Finding a confidence interval for parameter of normal distribution We observe a simple random sample $(X_1,\ldots , X_n$), where the distribution of observed property $X$ is given by its density: $f(x)=\frac{1}{x\sigma \sqrt{2\pi}}e^{-\frac{(\ln x-m)^2}{2\sigma^2}}, x>0$.
Find a 80% confidence interval for $\sigma$, if we know:
Mean value of logarithms of the values from the dataset that we get deviates from its expected value no more than $\frac{\sigma}{2}$, with probability 0.98758;
If $(x_1, \ldots ,x_n)$ is our dataset (a realisation of our sample), $\sum_{i=1}^n \ln x_i=3.139797$ and $\sum_{i=1}^n (\ln x_i)^2=35.63178$.
My attempt:
It's not hard to see that, if $Y=\ln X$, then $Y$ has $\mathscr{N}(m,\sigma^2)$ distribution. Now we work with $Y$. Let
$$\overline{Y}_n=\frac{1}{n}\sum_{i=1}^n Y_i$$
$$\overline{S}_n^2=\frac{1}{n}\sum_{i=1}^n (Y_i-\overline{Y}_n)^2$$.
We know that $\frac{n\overline{S}_n^2}{\sigma^2}$ has $\chi_{n-1}^2$ distribution.
If we look for confidence interval in the form:
$$p\{U_n\leq \sigma \leq V_n\}=0.8$$, we can actually demand $$p\{\frac{n\overline{S}_n^2}{V_n^2}\leq  \frac{n\overline{S}_n^2}{\sigma^2} \leq \frac{n\overline{S}_n^2}{U_n^2} \}=0.8$$.
We can demand:
$$p\{\frac{n\overline{S}_n^2}{V_n^2}>\frac{n\overline{S}_n^2}{\sigma^2}\}=p\{\frac{n\overline{S}_n^2}{U_n^2}<\frac{n\overline{S}_n^2}{\sigma^2}\}=\frac{1-0.8}{2}=0.1$$.
This gives us (from what we know about $\chi_{n-1}^2$ distribution):
$$V_n^2=\frac{n\overline{S}_n^2}{\chi_{n-1,0.9}^2}$$
$$U_n^2=\frac{n\overline{S}_n^2}{\chi_{n-1,0.1}^2}$$.
I assume that the info given in this problem means that one confidence interval for $\frac{1}{n}\sum_{i=1}^n \ln x_i-m$, with confidence level $0.98758$ is $[-\frac{\sigma}{2}, \frac{\sigma}{2}]$. But I don't know how to move on from here.
We also know that for:
$$\overline{S}_*^2=\frac{1}{n}\sum_{i=1}^n (Y_i-m)^2$$
$\frac{n\overline{S}_*^2}{\sigma^2}$ has $\chi_n^2$ distribution. I thought maybe we should use this because of this CI that has been given to us, but I don't know how to do that.
 A: We will use the pivotal $W=\frac{\sum_{i=1}^n (Y_i-\bar Y)^2}{\sigma^2}$ which has the $\text{Chi Squared}(n-1)$ distribution.
The probability $$P(c_1<W<c_2)=.8$$ is given by any $\gamma_2>\gamma_1$ such that  $\gamma_2-\gamma_1=.8$. Let the cdf of $W$ be $G$, then
$$P(G^{-1}(\gamma_1)<W<G^{-1}(\gamma_2))=.8$$
and so the constants $c_1, c_2$ are determined. Now we simply need to solve:
$$P(c_1<\frac{\sum_{i=1}^n(Y_i-\bar Y)^2}{\sigma^2}<c_2)=.8\\
P(\frac{\sum_{i=1}^n(Y_i-\bar Y)^2}{c_2}<\sigma^2<\frac{\sum_{i=1}^n(Y_i-\bar Y)^2}{c_1})=.8\\
P(\sqrt{\frac{\sum_{i=1}^nY_i^2-\frac{(\sum_{i=1}^n Y_i)^2}n}{G^{-1}(\gamma_2)}}<\sigma<\sqrt{\frac{\sum_{i=1}^nY_i^2-\frac{(\sum_{i=1}^n Y_i)^2}n}{G^{-1}(\gamma_1)})}=.8$$
We're also given that $P(|\bar Y-m|<\frac \sigma 2)=.98758$. We can use this to find $n$, the sample size:
$$P(\frac{|\bar X-m|}\sigma\sqrt n<\frac{\sqrt n}2)=.98758\\
P(|Z|<\frac{\sqrt n}2)=.98758\\
\frac {\sqrt n}2\ge \Phi^{-1}((1+.98758)/2)\\
n\ge[2\Phi^{-1}(.99379)]^2$$
So the sample size is $n=25$. Now we can choose any $\gamma_2,\gamma_1$, such as $\gamma_2=.9$ and $\gamma_1=.1$. But since the $\chi^2$ distribution is not symmetric is it not obvious this is the best choice (e.g. leading to the smallest interval). But it is one choice.
$G^{-1}(.9)=33.20$ (or $\chi^2_{24,.9}$ if you like) and $G^{-1}(.1)=10.86$. One possible 80% confidence interval for $\sigma$ is then $$\left(\sqrt{\frac{35.63-\frac{3.14^2}{25}}{33.20}}, \sqrt{\frac{35.63-\frac{3.14^2}{25}}{10.86}}\right)=(1.030201, 1.801258)$$
