how to test if one variable accounts for more variance in a regression

[Rachel Jones]

Hi all,

I am running two simple regression models with a DV (x) and IV (y&z).

So it looks like:
reg x y
reg x z

How can I compare if the variable y and z explains more of the variance? They both significant predict the IV. And one has a larger t-value. But how do I test if the t-values in the regressions are significantly different from each other?

Would an lrtest work? But I just want to test the variables, not the models, I think. Or would I put all 3 variables in one model?

Thank you!

[Carlo Lazzaro]

Rachel:
this seems to be a job for -suest-:

. use "C:\Program Files\Stata17\ado\base\a\auto.dta"
(1978 automobile data)

. regress price mpg

      Source |       SS           df       MS      Number of obs   =        74
-------------+----------------------------------   F(1, 72)        =     20.26
       Model |   139449474         1   139449474   Prob > F        =    0.0000
    Residual |   495615923        72  6883554.48   R-squared       =    0.2196
-------------+----------------------------------   Adj R-squared   =    0.2087
       Total |   635065396        73  8699525.97   Root MSE        =    2623.7

------------------------------------------------------------------------------
       price | Coefficient  Std. err.      t    P>|t|     [95% conf. interval]
-------------+----------------------------------------------------------------
         mpg |  -238.8943   53.07669    -4.50   0.000    -344.7008   -133.0879
       _cons |   11253.06   1170.813     9.61   0.000     8919.088    13587.03
------------------------------------------------------------------------------


. estimates store A

. regress price trunk

      Source |       SS           df       MS      Number of obs   =        74
-------------+----------------------------------   F(1, 72)        =      7.89
       Model |  62747229.9         1  62747229.9   Prob > F        =    0.0064
    Residual |   572318166        72  7948863.42   R-squared       =    0.0988
-------------+----------------------------------   Adj R-squared   =    0.0863
       Total |   635065396        73  8699525.97   Root MSE        =    2819.4

------------------------------------------------------------------------------
       price | Coefficient  Std. err.      t    P>|t|     [95% conf. interval]
-------------+----------------------------------------------------------------
       trunk |   216.7482   77.14554     2.81   0.006     62.96142     370.535
       _cons |   3183.504   1110.728     2.87   0.005     969.3088    5397.699
------------------------------------------------------------------------------

. estimates store B

. suest A B

Simultaneous results for A, B                               Number of obs = 74

------------------------------------------------------------------------------
             |               Robust
             | Coefficient  std. err.      z    P>|z|     [95% conf. interval]
-------------+----------------------------------------------------------------
A_mean       |
         mpg |  -238.8943   57.08197    -4.19   0.000     -350.773   -127.0157
       _cons |   11253.06   1366.933     8.23   0.000     8573.922     13932.2
-------------+----------------------------------------------------------------
A_lnvar      |
       _cons |   15.74465   .2176038    72.35   0.000     15.31815    16.17114
-------------+----------------------------------------------------------------
B_mean       |
       trunk |   216.7482   57.53296     3.77   0.000     103.9857    329.5108
       _cons |   3183.504   723.6459     4.40   0.000     1765.184    4601.824
-------------+----------------------------------------------------------------
B_lnvar      |
       _cons |   15.88854   .2315864    68.61   0.000     15.43464    16.34244
------------------------------------------------------------------------------

. test [A_mean]mpg=[B_mean]trunk

 ( 1)  [A_mean]mpg - [B_mean]trunk = 0

           chi2(  1) =   20.60
         Prob > chi2 =    0.0000

.

