Kelvin Gregory

2022-03-02

A clarification on the level of confidence for a Chi-squared distribution
Suppose we have ${\chi }_{1-\alpha ,29}^{2}=45.22$, where the Chi-squared distribution has 29 degrees of freedom and $\left(1-\alpha \right)\mathrm{%}$ level of confidence. How then do we compute $1-\alpha$? The notation of ${\chi }^{2}$. I've found in books and online sources all come in the form of ${\chi }_{c}^{2}$, where c is the degrees of freedom.
I know this question seems trivial, but I'm still relatively new to Statistics.. Some clarification will be great!

ofisu2n3

Recall that for $X\sim \mathrm{ChiSquare}\left(\nu \right)$, that is to say ${\chi }_{\nu }^{2}$, we have
${f}_{X}\left(x\right)=\frac{{x}^{\frac{\nu }{2}-1}{e}^{-\frac{x}{2}}}{{2}^{\frac{\nu }{2}}\mathrm{\Gamma }\left(\frac{\nu }{2}\right)},\phantom{\rule{1em}{0ex}}x>0,$
hence, $Pr\left[X\le {\chi }_{1-\alpha ,\nu }^{2}\right]=1-\alpha .$
In other words, ${\chi }_{1-\alpha ,\nu }^{2}$ is the $1-\alpha$ quantile of the chi-square distribution with ν degrees of freedom. Given this value and ν, the computation for α involves computing an integral; in your case, numeric evaluation of ${\int }_{x=0}^{45.22}{f}_{X}\left(x\right),dx\approx 0.971992$,
or $a\approx 0.028$.

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