The general asymptotic result for the asymptotic distribution of the sample variance is (see this post)
$$\sqrt n(\hat v - v) \xrightarrow{d} N\left(0,\mu_4 - v^2\right)$$
where here, I have used the notation $v\equiv \sigma^2$ to avoid later confusion with squares, and where $\mu_4 = \mathrm{E}\left((X_i -\mu)^4\right)$. Therefore by the continuous mapping theorem
$$\frac {n(\hat v - v)^2}{\mu_4 - v^2} \xrightarrow{d} \chi^2_1 $$
Then, accepting the approximation,
$$P\left(\frac {n(\hat v - v)^2}{\mu_4 - v^2}\leq \chi^2_{1,1-a}\right)=1-a$$
The term in the parenthesis will give us a quadratic equation in $v$ that will include the unknown term $\mu_4$. Accepting a further approximation, we can estimate this from the sample. Then we will obtain
$$P\left(Av^2 + Bv +\Gamma\leq 0 \right)=1-a$$
The roots of the polynomial are
$$v^*_{1,2}= \frac {-B \pm \sqrt {B^2 -4A\Gamma}}{2A}$$
and our $1-a$ confidence interval for the population variance will be
$$\max\Big\{0,\min\{v^*_{1,2}\}\Big\}\leq \sigma^2 \leq \max\{v^*_{1,2}\}$$
since the probability that the quadratic polynomial is smaller than zero, equals (in our case, where $A>0$) the probability that the population variance lies in between the roots of the polynomial.
Monte Carlo Study
For clarity, denote $\chi^2_{1,1-a}\equiv z$.
A little algebra gives us that
$$A = n+z, \;\;\ B = -2n\hat v,\;\; \Gamma = n\hat v^2 -z \hat \mu_4$$
which leads to
$$v^*_{1,2}= \frac {n\hat v \pm \sqrt {nz(\hat \mu_4-\hat v^2)+z^2\hat \mu_4}}{n+z}$$
For $a=0.05$ we have $\chi^2_{1,1-a}\equiv z = 3.84$
I generated $10,000$ samples each of size $n=100$ from a Gamma distribution with shape parameter $k=3$ and scale parameter $\theta = 2$. The true mean is $\mu = 6$, and the true variance is $v=\sigma^2 =12$.
Results:
The sample distribution of the sample variance had a long road ahead to become normal, but this is to be expected for the small sample size chosen. Its average value though was $11.88$, pretty close to the true value.
The estimation bound was smaller than the true variance, in $1,456$ samples, while the lower bound was greater than the true variance only $17$ times. So the true value was missed by the $CI$ in $14.73$% of the samples, mostly due to undershooting, giving a confidence level of $85$%, which is a $~10$ percentage points worsening from the nominal confidence level of $95$%.
On average the lower bound was $7.20$, while on average the upper bound was $15.68$.
The average length of the CI was $8.47$. Its minimum length was $2.56$ while its maximum length was $34.52$.
Yes, it is true. In the language of statistics, we would say that if you have no knowledge of the population mean, then the quantity
$$\frac{1}{n-1} \sum_{i=1}^n \left(x_i-\bar{x} \right)^2$$
is unbiased, which simply means that it estimates the population variance correctly on average. But if you do know the population mean, there is no need to use an estimate for it- this is what the $\bar{x}$ serves for-and the finite-sample correction that comes with it.
In fact, it can be shown that the quantity
$$\frac{1}{n} \sum_{i=1}^n \left(x_i-\mu \right)^2$$
is not only unbiased but also has lower variance than the quantity above. This is quite intuitive as part of the uncertainty has now been removed. So we use this one in this situation.
It is worth noting that the estimators will differ very little in large sample sizes and hence they are asymptotically equivalent.
Best Answer
Suppose you have a random sample of size $n$ from the population $\mathsf{Norm}(\mu, \sigma),$ where $\sigma$ is not known and $\mu$ is known.
Let $V = \frac 1n\sum_{i=1}^n (X_i - \mu)^2.$
Then $V$ is a better estimate of the population variance $\sigma^2$ than is $S^2=\frac{1}{n-1}\sum_{i=1}^n (X_i - \bar X)^2,$ where $\bar X =\frac 1 n \sum_{i=1}^n X_i.$
Also, a 95% CI for $\sigma^2$ tends to be narrower if we use $V$ than if we use $S^2.$ [Samples can vary, so this CI is not always narrower.]
In particular, a 95% CI for $\sigma^2$ is based on the relationship $\frac{nV}{\sigma^2} \sim \mathsf{Chisq}(\nu = n).$
Example: Suppose I have the sample
xof size $n = 50$ from $\mathsf{Norm}(\mu = 20, \sigma = 3),$ where I assume $\mu$ is known and $\sigma$ is not.The formula for this confidence interval is $\left(\frac{50V}{U}, \frac{50V}{L}\right),$ where $L$ and $U$ cut probabilities $0.025$ from the lower and upper tails, respectively, of $\mathsf{Chisq}(\nu=50).$ For the data of my example, the CI is $(7.49\, 16.52)$ of width $9.04.$
By contrast, the 95% CI for $\sigma^2$ based on $S^2,$ where $\mu$ is estimated by $\bar X,$ uses the relationship $\frac{(n-1)S^2}{\sigma^2}\sim\mathsf{Chisq}(\nu=49).$
For the data of my example, the CI is $(7.55,\, 16.80)$ of width $9.25 > 9.04.$