4
$\begingroup$

I have an observational quasi-experimental study, where I try to estimate the effect of a "treatment" (participation in a programme) on a continuous outcome.

Participants (some two-thirds of all participants) are matched with non-participants on a few background characteristics. To estimate the effects (difference-in-difference), I use a multiple regression. With this method I get an estimated effect size of approximately 0,10, not significant (standard error 0.09). In the original data set I had more than 6000 treated and around 200,000 untreated in a "comparision group".

It was suggested that the actual "treatment" was too small (it ranges from 1 to 9 visits to a counselling provider) and it was suggested that the threshold for treatment should be moved up to "more than one visit".

This suggestion limits the number of treated quite badly (The number of matched controls also decreases). Defining treated as observations with more than one visit reduces the number of observations to 740. (Because of a skewed distribution in number of visits, and necessary qualifications in what constitutes a usable "treatment spell"). I am quite worried about the power of the renewed estimate. I would like to reject the suggestion, citing a further reduction in power because of the small effect size and the reduction in the sample. But how would I calculate how many observations I need to "keep the power" in this regression setting (as rebuttal)? Just calculating difference in group means does not control for secular drift or other confounders.

Please, any help appreciated. I hope I have explained my problem adequately $$ \ln Y_t = \alpha +\beta(Treated*After) + \gamma_1*Treated + \gamma_2*After +\gamma_3X $$ where $\beta$ is effect size and Treated=1 if treated (0 otherwise) After = 1 if observation period is after treatment (0 if before treatment) X a set of other explanatory variables

PS just redoing the estimations gives a new effect size of 0,064 , stderr 0,071 . More treatment, smaller average effect. well fancy that!

$\endgroup$
5
$\begingroup$

You could calculate the minimum detectable effect (MDE) for the average treatment effect (ATE) under the assumption that your outcome $Y$ is normally distributed. Then $$\text{MDE} = \sqrt{\frac{\widehat{\text{Var}(Y)}}{n}}\sqrt{\frac{1}{p(1-p)}}\left( q_{1-\frac{\alpha}{2}}+q_\lambda\right)$$ where $\widehat{\text{Var}(Y)}$ is the estimated variance of the outcome, $n$ is the sample size, $p$ is the fraction of program participants, $q_{1-\frac{\alpha}{2}}$ and $q_{\lambda}$ are the $1-\frac{\alpha}{2}^{th}$ and $\lambda^{th}$ quantiles of the standard normal distribution; $\alpha$ is the level and $\lambda$ is the desired power which are chosen by you. Typical choices are $\alpha = 0.05$ and $\lambda = 0.2$.

In terms of participation (number of treated relative to untreated), the MDE is the smallest when $p=0.5$, i.e. when you have the same number of treated and untreated individuals. The MDE also decreases when you increase the sample size $n$.

All of this is easily done in your case. The term $\sqrt{\frac{\widehat{\text{Var}(Y)}}{n}}$ is the standard error of your treatment parameter, $p$ you can calculate from your data, and the above quantiles of the standard normal distribution are $q_{1-\frac{\alpha}{2}} = 1.96$ and $q_{\lambda} = 0.85$ for $\alpha = 0.05$ and $\lambda = 0.2$. Then if you find that your treatment effect is smaller than the MDE, $\beta < \text{MDE}$ you are under-powered. The only solutions in this case are to increase the sample size, increase the number of treated to match the number of untreated, or accept a higher level/lower power.

$\endgroup$
4
  • $\begingroup$ Thanks for your reply - but wow! I have previously only considered power theoretically. Basically, it seems that there is a need for incredibly low variability in data to reliably detect effect sizes in difference-in-difference settings. I would see that a 10 percent increase in the outcome relative to the comparision group as a large effect - but even in a totally balanced sample I need a standard error of 0.018 to have sufficient power at beta=0.2! $\endgroup$ – Appolonia K Sep 28 '14 at 7:14
  • $\begingroup$ That's why such power calculations are often done before running experiments, for instance. Sometimes researchers already find at that stage that their study will be under-powered and that it is not worth running the experiment (or they need to increase the sample size or better balanced the treated to untreated ratio). But I'm glad this was useful for you :-) $\endgroup$ – Andy Sep 28 '14 at 8:12
  • $\begingroup$ In my line of work we mostly do ex-post evaluations, which means that the number of treated is fixed. What we can do is use a matching method that better balances matching on pre-treatment observables. That usually gives a better balance between treated and untreated. Outcomes also suffer from high variability (example above concers firm turnover), so power tends to be low. I need to read up on this and consider my recommendations for future studies... Evaluating small programs with potentially small program effects may be a fools errand. Won't stop me from getting those commissions, though $\endgroup$ – Appolonia K Sep 28 '14 at 9:25
  • 1
    $\begingroup$ @Andy could you provide a derivation for this formula? I'm trying to work out why it works but I can't quite get there. $\endgroup$ – crf Aug 19 '15 at 1:22
0
$\begingroup$

Andy's answer is correct, and I just wanted to mention that this is very easily done, along with many variants, in the free program G*Power. Best of luck.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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