Editing Student's t-distribution (section)

====Confidence intervals====
Suppose the number ''A'' is so chosen that

:<math>\ \operatorname{\mathbb P}\left\{\ -A < T < A\ \right\} = 0.9\ ,</math>

when {{mvar|T}} has a {{mvar|t}}&nbsp;distribution with {{nobr|{{math|''n'' − 1}} &thinsp;}} degrees of freedom. By symmetry, this is the same as saying that {{mvar|A}} satisfies

:<math>\ \operatorname{\mathbb P}\left\{\ T < A\ \right\} = 0.95\ ,</math>

so ''A'' is the "95th percentile" of this probability distribution, or <math>\ A = t_{(0.05,n-1)} ~.</math> Then

:<math>\ \operatorname{\mathbb P}\left\{\ -A < \frac{\ \overline{X}_n - \mu\ }{ S_n/\sqrt{n\ } } < A\ \right\} = 0.9\ ,</math>

where {{nobr|''S''{{sub|''n''}} }} is the sample standard deviation of the observed values. This is equivalent to

:<math>\ \operatorname{\mathbb P}\left\{\ \overline{X}_n - A \frac{ S_n }{\ \sqrt{n\ }\ } < \mu < \overline{X}_n + A\ \frac{ S_n }{\ \sqrt{n\ }\ }\ \right\} = 0.9.</math>

Therefore, the interval whose endpoints are

:<math>\ \overline{X}_n\ \pm A\ \frac{ S_n }{\ \sqrt{n\ }\ }\ </math>

is a 90% [[confidence interval]] for μ. Therefore, if we find the mean of a set of observations that we can reasonably expect to have a normal distribution, we can use the {{mvar|t}}&nbsp;distribution to examine whether the confidence limits on that mean include some theoretically predicted value – such as the value predicted on a [[null hypothesis]].

It is this result that is used in the [[Student's t-test|Student's {{mvar|t}}&nbsp;test]]s: since the difference between the means of samples from two normal distributions is itself distributed normally, the {{mvar|t}}&nbsp;distribution can be used to examine whether that difference can reasonably be supposed to be zero.

If the data are normally distributed, the one-sided {{nobr|{{math|(1 − ''α'')}} upper}} confidence limit (UCL) of the mean, can be calculated using the following equation:

:<math>\mathsf{UCL}_{1-\alpha} = \overline{X}_n + t_{\alpha,n-1}\ \frac{ S_n }{\ \sqrt{n\ }\ } ~.</math>
 
The resulting UCL will be the greatest average value that will occur for a given confidence interval and population size. In other words, <math>\overline{X}_n</math> being the mean of the set of observations, the probability that the mean of the distribution is inferior to {{nobr|UCL{{sub|{{math|1 − ''α''}} }} }} is equal to the confidence {{nobr|level {{math|1 − ''α''}} .}}