numpy.random.standard_t¶
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numpy.random.standard_t(df, size=None)¶
- Draw samples from a standard Student’s t distribution with df degrees of freedom. - A special case of the hyperbolic distribution. As df gets large, the result resembles that of the standard normal distribution ( - standard_normal).- Parameters: - df : int or array_like of ints - Degrees of freedom, should be > 0. - size : int or tuple of ints, optional - Output shape. If the given shape is, e.g., - (m, n, k), then- m * n * ksamples are drawn. If size is- None(default), a single value is returned if- dfis a scalar. Otherwise,- np.array(df).sizesamples are drawn.- Returns: - out : ndarray or scalar - Drawn samples from the parameterized standard Student’s t distribution. - Notes - The probability density function for the t distribution is  - The t test is based on an assumption that the data come from a Normal distribution. The t test provides a way to test whether the sample mean (that is the mean calculated from the data) is a good estimate of the true mean. - The derivation of the t-distribution was first published in 1908 by William Gosset while working for the Guinness Brewery in Dublin. Due to proprietary issues, he had to publish under a pseudonym, and so he used the name Student. - References - [R272] - (1, 2) Dalgaard, Peter, “Introductory Statistics With R”, Springer, 2002. - [R273] - Wikipedia, “Student’s t-distribution” http://en.wikipedia.org/wiki/Student’s_t-distribution - Examples - From Dalgaard page 83 [R272], suppose the daily energy intake for 11 women in Kj is: - >>> intake = np.array([5260., 5470, 5640, 6180, 6390, 6515, 6805, 7515, \ ... 7515, 8230, 8770]) - Does their energy intake deviate systematically from the recommended value of 7725 kJ? - We have 10 degrees of freedom, so is the sample mean within 95% of the recommended value? - >>> s = np.random.standard_t(10, size=100000) >>> np.mean(intake) 6753.636363636364 >>> intake.std(ddof=1) 1142.1232221373727 - Calculate the t statistic, setting the ddof parameter to the unbiased value so the divisor in the standard deviation will be degrees of freedom, N-1. - >>> t = (np.mean(intake)-7725)/(intake.std(ddof=1)/np.sqrt(len(intake))) >>> import matplotlib.pyplot as plt >>> h = plt.hist(s, bins=100, normed=True) - For a one-sided t-test, how far out in the distribution does the t statistic appear? - >>> np.sum(s<t) / float(len(s)) 0.0090699999999999999 #random - So the p-value is about 0.009, which says the null hypothesis has a probability of about 99% of being true. - (Source code, png, pdf) 