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I have found somewhere a mention to the possibility of using dummies variables instead of the Chow test to test whether the coefficients in two linear regressions on different data sets are equal.

Could someone please direct me to a reference where I could study the details?

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  • $\begingroup$ which Chow test do you mean? there are at least four of them out there? are we talking about parameter constancy tests? $\endgroup$
    – Aksakal
    Apr 12 '14 at 14:33
  • $\begingroup$ Yes, I mean the one about parameter constancy. Below, Stat interprets correctly the meaning of my question. $\endgroup$
    – Paul Smith
    Apr 12 '14 at 17:17
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Let's first create a fake data with a break-point at 3.

> x=seq(1,5,length=100)
> y=numeric(100)
> y[1:50]=2*x[1:50]
> y[51:100]=rep(2*x[51],50)
> z=rnorm(100,0,.15)
> y=y+z
> plot(x,y)

enter image description here

Now I am gonna perform a Chow test using package strucchange in R to test if 3 is a break-point or not.

> require(strucchange)
> sctest(y ~ x, type = "Chow", point = 3)

        Chow test

data:  y ~ x
F = 3.4086, p-value = 0.03714

So based on this test, the point x=3 is a break-point. Now I will create a dummy variable dum.x and define it as 0 when $x>=3$ and 1 otherwise.

> dum.x=rep(1,100)
> dum.x[x>=3]=0

Next I fit a linear regression using the dummy variable I created with an interaction term and take the summary. So the model I am fitting here is $Y=\beta_0+\beta_1x+\beta_2dum.x+\beta_3x \times dum.x$.

> M=lm(y~x*dum.x)
> summary(M)

Call:
lm(formula = y ~ x * dum.x)

Residuals:
     Min       1Q   Median       3Q      Max 
-0.35089 -0.09929 -0.01161  0.08907  0.40424 

Coefficients:
            Estimate Std. Error t value Pr(>|t|)    
(Intercept)  6.32411    0.14728   42.94   <2e-16 ***
x           -0.07234    0.03634   -1.99   0.0494 *  
dum.x       -6.32203    0.16544  -38.21   <2e-16 ***
x:dum.x      2.06979    0.05140   40.27   <2e-16 ***
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 0.1498 on 96 degrees of freedom
Multiple R-squared:  0.9877,    Adjusted R-squared:  0.9874 
F-statistic:  2579 on 3 and 96 DF,  p-value: < 2.2e-16

Note that when $x\geq 3$, then dum.x=0 so $$Y=\beta_0+\beta_1x,$$ and when $x<3$ then dum.x=1 so $$Y=\beta_0+\beta_1x+\beta_2+\beta_3x=(\beta_0+\beta_2)+(\beta_1+\beta_3)x.$$ This means that I am actually changing both the intercept and slope by dividing my dataset and fitting above linear regression. According to the summary output the p-values for dum.x and x:dum.x are less than 0.05. So we reject $H_0:\beta_3=0$ vs. $H_1:\beta_3\ne 0$ at 5% sig. level. This means that the slope is changing at $x=3$ that confirms the Chow test we had before. Finally lets try to change our $x$ in a way that we don't have any break-point at 3.

> y=2*x+rnorm(100)
> plot(x,y) 

enter image description here

Using the Chow test again, we have:

> sctest(y ~ x, type = "Chow", point = 3)

        Chow test

data:  y ~ x
F = 2.1406, p-value = 0.1232

Therefore, x=3 is not a break-point as expected. Lets fit a linear model using our dummy variable:

> M2=lm(y~x*dum.x)
> summary(M2)

Call:
lm(formula = y ~ x * dum.x)

Residuals:
     Min       1Q   Median       3Q      Max 
-2.50938 -0.64484 -0.03025  0.67947  2.21949 

Coefficients:
            Estimate Std. Error t value Pr(>|t|)    
(Intercept)   1.1014     0.9875   1.115    0.267    
x             1.7388     0.2437   7.135 1.83e-10 ***
dum.x        -1.5082     1.1093  -1.360    0.177    
x:dum.x       0.3508     0.3446   1.018    0.311    
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 1.005 on 96 degrees of freedom
Multiple R-squared:  0.8595,    Adjusted R-squared:  0.8551 
F-statistic: 195.8 on 3 and 96 DF,  p-value: < 2.2e-16

As you can see from summary output, neither the slope nor the intercept is changing at 3 and again confirms the Chow test.

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  • $\begingroup$ I looked at the sctest documentation, it doesn't even mention Chow test :) A general comment on dummy variable tests: in econometrics the datasets are usually not that big, and even the base model struggles to have enough observations to estimate the parameters. So when you add a dummy and all its interactions, often there's too many parameters to estimate, and most of them come not significant. $\endgroup$
    – Aksakal
    Apr 12 '14 at 17:27

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