in a book of mine, the following is stated:

Consider a random sample $X_1,\dots, X_n \sim N(\mu,\sigma^2)$ i.i.d. with unknown meand and variance and a future observation $X_{n+1} \sim N(\mu,\sigma^2)$ being indep. from the random sample. Then the $(1-\alpha)100\%$ prediction interval is given by

\begin{align}P(L < X_{n+1} < U) = 1-\alpha\end{align},

where $L$ and $U$ are the lower and upper confidence limits. Now it can be shown that $L = \bar{x} - t_{\alpha/2}^{(n-1)}s\sqrt{1+1/n}$ and $U = \bar{x} + t_{\alpha/2}^{(n-1)}s\sqrt{1+1/n}$, where $\bar{x}$ is the observed sample mean, $s$ is the observed sample standard deviation and $t_{\alpha/2}^{(n-1)}$ is the upper $\alpha/2$ percentile from a $t$-distr. with $(n-1)$ degrees of freedom.

Now they say that $L$ and $U$ are estimators of $F^{-1}(\alpha/2)$ and $F^{-1}(1-\alpha/2)$ respectively, so they denote $L = \hat{F}^{-1}(\alpha/2)$ and $U= \hat{F}^{-1}(1-\alpha/2)$.

Now, my problem:
They say that $(1-\gamma)100\%$ condfidence intervals for $F^{-1}(\alpha/2)$ are given by

\begin{align} \hat{F}^{-1}(\alpha/2) \pm \Phi(1-\gamma/2)\sqrt{((2+\Phi(1-\alpha)^2)/2}\frac{s}{\sqrt{n}} \end{align}

and I have no idea how to show that this is true. Does anyone have an idea?


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