Interval estimate of variance? Let's say we know the mean of a given distribution. Does this affect the interval estimate of the variance of a random variable (which is otherwise computed using the sample variance) ? As in, can we obtain a smaller interval for the same confidence percent?
 A: The answer to your question depends on the distribution of the population. Suppose the population is normal with
population mean $\mu$ and population variance $\sigma^2.$
Then what you suspect is true.
In the usual case, both $\mu$ and $\sigma^2$ are unknown.
You find $\bar X$ to estimate the population mean and then
find $S^2 = \frac{\sum(X_i - \bar X)^2}{n-1}$ to estimate $\sigma^2.$
Then 
$$ (n-1)S^2/\sigma^2 \sim Chisq(df = n-1),$$
and a 95% CI for $\sigma^2$ is
$$((n-1)S^2/U, (n-1)S^2/L),$$
where $L$ and $U$ cut probability 2.5% from the lower and
upper tails, respectively, of $Chisq(n-1)$.
By contrast, if $\mu = \mu_0$ is known, then the usual estimate
of $\sigma^2$ is $V = \frac{\sum(X_i - \mu_0)^2}{n}.$ Then
$$nV/\sigma^2 \sim Chisq(df =n),$$
and a 95% CI for $\sigma^2$ is
$$(nV/B,\; nV/A),$$
where $A$ and $B$ cut 2.5% from the lower and upper tails,
respectively, of $Chisq(n).$
The latter interval is shorter on average; intuitively, because it is based
on the additional information that $\mu = \mu_0.$ In a particular
instance it need not be shorter because of the randomness of
$\bar X, S^2,$ and $V$.
In particular, if $n = 5$ and $\sigma^2 = 1$, then the
average length of the above CI is 5.63 in the case where
$\mu$ is known and 7.90 in the case where $\mu$ is unknown.
Computations in R are as follows:
 5*diff(1/qchisq(c(.975,.025),5))
 ## 5.62568
 4*diff(1/qchisq(c(.975,.025),4))
 ## 7.898361

A: In my previous Answer I state that CIs for $\sigma^2$ based on $S^2$ when $\mu$ is unknown are, on average, longer than
 CIs for $\sigma^2$ based on $V$ when $\mu$ is known. In a Comment
you raise the interesting question how often and in what circumstances
the former type of CI actually turns out to be shorter than the latter.
This is an intricate relationship that depends on random variables
$\bar X, S^2,$ and $V$. Perhaps it is best to take a first look
at it in terms of a simulation. Below, I have simulated
$m = 40,000$ samples of size $n = 5$ from a population distributed
$Norm(\mu = 100,\, \sigma=15).$ These are put into an $m \times n$ matrix DTA. For each row, $\bar X$, $S^2$, and $V$ are found,
along with the actual length of each style of CI. 
The first few lines
of numerical output serve as a reality check that the estimates
are as they should be, based on the parameters $\mu = 100$ and $\sigma^2 = 225.$ It turns out that about 16% of the CI pairs
have lengths 'contrary' to the overall expected values.
In the 'matrix' plot of $\bar X$, $S^2$, and $V,$ the points
corresponding to the contrary samples are shown in red. It seems
that these are the cases in which $\bar X$ is far from $\mu$ 
compared with a $relatively$ small estimate $S^2$ or $V$ of $\sigma^2.$ Perhaps more directly, they are cases in which $V$ is
relatively large compared with $S^2.$ The curves of demarcation in the figures
between 'contrary' cases and 'anticipated' cases are quite crisp--suggesting that there are some exact analytic relationships. I will leave you to investigate that idea. (I am not aware that this particular question has been explored previously, but perhaps it has;
I am not aware of everything.)
[I include the R program for
the simulation, in case it is of interest to you. Obviously, another
run of the same program will give slightly different results.]
 m = 40000;  n = 5;  mu = 100;  sg = 15
 DTA = matrix(rnorm(m*n, mu, sg), nrow=m)
 x.bar = rowMeans(DTA);  s.sq = apply(DTA, 1, var)
 v = rowSums((DTA-mu)^2)/n
 LU = diff(1/qchisq(c(.975,.025),n-1))
 AB = diff(1/qchisq(c(.975,.025), n))
 ci1.len = (n-1)*s.sq*LU;  ci2.len = n*v*AB
 cond = (ci1.len < ci2.len)
 mean(x.bar); mean(s.sq); mean(v)
 ## 100.0113  # good est of mu = 100
 ## 225.4049  # close to 225
 ## 225.3743  # ditto
 mean(ci1.len); mean(ci2.len); mean(cond)
 ## 1780.329
 ## 1267.883
 ## 0.15795   # fraction of 'contrary' samples
 farb = rep("black", m); farb[cond] = "red"
 pairs(cbind(x.bar,s.sq,v), col=farb, pch=".")


