Editing Student's t-distribution (section)

=====Derivation=====
Suppose ''X''<sub>1</sub>, ..., ''X''<sub>''n''</sub> are [[statistical independence|independent]] realizations of the normally-distributed, random variable ''X'', which has an expected value ''μ'' and [[variance]] ''σ''<sup>2</sup>. Let

:<math>\overline{X}_n = \frac{1}{n}(X_1+\cdots+X_n)</math>

be the sample mean, and

:<math>s^2 = \frac{1}{n-1} \sum_{i=1}^n \left(X_i - \overline{X}_n\right)^2</math>

be an unbiased estimate of the variance from the sample.  It can be shown that the random variable

: <math>V = (n-1)\frac{s^2}{\sigma^2} </math>

has a chi-squared distribution with <math>\nu = n - 1</math> degrees of freedom (by [[Cochran's theorem]]).<ref>{{cite journal|authorlink1=William Gemmell Cochran | last1=Cochran |first1=W. G.|date=1934|title=The distribution of quadratic forms in a normal system, with applications to the analysis of covariance|journal=[[Mathematical Proceedings of the Cambridge Philosophical Society]]|volume=30|issue=2|pages=178–191|bibcode=1934PCPS...30..178C|doi=10.1017/S0305004100016595|s2cid=122547084 }}</ref>  It is readily shown that the quantity

:<math>Z = \left(\overline{X}_n - \mu\right) \frac{\sqrt{n}}{\sigma}</math>

is normally distributed with mean 0 and variance 1, since the sample mean <math>\overline{X}_n</math> is normally distributed with mean ''μ'' and variance ''σ''<sup>2</sup>/''n''.  Moreover, it is possible to show that these two random variables (the normally distributed one ''Z'' and the chi-squared-distributed one ''V'') are independent. Consequently{{clarify|date=November 2012}} the [[pivotal quantity]]

:<math display="inline">T \equiv \frac{Z}{\sqrt{V/\nu}} = \left(\overline{X}_n - \mu\right) \frac{\sqrt{n}}{s},</math>

which differs from ''Z'' in that the exact standard deviation ''σ'' is replaced by the sample standard error ''s'', has a Student's ''t''-distribution as defined above. Notice that the unknown population variance ''σ''<sup>2</sup> does not appear in ''T'', since it was in both the numerator and the denominator, so it canceled. Gosset intuitively obtained the probability density function stated above, with <math>\nu</math> equal to ''n''&nbsp;−&nbsp;1, and Fisher proved it in 1925.<ref name="Fisher 1925 90–104"/>

The distribution of the test statistic ''T'' depends on <math>\nu</math>, but not ''μ'' or ''σ''; the lack of dependence on ''μ'' and ''σ'' is what makes the ''t''-distribution important in both theory and practice.