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    $\begingroup$ Are you trying to suggest that sample sizes in the thousands, with likely parameter values near $1/2$, are not large for this purpose? $\endgroup$ Commented May 3, 2016 at 1:03
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    $\begingroup$ For this case, I think you could use Dan's method but compute the p value in an exact way (binomial) and approxiamte way (normal Z>Φ−1(1−α/2)Z>Φ−1(1−α/2) and Z<Φ−1(α/2) ) to compare whether they are close enough. $\endgroup$ Commented May 6, 2016 at 4:39
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    $\begingroup$ +1 Not because sample sizes weren't large enough, but because the answer fits the title question and answers it for any sample size - therefore being useful for readers arriving here guided by title text (or Google) and having an smaller sample size in mind. $\endgroup$ Commented Jul 10, 2021 at 10:45
  • $\begingroup$ @whuber which parameters are you referring to? p is problem specific and does not depend on the sample size. It will, in general, not be 0.5. $\endgroup$ Commented Feb 12 at 12:16
  • $\begingroup$ @Mayou36 The implied model in the question requires three parameters. "Likely values near $1/2$" is what this answer seems to be implying, not what I am supposing or asserting. $\endgroup$ Commented Feb 12 at 15:47