Deriving an expression for a confidence interval for σ^2 using the asymptotic distribution of √n(σ̂^2−σ^2) We have We have $X1,…,Xn i.i.d N(μ,σ^2) $where $μ$ is known and $σ^2$ isn't known. 
 $σ̂^2=(\frac{1}{n})∑(X_i−μ)^2$.
First of all what I did, I derived an equitailed 95% confidence interval for $σ^2$. But then if given that $σ̂^2$ is a sample mean, I found the asymptotic distribution of $√n(σ̂^2−σ^2)$. (Thank you for everyone who helped me on here to see where I had gone wrong!). I now want to find an alternative confidence interval for $σ^2$ using the asymptotic distribution of $√n(σ̂^2−σ^2)$. (Which was $N(0, 2σ^4)$.
I know that $(\frac{√n(σ̂^2−σ^2)}{(σ^2√2})$ $-> N(0,1)$, so my initial plan was to use this as a pivot. However thinking about it now, $σ^2$ is unknown so I think my pivot must be wrong. How would you go about deriving this confidence interval?
Thank you!
 A: $(\frac{\sqrt n(σ̂^2−σ^2)}{σ^2√2}) \sim N(0,1)$ ==> $\Pr(-Z <\frac{\sqrt n(σ̂^2−σ^2)}{σ^2√2} < Z) = \alpha$, where $Z$ is Z-score for $1-\frac 12\alpha$.
You can get a solution of $\sigma^2$ from $-Z <\frac{\sqrt n(σ̂^2−σ^2)}{σ^2√2}$, and another one from $\frac{\sqrt n(σ̂^2−σ^2)}{σ^2√2} < Z$.
From 
$$\frac{\sqrt n(σ̂^2−σ^2)}{σ^2√2} < Z$$
$${\sqrt n(σ̂^2−σ^2)} < Z{σ^2√2}$$
$${\sqrt nσ̂^2} < Z{σ^2√2}+\sqrt nσ^2$$
$${\sqrt nσ̂^2} < (√2Z+\sqrt n)σ^2$$
$$σ^2 > \frac {\sqrt nσ̂^2} {√2Z+\sqrt n}$$
Simulation using SAS: Sample size =1000, repeat 1000 times.
  data temp; do i = 1 to 1000; do j = 1 to 1000; x=rannor(665654); output; end; end; run;
  proc means noprint; var x; by i; output out= var var=var; run;
  data var; set var; l= var*sqrt(1000)/(sqrt(2)*(1.96) + sqrt(1000));
      u= var*sqrt(1000)/(sqrt(2)*(-1.96) + sqrt(1000)); cover = (l<=1 <=u);
  run;
  proc freq data=var; tables cover; run;

     cover    Frequency     Percent     
    ----------------------------------
       0          60        6.00      
       1         940       94.00      
    ----------------------------------

Set sample size to 10000 and repeat 10000 times (takes long time) The CI cover true value 1 9484 times 
A: You can check this open-access paper in The Quantitative Methods for Psychology which lists many confidence intervals, including for standard deviations and variances.
(link corrected).
