Let us consider a random sample $x_1,x_2,\\dots,x_N$ drawn from a normal distribution with unknown mean $\\mu$ and unknown variance $\\sigma^2$. The best point estimates for the mean and the variance are given by the sample mean\\begin{equation}
\n\\hat\\mu = \\frac{1}{N}\\sum_{i=1}^{N} x_i
\n\\end{equation}
\nand the sample variance
\n\\begin{equation}
\n\\hat\\sigma^2 = \\frac{1}{N-1}\\sum_{i=1}^{N} (x_i-\\hat\\mu)^2\\,,
\n\\end{equation}
\nrespectively. The sample mean and the sample variance and interval estimates for the mean and the variance may be computed easily using Python.<\/p>\n
#!\/usr\/bin\/env python\n# -*- coding: utf-8 -*-\n\nfrom pylab import *\nimport scipy.stats \n\nN=1000. # sample size\ngamma=0.95 # confidence level\n\nmu=10. # true mean\nsigma=2. # true standard diviation\nx=randn(N)*sigma+mu # surrogate data\n\nmu_hat=mean(x) # sample mean\nsigma_hat=std(x, ddof=1) # sample standard deviation\n\nprint('sample mean mu_hat : %f' % mu_hat)\nprint('sample standard deviation sigma_hat : %f' % sigma_hat)\nl=scipy.stats.t.ppf( (1-gamma)\/2, N-1) # lower percentile\nu=scipy.stats.t.ppf( 1-(1-gamma)\/2, N-1) # upper percentile\nprint('confidence interval mu_hat : (%f, %f)' % \n (mu_hat+l*sigma_hat\/sqrt(N), mu_hat+u*sigma_hat\/sqrt(N)))\nl=scipy.stats.chi2.ppf( (1-gamma)\/2, N-1) # lower percentile\nu=scipy.stats.chi2.ppf( 1-(1-gamma)\/2, N-1) # upper percentile\nprint('confidence interval sigma_hat : (%f, %f)' % \n ( sqrt((N-1)\/u)*sigma_hat, sqrt((N-1)\/l)*sigma_hat))\n<\/pre>\n","protected":false},"excerpt":{"rendered":"