from scipy import stats import scipy as sp import numpy as np s = np.array([1,2,3,4,4,4,5,5,5,5,4,4,4,6,7,8]) n, min_max, mean, var, skew, kurt = stats.describe(s) std=math.sqrt(var) #note these are sample standard deviations #and sample variance values #to get population values s.std() and s.var() will work #The location (loc) keyword specifies the mean. #The scale (scale) keyword specifies the standard deviation. # We will assume a normal distribution R = stats.norm.interval(0.05,loc=mean,scale=std) >>> R (4.33017855099411, 4.54482144900589)
Monday, December 3, 2012
Confidence Intervals in Python
Subscribe to:
Post Comments (Atom)
With that you get the 5% confidence intervals, probably not what you were looking for. Should be:
ReplyDeleteR = stats.norm.interval(0.95,loc=mean,scale=std)
This comment has been removed by the author.
DeleteActually, the above is also incorrect. Should be:
ReplyDeleteR = stats.norm.interval(0.95,loc=mean,scale=std/math.sqrt(len(s)))
Also, you could use the same approach with the t-distribution,
which is more appropriate as the number of values is small:
R = stats.t.interval(0.95,len(s)-1,loc=mean,scale=std/math.sqrt(len(s)))
Anyway, a neat way to calculate the confidence intervals...
thanx!
Delete