Monday, December 10, 2012
Checking for differences in estimates by group
Stata do file
* Imagine that you have two groups in your sample and you are wondering if both groups respond the explanatory variables in the same manner.
set obs 1000
gen x = rnormal()
gen y = 2 + 3*x + rnormal()*2 if group==0
replace y = 1 + 0*x + rnormal()*2 if group==1
* There is several ways to test if the two sets of estimators are the same.
* One method is using interaction dummies.
gen x_group = x*group
* The technique is to include an interaction term in the model for each coefficient that you would like to check if it is equal.
* Group counts as an interaction term between the constant and the group indicator.
* x_group is the interaction between the x variable and the group variable.
reg y group x_group x
* In order to interpret the coefficients in the above regression.
* First set group equal to zero.
* In that case all that is in the regression is: y = 1*b0 + x*b1 which is the estimator for the group 0.
* We can see the constant estimate is close to 2 and the coefficient on x is close to 3 which are our values.
* Now the tricky thing is to interpret the remaining coefficients.
* Because the group coefficient is effectively interacted with the constant then in order to estimate the constant in the group=1 set just add _b[group- to _b[cons].
di "Group 2 constant estimate is " _b[group]+ _b[_cons] " which we can see is close to 1"
* We do a similar procedure with the slope.
di "Group 2 slope estimate is " _b[x]+ _b[x_group] " which we can see is close to 0"
* The nice thing about this formulation is that the coefficients are already included in the estimation.
* Thus we can easily test the hypothesis that both are the same manner.
* If both groups were the same then they would have the same mean (same constants) and respond the same the the treatment x.
* An alternative way of testing each coefficient individually is:
reg y x if group == 0
local b_cong0 = _b[_cons]
local b_xg0 = _b[x]
local se_cong0 = _se[_cons]
local se_xg0 = _se[x]
reg y x if group == 1
* The t-test that they are the same is b[group=0]-b[group=1]=0
* var(b[group=0]-b[group=1])= var(b[group=0]) + var(b[group=1])
local joint_var_cons = `se_cong0'^2 + _se[_cons]^2
* And the t test is (b[group=0]-b[group=1])/(var(b[group=0]) + var(b[group=1]))^.5
di "t_constants = " (`b_cong0'- _b[_cons] )/(`joint_var_cons')^.5
local joint_var_x= `se_xg0'^2 + _se[x]^2
di "t_constants = " (`b_xg0'-_b[x])/`joint_var_x'^.5
* This formation assumes that the estimates vary independently.
* If done properly the t-stats are almost indentical between formulations.
* By the way, please fill out my survey posted on the last post.
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Thanks for the interesting post. I was wondering why you have used rnormal()*2 to generate the error term. Why not just rnormal()?ReplyDelete
It is a somewhat arbitrary choice. I just wanted the error to have a variance of 4 so that there is less explained variance. I could have specified it similarly as rnormal(1,2) however I don't like that formulation because the normal distribution sometimes has the second parameter the variance and sometimes the standard deviation. rnormal()*2 is less ambiguous.Delete
Thanks for this post!ReplyDelete