Welch's t-test: Difference between revisions

Content added Content deleted

Inline

@@ Line 28: / Line 28: @@
 <math>
- \Beta(x,y) = \dfrac{\Gamma(x)\,\Gamma(y)}{\Gamma(x+y)} =  \exp((\ln(\Gamma(x)) + \ln(\Gamma(y)) - \ln(\Gamma(x+y)))
+ \Beta(x,y) = \dfrac{\Gamma(x)\,\Gamma(y)}{\Gamma(x+y)} =\exp(\ln\dfrac{\Gamma(x)\,\Gamma(y)}{\Gamma(x+y)}) = \exp((\ln(\Gamma(x)) + \ln(\Gamma(y)) - \ln(\Gamma(x+y)))
 \!</math>
 <math> p </math> can be calculated in terms of gamma functions and integrals more simply:
-<math> p=1-\frac{1}{2}\times\frac{\int_0^\frac{\nu}{t^2+\nu} r^{\frac{\nu}{2}-1}\,(1-r)^{-0.5}\,\mathrm{d}r}{\exp((\ln(\Gamma(x)) + \ln(\Gamma(y)) - \ln(\Gamma(x+y)))} </math>
+<math> p=1-\frac{1}{2}\times\frac{\int_0^\frac{\nu}{t^2+\nu} r^{\frac{\nu}{2}-1}\,(1-r)^{-0.5}\,\mathrm{d}r}{\exp((\ln(\Gamma(\frac{\nu}{2})) + \ln(\Gamma(0.5)) - \ln(\Gamma(\frac{\nu}{2}+0.5)))} </math>
 which simplifies to
-<math> p = 1-\frac{1}{2}\times\frac{\int_0^\frac{\nu}{t^2+\nu} \frac{r^{\frac{\nu}{2}-1}}{\sqrt{1-r}}\,\mathrm{d}r}{ \exp((\ln(\Gamma(x)) + \ln(\Gamma(y)) - \ln(\Gamma(x+y))) }</math>
+<math> p = 1-\frac{1}{2}\times\frac{\int_0^\frac{\nu}{t^2+\nu} \frac{r^{\frac{\nu}{2}-1}}{\sqrt{1-r}}\,\mathrm{d}r}{ \exp((\ln(\Gamma(\frac{\nu}{2})) + \ln(\Gamma(0.5)) - \ln(\Gamma(\frac{\nu}{2}+0.5))) }</math>
 =={{header|C}}==