Thread: contingency tables
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Old June 27th, 2009, 11:37 AM
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To calculate the value yourself, you have to solve the following integral (which is not too bad in this particular case):


\frac {1}{2^{4/2} \Gamma (4/2)} \int^{\infty}_{8.30972} x^{4/2 -1}e^{-x/2} \ dx = \frac {1}{4 \Gamma (2)} \int^{\infty}_{8.30972} xe^{-x/2} \ dx


= \frac {1}{4} \int^{\infty}_{8.30972} xe^{-x/2} \ dx \ (\text{since} \ \Gamma{2} = \Gamma (1+1) = 1 \Gamma(1) = 1*1 = 1 )


= \frac{\text{-}xe^{-x/2}}{2} \Big|^{\infty}_{8.30972} + \frac{1}{2}\int^{\infty}_{8.30972} e^{-x/2} \ dx \ (\text{ integration by parts} )

= \frac {\text{-}xe^{-x/2}}{2} - e^{-x/2} \Big|^{\infty}_{8.30972}

0 - \Big(\frac{\text{-}(8.30972)e^{-8.30972/2}}{2} - e^{-8.30972/2} \Big) \approx 0.08087
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