[Rachel Jones]

Rachel:
this seems to be a job for -suest-:

. use "C:\Program Files\Stata17\ado\base\a\auto.dta"
(1978 automobile data)

. regress price mpg

Source | SS df MS Number of obs = 74
-------------+---------------------------------- F(1, 72) = 20.26
Model | 139449474 1 139449474 Prob > F = 0.0000
Residual | 495615923 72 6883554.48 R-squared = 0.2196
-------------+---------------------------------- Adj R-squared = 0.2087
Total | 635065396 73 8699525.97 Root MSE = 2623.7

------------------------------------------------------------------------------
price | Coefficient Std. err. t P>|t| [95% conf. interval]
-------------+----------------------------------------------------------------
mpg | -238.8943 53.07669 -4.50 0.000 -344.7008 -133.0879
_cons | 11253.06 1170.813 9.61 0.000 8919.088 13587.03
------------------------------------------------------------------------------


. estimates store A

. regress price trunk

Source | SS df MS Number of obs = 74
-------------+---------------------------------- F(1, 72) = 7.89
Model | 62747229.9 1 62747229.9 Prob > F = 0.0064
Residual | 572318166 72 7948863.42 R-squared = 0.0988
-------------+---------------------------------- Adj R-squared = 0.0863
Total | 635065396 73 8699525.97 Root MSE = 2819.4

------------------------------------------------------------------------------
price | Coefficient Std. err. t P>|t| [95% conf. interval]
-------------+----------------------------------------------------------------
trunk | 216.7482 77.14554 2.81 0.006 62.96142 370.535
_cons | 3183.504 1110.728 2.87 0.005 969.3088 5397.699
------------------------------------------------------------------------------

. estimates store B

. suest A B

Simultaneous results for A, B Number of obs = 74

------------------------------------------------------------------------------
| Robust
| Coefficient std. err. z P>|z| [95% conf. interval]
-------------+----------------------------------------------------------------
A_mean |
mpg | -238.8943 57.08197 -4.19 0.000 -350.773 -127.0157
_cons | 11253.06 1366.933 8.23 0.000 8573.922 13932.2
-------------+----------------------------------------------------------------
A_lnvar |
_cons | 15.74465 .2176038 72.35 0.000 15.31815 16.17114
-------------+----------------------------------------------------------------
B_mean |
trunk | 216.7482 57.53296 3.77 0.000 103.9857 329.5108
_cons | 3183.504 723.6459 4.40 0.000 1765.184 4601.824
-------------+----------------------------------------------------------------
B_lnvar |
_cons | 15.88854 .2315864 68.61 0.000 15.43464 16.34244
------------------------------------------------------------------------------

. test [A_mean]mpg=[B_mean]trunk

( 1) [A_mean]mpg - [B_mean]trunk = 0

chi2( 1) = 20.60
Prob > chi2 = 0.0000

.

Hi Carlo thank you so much for sharing this!

I read more about the suest command and it seems super helpful. Do you have any recommended reading on the test? I just want to make sure I understand the interpretation. Is this output saying that the trunk variable explains a significantly higher amount of the model variance compared to mpg?

Thank you!

[Rachel Jones]

. test[m1_mean]X1=[m2_mean]X2

( 1) [m1_mean]X1 - [m2_mean]X2 = 0

chi2( 1) = 49.94
Prob > chi2 = 0.0000

This was my output!

So does this mean X1 is significantly higher than X2?

Also, do I need to use the lincom instead of test since my X1 and X2 are not measured the same way?

Thank you!

[Carlo Lazzaro]

Rachel:
1)yes;
2) -lincom- cannot change/fix the way your variable were measured.

[Felix Bittmann]

Looks like a job for dominance analysis.

https://psycnet.apa.org/doiLanding?d...2-989X.8.2.129

ssc install domin, replace
help domin

[Rachel Jones]

Looks like a job for dominance analysis.

https://psycnet.apa.org/doiLanding?d...2-989X.8.2.129

ssc install domin, replace
help domin

This looks great!

Is it really as easy as just doing:

domin x1 y1 y2

and then when the output says x1 dominates x2, that is the variable that explains more variance? I got the same results with the suest method above

[Felix Bittmann]

Yes, its quite simple but you should read the extensive documentation of the command and the paper I linked you so you see what these measures of "dominance" actually mean. I usually rely on the R2 statistics as these measures of dominance over other variables becomes complex when many variables are involved.