![]() from scipy import statsĬonf_int_b = (0.68, loc=mean, scale=sigma / np.sqrt(len(a))) (mu,std,size) returns an array centered on mu with a standard deviation of std (in the docs, this is defined as Standard deviation (spread or “width”) of the distribution.). I tested out your methods using an array with a known confidence interval. Note that the difference between the confIntMean() and st.norm.interval() intervals is relatively small here len(a) = 16 is not too small. In : st.norm.interval(0.68, loc=np.mean(a), scale=st.sem(a)) Here the output for Gabriel's example: In : a = np.array() ![]() GraphPad's confidence interval of a mean page has "user level" info on the sample size dependency. This code (based on shasan's answer) matches their confidence intervals: import numpy as np, scipy.stats as st E.g., more than 6-fold for n=2 compared to a large n. I just checked how R and GraphPad calculate confidence intervals, and they increase the interval in case of small sample size (n). Since this value for sigma is the unbiased estimator for the population standard deviation. If you take a sample and want to estimate the population mean and standardĭeviation, you should use mean, sigma = a.mean(), a.std(ddof=1) The population is normally distributed - those are not automatic givens! Sample mean and std deviation, mean, sigma = a.mean(), a.std()īe careful to note that there is no guarantee that these willĮqual the population mean and standard deviation and that we are assuming If you take a sample from a distribution and compute the Prints 68.03% of the single draws are in conf_int_aīeware that if you define conf_int_b with the estimates for mean and sigmaīased on the sample a, the mean may not fall in conf_int_b with the desired Print(' of the single draws are in conf_int_a' ![]() Here is some example code, based on Tom's code, which demonstrates the claims made above: import numpy as npĬonf_int_a = (0.68, loc=mean, scale=sigma) This is the origin of the sqrt(N) in the denominator. So the variance of the mean equals (variance of the sum)/N**2 = N * sigma**2 / N**2 = sigma**2 / NĪnd so the standard deviation of the mean (which is the square root of the variance) equals sigma/sqrt(N). When you multiply a random variable (like the sum) by a constant, the variance is multiplied by the constant squared. ![]() The mean is equal to the sum divided by 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. 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). With mean mu and std deviation sigma is (0.68, loc=mu, scale=sigma/sqrt(N)) The 68% confidence interval for the mean of N draws from a normal distribution Mean mu and std deviation sigma is (0.68, loc=mu, scale=sigma) The 68% confidence interval for a single draw from a normal distribution with ![]()
0 Comments
Leave a Reply. |
AuthorWrite something about yourself. No need to be fancy, just an overview. ArchivesCategories |