Hello,

I have two interaction terms, which I would like to compare statistically. Say, some variable X1 diminishes the relationship between X2 and Y to a greater extent than the relationship between X3 and Y. All X1, X2, and X3 are standardized.

Can I use the test command to compare two moderation terms like I do below (test _b[c.x1_std#c.x2_std]=_b[c.x1_std#c.x3_std])?

Thank you.

Code:
. webuse regress, clear

. egen x1_std = std(x1)

. egen x2_std = std(x2)

. egen x3_std = std(x3)

. reg y c.x1_std##c.x2_std c.x1_std##c.x3_std, noomitted
note: x1_std omitted because of collinearity

      Source |       SS           df       MS      Number of obs   =       148
-------------+----------------------------------   F(5, 142)       =     70.48
       Model |  3483.26677         5  696.653355   Prob > F        =    0.0000
    Residual |  1403.65215       142  9.88487427   R-squared       =    0.7128
-------------+----------------------------------   Adj R-squared   =    0.7027
       Total |  4886.91892       147  33.2443464   Root MSE        =     3.144

------------------------------------------------------------------------------
           y |      Coef.   Std. Err.      t    P>|t|     [95% Conf. Interval]
-------------+----------------------------------------------------------------
      x1_std |   .6408871   .4571217     1.40   0.163     -.262756     1.54453
      x2_std |   1.956235   .4744294     4.12   0.000     1.018377    2.894092
             |
    c.x1_std#|
    c.x2_std |  -.5591356   .5141205    -1.09   0.279    -1.575455    .4571836
             |
      x3_std |  -5.296306   .4173887   -12.69   0.000    -6.121404   -4.471207
             |
    c.x1_std#|
    c.x3_std |  -1.161696   .3973641    -2.92   0.004    -1.947209   -.3761817
             |
       _cons |   20.02874   .3886223    51.54   0.000     19.26051    20.79698
------------------------------------------------------------------------------

. test _b[c.x1_std#c.x2_std]=_b[c.x1_std#c.x3_std]

 ( 1)  c.x1_std#c.x2_std - c.x1_std#c.x3_std = 0

       F(  1,   142) =    0.53
            Prob > F =    0.4658