Likely wrong Show more
I'm not sure, but given how I recently heard p-values explained, I think that if I know the p value required (I might have them mixed up, .9 or .1), the required outcome of a draw ($0.02), and the size of the distribution (number of seconds in a day, my unit), I can estimate this. It'd be [correct p value]*[seconds in a day]*0.02
Scholar Social is a microblogging platform for researchers, grad students, librarians, archivists, undergrads, academically inclined high schoolers, educators of all levels, journal editors, research assistants, professors, administrators—anyone involved in academia who is willing to engage with others respectfully. Read more ...