Editing Prediction interval (section)

==== Unknown mean, unknown variance ====
Combining the above for a normal distribution <math>N(\mu,\sigma^2)</math> with both ''μ'' and ''σ''<sup>2</sup> unknown yields the following ancillary statistic:<ref>{{Harvtxt|Geisser|1993|loc=[https://books.google.com/books?id=wfdlBZ_iwZoC Example 2.2, p. 9–10]}}</ref>
:<math>\frac{X_{n+1}-\overline{X}}{s\sqrt{1+1/n}} \sim T_{n-1}</math>
This simple combination is possible because the sample mean and sample variance of the normal distribution are independent statistics; this is only true for the normal distribution, and in fact characterizes the normal distribution.

Solving for <math>X_{n+1}</math> yields the prediction distribution 
:<math>\overline{X} + s\sqrt{1+1/n} \cdot T_{n-1}</math>

The probability of <math>X_{n+1}</math> falling in a given interval is then:
:<math>\Pr\left(\overline{X}-T_{n-1,a} s\sqrt{1+(1/n)}\leq X_{n+1}   \leq\overline{X}+T_{n-1,a} s\sqrt{1+(1/n)}\,\right)=p</math>

where ''T<sub>n-1,a</sub>'' is the 100((1&nbsp;−&nbsp;''p'')/2)<sup>th</sup> [[percentile]] of [[Student's t-distribution]] with ''n''&nbsp;&minus;&nbsp;1 degrees of freedom.  Therefore, the numbers

:<math>\overline{X} \pm T_{n-1,a} s \sqrt{1+(1/n)}</math>

are the endpoints of a 100(1&nbsp;−&nbsp;''p'')% prediction interval for <math>X_{n+1}</math>.