numpy.random.standard_t¶

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
).Note
New code should use the
standard_t
method of adefault_rng()
instance instead; see randomquickstart. Parameters
 dffloat or array_like of floats
Degrees of freedom, must be > 0.
 sizeint or tuple of ints, optional
Output shape. If the given shape is, e.g.,
(m, n, k)
, thenm * n * k
samples are drawn. If size isNone
(default), a single value is returned ifdf
is a scalar. Otherwise,np.array(df).size
samples are drawn.
 Returns
 outndarray or scalar
Drawn samples from the parameterized standard Student’s t distribution.
See also
Generator.standard_t
which should be used for new code.
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 tdistribution 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
 1
Dalgaard, Peter, “Introductory Statistics With R”, Springer, 2002.
 2
Wikipedia, “Student’s tdistribution” https://en.wikipedia.org/wiki/Student’s_tdistribution
Examples
From Dalgaard page 83 [1], suppose the daily energy intake for 11 women in kilojoules (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, N1.
>>> t = (np.mean(intake)7725)/(intake.std(ddof=1)/np.sqrt(len(intake))) >>> import matplotlib.pyplot as plt >>> h = plt.hist(s, bins=100, density=True)
For a onesided ttest, how far out in the distribution does the t statistic appear?
>>> np.sum(s<t) / float(len(s)) 0.0090699999999999999 #random
So the pvalue is about 0.009, which says the null hypothesis has a probability of about 99% of being true.