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Apparently it is not only for "p > 0 .5" because I plugged it into
Excel and checked it on .01, .05, .1, .9, .95, and .99 and got correct
z-scores but noticed they're for *population* statistics for
*normally distributed* data. You're dealing with samples so
the # of SDs away from the mean (and therefore z-scores) will vary
depending on sample size. Not to mention that the data aren't
normally distributed to begin with. You could apply Chebyshev's
theorem which is nonparametric, but it's also much more blunt. I
wouldn't worry about two digit accuracy, lol.
--- In amibroker@xxxxxxxxxxxxxxx, "palsanand" <palsanand@xxxx> wrote:
<snip>
>
> Schmeiser (1979) came up with the following simple formula for p >
> 0.5:
>
> z = {p ^ 0.135 - (1-p) ^ 0.135} / 0.1975
>
> According to a table in Shore (1982), it is accurate to two digits
at
> p = 0, 0.4, 0.8, ..., which may be good enough.
<snip>
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