We observe a simple random sample $(X_1,ldots , X_n$), where the distribution of observed property $X$ is given by its density: $f(x)=frac{1}{xsigma sqrt{2pi}}e^{-frac{(ln x-m)^2}{2sigma^2}}, x>0$.

Determine an 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 $frac{noverline{S}_n^2}{sigma^2}$ has $chi_{n-1}^2$ distribution.

If we look for confidence interval in the form:

$$p{U_nleq sigma leq V_n}=0.8$$, we can actually demand $$p{frac{noverline{S}_n^2}{V_n^2}leq frac{noverline{S}_n^2}{sigma^2} leq frac{noverline{S}_n^2}{U_n^2} }=0.8$$.

We can demand:

$$p{frac{noverline{S}_n^2}{V_n^2}>frac{noverline{S}_n^2}{sigma^2}}=p{frac{noverline{S}_n^2}{U_n^2}<frac{noverline{S}_n^2}{sigma^2}}=frac{1-0.8}{2}=0.1$$.

This gives us (from what we know about $chi_{n-1}^2$ distribution):

$$V_n^2=frac{noverline{S}_n^2}{chi_{n-1,0.9}^2}$$

$$U_n^2=frac{noverline{S}_n^2}{chi_{n-1,0.1}^2}$$.

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:

$$overline{S}_*^2=frac{1}{n}sum_{i=1}^n (Y_i-m)^2$$

$frac{noverline{S}_*^2}{sigma^2}$ 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.