For iid data {$X_n$} with mean $\mu$, variance $\sigma^2$, finite 4th order moment $\mu_4 = E(X_1-\mu)^4$. Let the estimator of $\sigma^2$ be $\sigma_n^2 =n^{-1}\sum_{t=1}^nX_t^2-(\bar X_n)^2,$ where $\bar X_n=n^{-1}\sum_{t=1}^nX_t$. By CLT we have \begin{align} \sqrt{n}(\sigma_n^2-\sigma^2) &\stackrel{D}{\to}N(0,\mu_4-\sigma^4) \\\\ \sqrt{n}(\sigma_n^2/\sigma^2-1) &\stackrel{D}{\to}N(0,\frac{\mu_4}{\sigma^4}-1). \end{align}
let $\hat \mu_4$ be the sample version of $\mu_4$. Then from the 1st convergence result, we have that a 95% CI of $\sigma^2$ is $\sigma_n^2(1\pm1.96\sqrt{\frac{\hat\mu_4/\sigma_n^4-1}{n}})$. From the 2nd convergence result, we have that a 95% CI of $\sigma^2$ is $\sigma_n^2(1\pm1.96\sqrt{\frac{\hat\mu_4/\sigma_n^4-1}{n}})^{-1}$.
I think the second CI is the one commenly used. I'm wondering why the 1st CI isn't used?