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This question is based on a comment by John Kruschke in his BEST paper, pages 589-590.

Kruschke, John K. "Bayesian estimation supersedes the t test." Journal of Experimental Psychology: General 142.2 (2013): 573.

Consider...a case in which there is a windfall of data, perhaps caused by miscommunication so two research assistants collect data instead of only one. That is, the researcher intended to collect $N = 8$ per group, but the miscommunication produced $N = 16$ per group. Most analysts and all statistical software would use $N = 16$ per group to compute a p value. This is inappropriate, because the space of possible $t_{null}$ values from the null hypothesis should actually be dominated by the intended sampling scheme, not by a rare accidental quirk.

Why should we take $N=8$ instead of $N=16?$ Yes, I see that Kruschke says "the space of possible $t_{null}$ values from the null hypothesis should actually be dominated by the intended sampling scheme, not by a rare accidental quirk," but why?

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Every simulated sample in the sampling distribution is supposed to be generated in the same way as the actual sample. That's the whole point of a sampling distribution: To represent what would have happened if the data were generated the same way as the actual sample but from a hypothetical world (e.g., null hypothesis). Therefore, if the actual data had a random sample size (not a fixed sample size), the randomness of the sample size should be incorporated into simulated samples. The resulting sampling distribution can be thought of as a probabilistic mixture of fixed-$N$ sampling distributions, with each $N$ weighted by the probability that it happens.

You can read a little more in an article (I wrote); e.g., "It is important to understand that the $p$ value would be different if the stopping or testing intentions were different. For example, if data collection were stopped because time ran out instead of because $N$ reached 18, then the sample size would be a random number and the sampling distribution would be different, hence the $p$ value would be different. When $N$ is a random value, the sampling distribution is a probabilistic mixture of different fixed-$N$ sampling distributions, and the resulting mixture is (in general) not the same as any one of the fixed-$N$ distributions."

I should point out that my perspective on this issue is unconventional, but I think also appropriate (otherwise I wouldn't promote it :-). For extended numerical examples of random-$N$ sampling distributions, see Ch 11 of DBDA2E, and for a rebuttal of arguments than random-$N$ sampling can be treated as if it were fixed-$N$, see p. 313 of DBDA2E.

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  • $\begingroup$ I’ve got some reading and thinking to do, but what a cool response to get! $\endgroup$
    – Dave
    Commented Oct 18, 2020 at 14:51
  • $\begingroup$ I've run some simulations about this since you posted, and the situation appears to be worse than you have described. Not only is it improper to use $N=16$ when the intern collects twice as much data, but it appears that $N=8$ also isn't so great! $\endgroup$
    – Dave
    Commented May 19, 2021 at 12:06
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    $\begingroup$ John described this very well. It's important to contrast with the Bayesian approach that needn't concern itself with often hidden or undocumented sampling schemes. Bayes is about revealing the unknown underlying parameters that generated the one dataset in hand. $\endgroup$ Commented May 19, 2021 at 12:07
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    $\begingroup$ The conventional perspective is that you ought to condition on the realized value of a random sample size if it's ancillary to the parameters of interest. See stats.stackexchange.com/q/83934/17230 $\endgroup$
    – Scortchi
    Commented Aug 19 at 14:55

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