I have a 1-dimensional array of data:

a = np.array([1,2,3,4,4,4,5,5,5,5,4,4,4,6,7,8])

for which I want to obtain the 68% confidence interval (ie: the 1 sigma).

The first comment in this answer states that this can be achieved using scipy.stats.norm.interval from the scipy.stats.norm function, via:

from scipy import stats
import numpy as np
mean, sigma = np.mean(a), np.std(a)conf_int = stats.norm.interval(0.68, loc=mean, scale=sigma)

But a comment in this post states that the actual correct way of obtaining the confidence interval is:

conf_int = stats.norm.interval(0.68, loc=mean, scale=sigma / np.sqrt(len(a)))

that is, sigma is divided by the square-root of the sample size: np.sqrt(len(a)).

The question is: which version is the correct one?

The 68% confidence interval for a single draw from a normal distribution withmean mu and std deviation sigma is

stats.norm.interval(0.68, loc=mu, scale=sigma)

The 68% confidence interval for the mean of N draws from a normal distributionwith mean mu and std deviation sigma is

stats.norm.interval(0.68, loc=mu, scale=sigma/sqrt(N))

Intuitively, these formulas make sense, since if you hold up a jar of jelly beans and ask a large number of people to guess the number of jelly beans, each individual may be off by a lot -- the same std deviation sigma -- but the average of the guesses will do a remarkably fine job of estimating the actual number and this is reflected by the standard deviation of the mean shrinking by a factor of 1/sqrt(N).

If a single draw has variance sigma**2, then by the Bienaymé formula, the sum of N uncorrelated draws has variance N*sigma**2.

The mean is equal to the sum divided by N. When you multiply a random variable (like the sum) by a constant, the variance is multiplied by the constant squared. That is

Var(cX) = c**2 * Var(X)

So the variance of the mean equals

(variance of the sum)/N**2 = N * sigma**2 / N**2 = sigma**2 / N

and so the standard deviation of the mean (which is the square root of the variance) equals

sigma/sqrt(N).

This is the origin of the sqrt(N) in the denominator.

Here is some example code, based on Tom's code, which demonstrates the claims made above:

import numpy as np
from scipy import statsN = 10000
a = np.random.normal(0, 1, N)
mean, sigma = a.mean(), a.std(ddof=1)
conf_int_a = stats.norm.interval(0.68, loc=mean, scale=sigma)print('{:0.2%} of the single draws are in conf_int_a'.format(((a >= conf_int_a[0]) & (a < conf_int_a[1])).sum() / float(N)))M = 1000
b = np.random.normal(0, 1, (N, M)).mean(axis=1)
conf_int_b = stats.norm.interval(0.68, loc=0, scale=1 / np.sqrt(M))
print('{:0.2%} of the means are in conf_int_b'.format(((b >= conf_int_b[0]) & (b < conf_int_b[1])).sum() / float(N)))

prints

68.03% of the single draws are in conf_int_a
67.78% of the means are in conf_int_b

Beware that if you define conf_int_b with the estimates for mean and sigma based on the sample a, the mean may not fall in conf_int_b with the desired frequency.

If you take a sample from a distribution and compute the sample mean and std deviation,

mean, sigma = a.mean(), a.std()

be careful to note that there is no guarantee that these will equal the population mean and standard deviation and that we are assuming the population is normally distributed -- those are not automatic givens!

If you take a sample and want to estimate the population mean and standard deviation, you should use

mean, sigma = a.mean(), a.std(ddof=1)

since this value for sigma is the unbiased estimator for the population standard deviation.