In a paper by Rouder et al (2009, p. 234), authors indicate that their formula for obtaining a default Bayes Factor from their equation (1) on page 231 can be used for any type of t-test.

But they mention that for the two independent samples t-test case, N (total sample size) for two samples (N1, N2) may be computed as "effective sample size":

N = (N1 x N2) / (N1 + N2)


1- Conceptually, what is effective sample size (does it have a fixed formula)?

2- Is this way of computing "effective sample size" in any way related to resolving imbalance in the design (i.e., base rate differences; e.g., one group has 17 subjects, the other group has 12 subjects)?

  • $\begingroup$ I have been looking at the paper page 231. Although they have some credible references I feel more comfortable with the paper by Kass and Raftery in the Journal of the American Statistical Association Volume 90 Number 140. This is a review article on Bayes Factors. The best I can make of this is that these test are looking at the odds favoring one hypothesis test over another in a Bayesian context. I am Puzzled by effective sample size and the formula given in your question. $\endgroup$ – Michael R. Chernick Dec 11 '16 at 10:42
  • $\begingroup$ Michael, thanks for your comment. I agree that Kass & Raftery (1995) is an elegant paper. My issue here is the Rouder et al's (2009) basis for use and definition of "effective sample size" for two-samples t-test? My conceptual understanding regarding "effective sample size" is from HERE. But I'm wondering how effective sample size used by Rouder et al (2009) as I indicated in my questions above? $\endgroup$ – rnorouzian Dec 11 '16 at 20:19
  • $\begingroup$ I would like to be helpful. I looked at "HERE" but I don't have time to look at all that discussion. My feeling was that when comparing two estimators (maybe one is efficient) and each estimator has a variance. Then the effect sample size would be the number required to make both variances equal. $\endgroup$ – Michael R. Chernick Dec 11 '16 at 21:55

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