Let $X_1, ..., X_n$ be a random sample from a normal distribution with an unknown $\mu$ and unknown $\sigma$. The sample mean of this sample is $\bar{X}$. The following graph shows the density curve of the sampling distribution of the sample mean:
The yellow shaded region represents where 95% of random sample means would fall and has an interval of $(\mu - L, \mu + L)$. Since there is a 95% chance that any random sample mean would fall in that interval, we can also say with 95% confidence that a given sample mean would contain the population mean within its confidence interval (CI). Finding the CI then for a sample mean simply requires centering the interval around the sample mean: $$(\mu - L + (\bar{X} - \mu), \mu + L + (\bar{X} - \mu)) = (\bar{X} - L, \bar{X} + L)$$
where $L = t_{\alpha/2} * \frac{s}{\sqrt(n)}$.
Unfortunately, this approach seems to fall apart when I try to apply it to finding the confidence interval for the sample variance. The sample variance has a sampling distribution of $\chi^2$ with $n - 1$ degrees of freedom. We can convert a sample variance to its corresponding $\chi^2$ value with the following formula:
$$\chi^2 = \frac{(n - 1)s^2}{\sigma^2}$$
Here is a plot of a chi-square distribution with $df = 4$. It can represent the distribution of the "chi-transformed" sample variances.
The red line is the mean of the distribution, the blue lines represent the $\chi^2$ values below which and above which lie 5% of the data. And the dotted green line represents the "chi-transformed" value of our theoretical sample variance.
I can no longer apply my "shift the interval" strategy because the distribution is skewed and also cannot go below 0. How do I intuit (mathematically and visually) the 90% confidence interval of the sample variance?