\(\frac{2(n-1)\overline{r^2}}{\chi_2^2} \leq \widehat{\sigma^2} \leq \frac{2(n-1)\overline{r^2}}{\chi_1^2}\)

I want to give confidence intervals in terms of \(\sigma\), not \(\sigma^2\).

If \([x, y]\) are the 95% confidence intervals for \(\sigma^2\) can I just take the square root and say that \([\sqrt{x}, \sqrt{y}]\) are 95% confidence intervals for \(\sigma\), or do I need to apply a correction factor for the concavity of the square root function?

If correction is required can you show me whether it applies only to the values, or does the confidence level itself require correction? (E.g., does the sqrt of 95% confidence intervals only represent 90% intervals?)