Coefficient of Variation for Poisson Process I am analyzing an engineering problem where the the coefficient of variation (CV) is being used to assess how reliably the quantity of a certain chemical can be measured. The context of this problem is as follows. The engineers have 16 stations and at each station they take about 100 samples. They measure the amount of a specific chemical in the samples. For each of the 16 sites, the sample mean and sdev of the amount of chemical in the samples are computed.
The engineers think there is an inherent problem with the how reliable the measurements are because the CV is very high. I believe the solution to their problem is simply that they are looking at a Poisson process. The basis for this belief is that the chemical measured is produced by scale in a pipe that "flakes off" occasionally (hence the measured quantity can be seen as the sum of a number of discrete events and the conditions for a Poisson process are most likely satisfied). 
This hypothesis is testable since for a Poisson distribution the sdev is equal to the square root of the mean. When I make a scatterplot of the sdev vs the sqrt of the mean (for the 16 stations), there is a strong positive linear relation (with a coeff of det about 73% and a p-value for the model about 0.00002)
My problem however is the following: when I look at the estimated slope for the regression model (sdev vs sqrt(Xbar)) it is about 11, although theory predicts it should be unity. I am wondering what would account for this? I do not know of another distribution that has the property that the sdev and the square root of the mean are proportional and I don't think a non-homogeneous Poisson process would produce my observed results.
Addition: when I try to fit the sdev and the mean with a linear model (to see if an exponential model might be appropriate) the fit is also very good (R^2 is about 70% and pvalue for the model is about 0.00005) but the slope again is off but not by as much (is 2.3, but should be unity). Which is at least better and perhaps the discrepancy can be explained by random error (the standard error is about 0.4, so 1 is close to being with a 95% CI for the slope). I'm not sure the model makes sense physically though.
Thanks for any help or suggestions anyone might have,
Matt
 A: The Poisson could come into this, but I think some of the ideas in your question are mistaken.
A Poisson process is a counting process -- it might be used as a model for the flaking events (so the number of such events could be well approximated by a Poisson) but the amount of material each time would be a random quantity. 
If you think that the amount at each event is i.i.d. from some distribution -- perhaps gamma or lognormal or Pareto, say -- then the process would be a compound Poisson process.
If $X_i$ is the amount released at the $i$th event then the total amount, $Y$ in time $t$ is:
$$Y(t) = \sum_{i=1}^{N(t)} X_i$$
where $N(t)$ is the number of Poisson events in time $t$.
The total amount of the material in some given interval would have a compound Poisson distribution. 
If $N\sim \text{Pois}(\lambda)$ then $Y\mid N = X_1+X_2+...+X_N$. 
There are other compound models using different count-distributions that the Poisson, of course, but the Poisson is a fairly commonly used one. It has some very simple properties.
Such distributions show up in a number of contexts, and simple facts (like how the mean and variance of the amount in some interval of time relate to the parameter of the Poisson counts and the mean and variance of amounts at each event ("size-distribution") are fairly readily derived via the law of total expectation and the law of total variance.
For the compound Poisson the mean and variance of the unconditional distribution of $Y$ are:
$E(Y) = \lambda\, \mu_X$ where $\lambda$ is the Poisson mean and $\mu_X$ is the mean of the size-distribution.
$\text{Var}(Y) = \lambda\, (\mu_X^2 + \sigma_X^2)$, taking $\sigma_X^2$ as the variance of the size-distribution.
This suggests that the coefficient of variation of a compound Poisson would be 
$c_Y = \sqrt{\frac{\mu_X^2 + \sigma_X^2}{\lambda\mu_X^2}} = \sqrt{\frac{1+c_X^2}{\lambda}} = c_N \,\sqrt{1+c_X^2} $
where $c_Y$, $c_X=\sigma_X/\mu_X$ and $c_N=\frac{1}{\sqrt{\lambda}}$ are the coefficients of variation of $Y$, $X$ and $N$ respectively.
This might explain why you would not expect to see slope of 1 in your plot.
