$\textbf{Quick question: }$
Suppose we have $\chi_{1-\alpha,29}^2=45.22$, where the Chi-squared distribution has 29 degrees of freedom and $(1-\alpha)\%$ 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!
Recall that for $X \sim \operatorname{ChiSquare}(\nu)$, that is to say $\chi_\nu^2$, we have $$f_X(x) = \frac{x^{\nu/2-1} e^{-x/2}}{2^{\nu/2} \Gamma(\nu/2)}, \quad x > 0,$$ hence $$\Pr[X \le \chi_{1-\alpha,\nu}^2] = 1-\alpha.$$ In other words, $\chi_{1-\alpha,\nu}^2$ is the $1-\alpha$ quantile of the chi-square distribution with $\nu$ degrees of freedom. Given this value and $\nu$, the computation for $\alpha$ involves computing an integral; in your case, numeric evaluation of $$\int_{x=0}^{45.22} f_X(x) \, dx \approx 0.971992,$$ or $\alpha \approx 0.028$.