Confidence Interval and Variance of Coefficient of Variation i'm currently struggling to find the confidence interval of a statistic.
I'm calculating the coefficient of variation for a specific sample. The coefficient of variation is
$$\frac{\sigma}{\mu}$$
I would like to construct a normal Confidence interval around the estimator.
I'm stuck when i'm trying to get the variance of my estimator.
$$ VAR[\frac{\sigma}{\mu}] = VAR[\frac{\sum(x_i-\bar{x})^2}{\sum(x_i)}] $$
I start expanding it and a lot of it resolves by itself. 
However, I'm very confused about the following quantities:
$$ VAR[\frac{\sum(x_i)^2}{\sum(x_i)}] $$ and $$ COV[\frac{\sum(x_i)^2}{\sum(x_i)},X] $$ 
can anyone lend me a hand here !
Thank you very much!
 A: Estimating CVs. The coefficient of variation (CV) $\kappa = \sigma/\mu.$ It can be
estimated by $\hat \kappa = K = S/\bar X,$ where $\bar X$ and $S$
are the sample mean and SD, respectively. For small $n,$ this estimate
is biased on the low side, but for moderate and large samples
the bias is small. Methods of finding confidence intervals (CIs)
for the CV depend on the nature of the underlying distribution.
Because the type of population distribution may be unknown, it may
be useful to use a nonparametric bootstrap CI for the $\kappa.$
Because the population may be skewed (especially right-skewed) in
practice, the bootstrap must anticipate skewness. 
Because I found the literature on CIs for the CV to be partly
hidden behind dollar barriers, and partly poorly explained, I'm
wondering if bootstrap CIs may be the best solution for your application. I gave
two examples of bootstrap CIs below, one using a sample from a 
normal population and one using a sample from a gamma population.
At least, you can compare these results with results from formulas
you may find in your Internet searches.
Bootstrap CIs. If we knew the distribution of $V = K - \kappa,$ we could find
bounds $L$ and $U$ cutting 2.5% from its lower and upper tails,
respectively to get $P(L < K - \kappa < U) = 0.95,$ from which
we would obtain the 95% CI $(K - U, K - L)$ for $\kappa.$
Not knowing the distribution of $V,$ we re-sample from our data
$X = (X_1, X_2, \dots, X_n).$ Iteratively we find re-samples
of size $n$ with replacement from $X,$ find $K^* = S^*/\bar X^*$
and then $V* = K^* - \kappa^*$ for each re-sample, where
the observed CV $K_{obs}$ from the original sample $X$ is used
for $\kappa^*.$ Finally, we get $L^*$ and $U^*$ by cutting 2.5%
from each tail of the $V^*$'s, the 'bootstrapped' values of $V$,
and use these estimated bounds to get the a 95% bootstrap CI. 
Examples of Bootstrap CIs. As a demonstration, I use a sample $X$ if $n = 100$ from 
$\mathsf{Norm}(\mu = 200, \sigma=25)$ with $\kappa = 0.125.$ 
In the outline above of the bootstrap procedure, $*$'s represented
quantities based on re-sampling. In the R program below we use .re
for the same purpose.
Note: It is important to understand that re-sampling does not
create additional information. Re-sampling exploits information in existing
data to do statistical analysis.
Normal. For the particular normal sample we used $K_{obs} = 0.118$, and
the 95% nonparametric bootstrap CI obtained is $(0.102, 0.135).$ 
Because bootstrap procedures involve random re-sampling, each run
of the program may give a slightly different CI, but not much
different with as many as $B = 10^5 = 100,000$ iterations.
x = rnorm(100,  200, 25)
k.obs = sd(x)/mean(x);  k.obs
## 0.1180088
B = 10^5;  v.re = numeric(B)
for(i in 1:B) {
  x.re = sample(x, 100, repl=T)
  k.re = sd(x.re)/mean(x.re)
  v.re[i] = k.re - k.obs }
UL = quantile(v.re, c(.975,.025))
k.obs - UL
##     97.5%      2.5% 
## 0.1018754 0.1350186 

Gamma. This bootstrap procedure is called 'nonparametric' because it does
not assume any particular type of distribution for the data. A
second sample of size $n = 100$ was taken from the distribution 
$\mathsf{Gamma}(shape=\alpha = 4, rate=\lambda=.1)$ with 
$\kappa = \sqrt{\alpha}/\alpha = 1/2.$ This sample has $K = 0.507$
and the 95% nonparametric bootstrap CI is $(0.442, 0.579).$
A second run of the bootstrap program with the same data gave
the CI $(0.442, 0.580).$
