Is there a python (scipy) function to determine parameters needed to obtain a target power?

I've managed to replicate the function using the below formula for n and the inverse survival function norm.isf from scipy.stats

from scipy.stats import norm, zscoredef sample_power_probtest(p1, p2, power=0.8, sig=0.05):    z = norm.isf([sig/2]) #two-sided t test    zp = -1 * norm.isf([power])     d = (p1-p2)    s =2*((p1+p2) /2)*(1-((p1+p2) /2))    n = s * ((zp + z)**2) / (d**2)    return int(round(n[0]))def sample_power_difftest(d, s, power=0.8, sig=0.05):    z = norm.isf([sig/2])    zp = -1 * norm.isf([power])    n = s * ((zp + z)**2) / (d**2)    return int(round(n[0]))if __name__ == '__main__':    n = sample_power_probtest(0.1, 0.11, power=0.8, sig=0.05)    print n  #14752    n = sample_power_difftest(0.1, 0.5, power=0.8, sig=0.05)    print n  #392

Some of the basic power calculations are now available in statsmodels

http://statsmodels.sourceforge.net/devel/stats.html#power-and-sample-size-calculationshttp://jpktd.blogspot.ca/2013/03/statistical-power-in-statsmodels.html

The blog article does not yet take the latest changes to the statsmodels code into account. Also, I haven't decided yet how many wrapper functions to provide, since many power calculations just reduce to the basic distribution.

>>> import statsmodels.stats.api as sms>>> es = sms.proportion_effectsize(0.5, 0.75)>>> sms.NormalIndPower().solve_power(es, power=0.9, alpha=0.05, ratio=1)76.652940372066908

In R stats

> power.prop.test(p1 = .50, p2 = .75, power = .90)     Two-sample comparison of proportions power calculation               n = 76.7069301141077             p1 = 0.5             p2 = 0.75      sig.level = 0.05          power = 0.9    alternative = two.sided NOTE: n is number in *each* group 

using R's pwr package

> library(pwr)> h<-ES.h(0.5,0.75)> pwr.2p.test(h=h, power=0.9, sig.level=0.05)     Difference of proportion power calculation for binomial distribution (arcsine transformation)               h = 0.5235987755982985              n = 76.6529406106181      sig.level = 0.05          power = 0.9    alternative = two.sided NOTE: same sample sizes 

Matt's answer for getting the needed n (per group) is almost right, but there is a small error.

Given d (difference in means), s (standard deviation), sig (significance level, typically .05), and power (typically .80), the formula for calculating the number of observations per group is:

n= (2s^2 * ((z_(sig/2) + z_power)^2) / (d^2)

As you can see in his formula, he has

n = s * ((zp + z)**2) / (d**2)

the "s" part is wrong. a correct function that reproduces r's functionality is:

def sample_power_difftest(d, s, power=0.8, sig=0.05):    z = norm.isf([sig/2])     zp = -1 * norm.isf([power])    n = (2*(s**2)) * ((zp + z)**2) / (d**2)    return int(round(n[0]))

Hope this helps.