Diagnostics for quasipoisson glm for continuous data I'm a little confused about how to use the quasipoisson family in the glm function. It was recommended by someone that I use it for my analysis, even though the data are continuous - and as such, I don't get the same warning messages (regarding the fact that my data are non-discrete) as I do when I use it for the Poisson family. Why is this? Is the quasipoisson family actually compatible with continuous data? I thought Gamma might be more suitable for a continuous dataset, but about 50% of my values are zeroes (this was the original reason quasipoisson was recommended).
I don't quite understand all of the information I've been given about lm's/glm's in general - do they assume anything about the distribution of y (the response variable), or just the distribution of the residuals, when a model has already come up with fitted values?
As I said earlier, I was given the typical advice of 'if the assumptions for the Poisson family aren't met, use quasipoisson.' ...but how can I test if the assumptions of the quasipoisson distribution are met (or Gamma, if I just add one to everything)?
I can't use a Kolmogorov-Smirnov test, as the function does not like repeated values (as I said, about half of the y-values are zero).
Would you suggest I just give up and use nonparametric stats? My project is already overdue and I have lots of other work to do!
I'm ok with sharing my data if that helps
 A: From what you stated, I think you may have been misled. 
In a standard Poisson GLM, you are assuming that your data follow 
$Y_i \sim Pois(\lambda_i)$
where $ \lambda_i = \exp(X^T \beta)$
$X_i$ are the covariates for subject $i$ and $\beta$ are the coefficients you are interested in estimating. 
This builds many assumptions into your model. One (and only one of many), is that $Var(Y_i) = E(Y_i)$. This assumption is relaxed in a quasi-Poisson model; it is rather assumed that $Var(Y_i) = \alpha E(Y_i)$, where $\alpha$ is a parameter to be estimated. This allows for over dispersion when $\alpha > 1$, i.e. more variance than allowed by the exact Poisson case. It's important to note that you're still assuming the variance is proportional to the mean (but of course that's less restrictive than being equal to the mean). 
This does not directly answer your issue of being zero-inflated, although it is probably a slight improvement compared with the standard Poisson GLM. As for what I think you should do, it is hard to say without knowing more about your data. If it's the case that you're just comparing response variables in different groups (as opposed to across different different levels of a continuous variable), then I would certainly endorse simple non-parametric methods rather than trying to check all the assumptions of different potential parametric models. 
