Convert Z-score (Z-value, standard score) to p-value for normal distribution in Python

问题:

How does one convert a Z-score from the Z-distribution (standard normal distribution, Gaussian distribution) to a p-value? I have yet to find the magical function in Scipy's stats module to do this, but one must be there.

回答1:

I like the survival function (upper tail probability) of the normal distribution a bit better, because the function name is more informative:

p_values = scipy.stats.norm.sf(abs(z_scores)) #one-sided  p_values = scipy.stats.norm.sf(abs(z_scores))*2 #twosided 

normal distribution "norm" is one of around 90 distributions in scipy.stats

norm.sf also calls the corresponding function in scipy.special as in gotgenes example

small advantage of survival function, sf: numerical precision should better for quantiles close to 1 than using the cdf

回答2:

I think the cumulative distribution function (cdf) is preferred to the survivor function. The survivor function is defined as 1-cdf, and may communicate improperly the assumptions the language model uses for directional percentiles. Also, the percentage point function (ppf) is the inverse of the cdf, which is very convenient.

>>> import scipy.stats as st >>> st.norm.ppf(.95) 1.6448536269514722 >>> st.norm.cdf(1.64) 0.94949741652589625 

回答3:

Aha! I found it: scipy.special.ndtr! This also appears to be under scipy.stats.stats.zprob as well (which is just a pointer to ndtr).

Specifically, given a one-dimensional numpy.array instance z_scores, one can obtain the p-values as

p_values = 1 - scipy.special.ndtr(z_scores) 

or alternatively

p_values = scipy.special.ndtr(-z_scores) 

回答4:

From formula:

import numpy as np import scipy.special as scsp def z2p(z):     """From z-score return p-value."""     return 0.5 * (1 + scsp.erf(z / np.sqrt(2))) 

文章来源: Convert Z-score (Z-value, standard score) to p-value for normal distribution in Python