2
$\begingroup$

Background information:

The investigation is under event study framework. The topic is on how the conversion to academies affects students' KS4 performance. The dataset is at school-level. Conversion before 2009 would be in the treatment group while schools converted after 2009 would be in the control group. I have school-level students' performance from year 2002 to year 2009. From year 2006 to year 2009 every year there were conversion taking place.

Therefore, in my dataset, there are two groups: the control group contains schools didn't convert to academies up to the year 2009; the treatment group is those converted to academies before (including year 2009) , but the conversion took place in different years.

Variable explanation:

year is the year when the data is observed

schstdks4_cappedpts is the average KT4 performance of a school

earlyconverters =1 for those converted to academies up to 2009; =0 otherwise

Suppose year=c is the year when the conversion took place;

afterconversion =1 if the $year\geqslant c$

Problem:

I was hoping to get DID coefficient by 

regress schstdks4_cappedpts i.year earlyconverters#afterconversion, r

What I found puzzling is:

Linear regression                               Number of obs     =        816
                                                F(9, 806)         =       7.81
                                                Prob > F          =     0.0000
                                                R-squared         =     0.0679
                                                Root MSE          =     .39251

-------------------------------------------------------------------------------------------------
                                |               Robust
            schstdks4_cappedpts |      Coef.   Std. Err.      t    P>|t|     [95% Conf. Interval]
--------------------------------+----------------------------------------------------------------
                           year |
                          2003  |  -.2799424   .2312072    -1.21   0.226    -.7337818    .1738969
                          2004  |  -.2039077   .2140511    -0.95   0.341     -.624071    .2162556
                          2005  |   -.200673    .208598    -0.96   0.336    -.6101324    .2087864
                          2006  |  -.1890817    .206715    -0.91   0.361    -.5948449    .2166816
                          2007  |  -.1416356   .2069297    -0.68   0.494    -.5478203    .2645492
                          2008  |  -.0700349   .2070475    -0.34   0.735    -.4764509     .336381
                          2009  |   -.017034   .2091049    -0.08   0.935    -.4274884    .3934205
                                |
earlyconverters#afterconversion |
                           0 1  |          0  (empty)
                           1 0  |   .1236718   .0343675     3.60   0.000     .0562113    .1911322
                           1 1  |   .1886763   .0400589     4.71   0.000     .1100443    .2673084
                                |
                          _cons |  -.3158101   .2065066    -1.53   0.127    -.7211644    .0895441

I don't understand why would I have coefficient for (earlyconverter==1)*(afterconversion=0), shouldn't this combination in theory==0?

If the model set up is correct, (earlyconverters==1#afterconversion==0) is supposed to measure?

Improve this question
$\endgroup$
1
  • 3
    $\begingroup$ In principle, this seems statistical at root. In practice, it is hard to imagine anyone wading through this who wasn't fluent in both difference-in-difference analyses and Stata [NB not STATA]. In practice, you'd be better asking this on Statalist, where many questions are asked on this territory. $\endgroup$
    – Nick Cox
    Commented Nov 20, 2017 at 14:48

1 Answer 1

Reset to default
1
$\begingroup$

This coefficient picks up any pre-existing differences between the treated group and the control that already exists prior to the treatment.

However, you don't really have a diff-in-diff specification here. You need to have

i.earlyconverters##i.afterconversion

so you include both the own effects and the DID interaction. NB the use of i. prefix in front of the categorical variables.

Improve this answer
$\endgroup$
5
  • $\begingroup$ I tried this command before...but I think this set-up would only work for cases such that the event happens at a particular point in time. For example, observations in the control group (as in earlyconverters==0) would also have a chance of having (afterconversion==1). But in this setting, this wouldn't happen because having earlyconverters==0 necessarily implies afterconversion==0... $\endgroup$
    – JoZ
    Commented Nov 20, 2017 at 18:40
  • $\begingroup$ @ChloeZhou You can have a time-varying treatment with this setup: you need a full set of time dummies plus the term I mentioned above (3 additional coefficients). The DID interaction is one for units and time periods subject to the policy and zero otherwise. See the first paragraph in Section 3 from [Wooldridge's 2007 NBER slides(nber.org/WNE/lect_10_diffindiffs.pdf) for more on this. $\endgroup$
    – dimitriy
    Commented Nov 20, 2017 at 18:53
  • $\begingroup$ I do have full set of time dummies in my regression, specified by i.year ... problem with my data is that if the observation has earlyconverters==0 , (afterconversion) would always ==0; similarly, observation with afterconversion==1 would always have earlyconverter==1; controlling for (earlyconverters) and (afterconversion) and then introduce the interactive dummies would have perfect colinearity problem.. $\endgroup$
    – JoZ
    Commented Nov 21, 2017 at 0:42
  • $\begingroup$ I think it would be better if I only use afterconversion*earlyconverter to estimate the DID.... I think to vilify this regression I only need to impose an additional assumption that treatment group before been treated is on average identical to the control group? $\endgroup$
    – JoZ
    Commented Nov 21, 2017 at 0:43
  • 1
    $\begingroup$ @ChloeZhou If I understand your data correctly, I am not sure if you can do DID with this data. $\endgroup$
    – dimitriy
    Commented Nov 21, 2017 at 0:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.