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Just wondering if the following is correct : $$\mathbb P\left(c_1\leqslant\chi^2_{n-p}\leqslant c_2\right)=1-\alpha\\\iff \left\{c_1=\chi^2_{n-p}(1-\alpha/2)\right\}\land\left\{c_2=\chi^2_{n-p}(\alpha/2)\right\}$$

$\text{Edit}$ : A bilateral confidence interval for $\:\sigma^2\:$ at level $\:(1-\alpha)\in\:]0,1[\:$ is $$\left[\frac{\widehat\sigma^2(n-p)}{c_2};\frac{\widehat\sigma^2(n-p)}{c_1}\right].$$

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The statement $$\mathbb P\left(c_1\leqslant\chi^2_{n-p}\leqslant c_2\right)=1-\alpha\iff \left\{c_1=\chi^2_{n-p}(1-\alpha/2)\right\}\land\left\{c_2=\chi^2_{n-p}(\alpha/2)\right\}$$ does not make any sense as $\chi^2_{n-p}$ is a continuous random variable, and $c_1=\chi^2_{n-p}(1-\alpha/2)$ occurs with zero probability.

To verify that, consider $c_1 = 0$, $c_2 = \infty$, and $\alpha=0$.

Math Lover
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  • Well I misunderstood your question then. I mean the Chi-Square is rather a statistic chosen at level $:\alpha:$ with $:n-p:$ degrees of freedom, so it's a constant. –  Feb 27 '18 at 17:22
  • What do you mean by $\mathbb P\left(c_1\leqslant\chi^2_{n-p}\leqslant c_2\right)=1-\alpha$? – Math Lover Feb 27 '18 at 17:24