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As the titles states, I would like to compare two coefficients in my multiple regression model but I'm not quite sure how.

Coefficients:
                 Estimate Std. Error t value Pr(>|t|)    
(Intercept)       68.9483    29.7439   2.318 0.024493 *  
Shots.PG          -0.5074     1.4696  -0.345 0.731334    
Shots.OT.PG        7.4992     3.1410   2.388 0.020707 *  
Dribbles.PG        0.6081     0.8121   0.749 0.457401    
Fouled.PG         -0.9856     0.8783  -1.122 0.267031    
Offsides.PG        1.0520     3.0728   0.342 0.733477    
Tackles.PG         0.2705     0.6721   0.402 0.689016    
Fouls.PG          -0.4230     0.7893  -0.536 0.594329    
Ints.PG            0.3414     0.5962   0.573 0.569451    
Shots.Allowed.PG  -3.3604     0.8063  -4.167 0.000119 ***

Above are the results I've obtained. At first glance I thought it was interesting Shots OT has double the impact of Shots Allowed but I see that their standard errors are significantly different so that worries me.

How would I go about comparing these two values?

Using linear.hypothesis() I get:

Linear hypothesis test

Hypothesis:
Shots.OT.PG  + 2 Shots.Allowed.PG = 0

Model 1: restricted model
Model 2: Points ~ Shots.PG + Shots.OT.PG + Dribbles.PG + Fouled.PG + Offsides.PG + 
    Tackles.PG + Fouls.PG + Ints.PG + Shots.Allowed.PG

  Res.Df    RSS Df Sum of Sq      F Pr(>F)
1     52 4488.5                           
2     51 4484.2  1    4.2107 0.0479 0.8277

How do I interpret this? Does this mean they are not different due to its large P Value. I am trying to find out whether or not the Shots OT has a larger effect on the Points total than the Shots Allowed PG

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Assuming md1 is lm model. You could use linear.hypothesis from car package. Test

linear.hypothesis(md1, "Shots.OT.PG = -2*Shots.Allowed.PG")

If your p-value is greater than 0.05 you could not reject the Null Hypothesis, in this case, in absolute values, Shots.OT.PG coefficient is not different from the double of Shots.Allowed.PG coefficient.

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  • $\begingroup$ Hey, I added my output for that in the body. Can you help me interpret that? $\endgroup$ – madsthaks Aug 4 '16 at 23:37
  • $\begingroup$ See the edit, you do not reject the null $\endgroup$ – Robert Aug 4 '16 at 23:43
  • $\begingroup$ But why would we double the Shots.Allowed.PG? Shouldn't we just check the difference between the actual values? $\endgroup$ – madsthaks Aug 5 '16 at 0:12
  • $\begingroup$ It depends on your hypothesis, see the OP. $\endgroup$ – Robert Aug 5 '16 at 0:18
  • $\begingroup$ Alright, and in the event we are able to reject the null hypothesis, we'll be able to see that they are in fact different. I dont know if that helps me make the inference I'm trying to make. When I look at the difference in slopes, it seems like Shots.OT.PG is twice and significant.. since we aren't able to reject the null hypothesis, does that mean they are most likely equally significant? $\endgroup$ – madsthaks Aug 5 '16 at 0:29
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The sofware you use should let you test for contrasts.

Alvernatively, if your model is $y = \beta_0 + \beta_1 X_1 + \beta_2 X2 + ... + remainder$, and you want to test $ H: \beta_1 - \beta_2 = 0 $, you can parametrized the model as $$y = \beta_0 + \beta_1 (X_1 + X_2) + (\beta_2 - \beta_1) X2 + ...remainder$$ and check the coefficient of $X_2$ in the second regression. But check if you can test contrast with the software you are using.

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  • $\begingroup$ How would you test such a model in R? $\endgroup$ – mat Apr 29 '18 at 10:13

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