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

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In R there is a very useful function that helps with determining parameters for a two sided t-test in order to obtain a target statistical power.

The function is called power.prop.test.

You can call it using:

power.prop.test(p1 = .50, p2 = .75, power = .90)

And it will tell you n the sample size needed to obtain this power. This is extremely useful in deterring sample sizes for tests.

Is there a similar function in the scipy package?

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, zscore

def 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

frequentist_ab_test, import numpy as np from numpy import pi, r_ import matplotlib.pyplot as plt x, y: fitfunc(p, x) - y # Distance to the target function p0 = [-15., 0.8, 0., -1.] # Initial guess for the parameters p1, success = optimize.leastsq(errfunc, p0[:], A clever use of the cost function can allow you to fit both set of data in one fit,  22 Is there a python (scipy) function to determine parameters needed to obtain a target power? Mar 5 '13 20 What's the best way to make a d3.js visualisation layout responsive?

Some of the basic power calculations are now available in statsmodels

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)

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 

sklearn.preprocessing.PowerTransformer, Click to sign-up and also get a free PDF Ebook version of the course. 0 and a standard deviation of 1, so-called standard, normal variables. The shapiro() SciPy function will calculate the Shapiro-Wilk on a given dataset. Sounds like a power/exponential distribution, not Gaussian. Name (required). Occasionally the need to check whether or not a number is a scalar (Python (long)int, Python float, Python complex, or rank-0 array) occurs in coding. This functionality is provided in the convenient function numpy.isscalar which returns a 1 or a 0.

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.

Fitting data - SciPy Cookbook, solve for any one parameter of the power of a two sample t-test. for t-test the keywords are: effect_size, nobs1, alpha, power, ratio. exactly one needs to be None , all others need numeric values The function uses scipy.optimize for finding the value that satisfies the If this fails to find a root, fsolve is used. Discrete distribution have mostly the same basic methods as the continuous distributions. However pdf is replaced the probability mass function pmf, no estimation methods, such as fit, are available, and scale is not a valid keyword parameter. The location parameter, keyword loc can still be used to shift the distribution.

You also have:

from statsmodels.stats.power import tt_ind_solve_power

and put "None" in the value you want to obtain. For instande, to obtain the number of observations in the case of effect_size = 0.1, power = 0.8 and so on, you should put:

tt_ind_solve_power(effect_size=0.1, nobs1 = None, alpha=0.05, power=0.8, ratio=1, alternative='two-sided')

and obtain: 1570.7330663315456 as the number of observations required. Or else, to obtain the power you can attain with the other values fixed:

tt_ind_solve_power(effect_size= 0.2, nobs1 = 200, alpha=0.05, power=None, ratio=1, alternative='two-sided')

and you obtain: 0.5140816347005553

A Gentle Introduction to Normality Tests in Python, and may help with the nodes variable and un-bunch the distribution slightly. perform the Yeo-Johnson transform and automatically determine the best parameters The power transform may make use of a log() function, which does not work on models with power transforms on the haberman dataset from numpy import  Alternatively, the distribution object can be called (as a function) to fix the shape, location and scale parameters. This returns a “frozen” RV object holding the given parameters fixed. Freeze the distribution and display the frozen pdf :

statsmodels.stats.power.tt_ind_solve_power, Frequency and the Fast Fourier Transform If you want to find the secrets of the Then, we calculate the length of the snippet and plot the audio (Figure 4-2). The DFT functionality in SciPy lives in the scipy.fftpack module. time it would take the signal to travel to, bounce off, and return from a target that is distance R away:. Occasionally the need to check whether or not a number is a scalar (Python (long)int, Python float, Python complex, or rank-0 array) occurs in coding. This functionality is provided in the convenient function sp.isscalar which returns a 1 or a 0.

Imbalanced Classification with Python: Better Metrics, Balance , The bracketing values of ∆φ required by Brent's method were chosen as 10-6 and In the second scenario, the analyst should first find the column that offers the target peak capacity in the ∆φ can be determined as was shown in the first scenario. The envelope demonstrates the limit of achievable separation power by a  You can simply pass a callable as the method parameter. The callable is called as method(fun, x0, args, **kwargs, **options) where kwargs corresponds to any other parameters passed to minimize (such as callback, hess, etc.), except the options dict, which has its contents also passed as method parameters pair by pair.

4. Frequency and the Fast Fourier Transform, The data required to answer the research question, quantity of product 0] exp = target['Quantity'] import scipy.stats as stat# perform normality test return sample# create function to calculate mean of the sample Statistical power is the probability of correctly rejecting a false null Bye-bye Python. Point estimates are estimates of population parameters based on sample data. For instance, if we wanted to know the average age of registered voters in the U.S., we could take a survey of registered voters and then use the average age of the respondents as a point estimate of the average age of the population as a whole.

  • I think it would be here if there is.
  • That function is also written in pure R so by calling it without () will show the source code. The port to numpy will be straight forward if it doesn't already exist.
  • Thanks @Justin this helped in creating the below.
  • Thanks @Raufio I used the page you linked to to find the isf function below.
  • Have you considered donating this to SciPy? It's surely a useful function to have.
  • You need to sign up at GitHub, then fork their repo, put your changes in and submit a pull request. (Unfortunately, the SciPy developer documentation is a bit of a mess at present...)
  • Thanks @larsmans I'm on github so I'll fork and do just this. Cheers
  • This looks really promising. Any chance you can address @erikwestlund's answer below?
  • I assume that: (d = difference of means) and (s = difference of std). But what are p1 and p2?
  • Something is wrong with this, the answers produced vary depending on whether you use R or Python, especially when you vary the ratio. Any ideas what's wrong?
  • It's R stats and Stata versus R pwr and statsmodels. See and associated mailing list thread for details. I don't remember where SAS is in this.