000     .1429948    .2949041
        ROLE |    -.18674   .2804738    -0.67   0.506    -.7364585    .3629785
    PERCFAIL |  -.0289526   .0054033    -5.36   0.000    -.0395428   -.0183623
   PRIORTH1a |     .88482   .5582126     1.59   0.113    -.2092566    1.978897
        RARE |  -.3888948   .4853113    -0.80   0.423    -1.340087    .5622978
     PROCOMP |    .024858   .0038338     6.48   0.000      .017344     .032372
    PROCOMP2 |   .0374457    .005405     6.93   0.000     .0268522    .0480392
        RnDm |   -.000147   .0000887    -1.66   0.097    -.0003207    .0000268
        SEa2 |  -2.105961   .8329737    -2.53   0.011    -3.738559   -.4733623
         FE3 |  -.1151262   .0459701    -2.50   0.012    -.2052259   -.0250264
------------------------------------------------------------------------------

. lincom SEa2-FE3

 ( 1)  [ceased]SEa2 - [ceased]FE3 = 0

------------------------------------------------------------------------------
      ceased |      Coef.   Std. Err.      z    P>|z|     [95% Conf. Interval]
-------------+----------------------------------------------------------------
         (1) |  -1.990835   .8360745    -2.38   0.017     -3.62951   -.3521587
------------------------------------------------------------------------------

[Clyde Schechter]

The interpretation is that the best estimate of the coefficient of SEA2 - the coefficient of FE3 is -1.99, with a confidence interval from -3.63 to -0.33. If you like significance tests, you would reject the null hypothesis that the coefficients are equal and would conclude that FE3's coefficient is greater (which in this regression means less negative).

[Antonio Garzon]

Thank you