Say I have a data vector X and I want to test if they come from a distribution say Poisson. Obviously, I can't specify the Lambda of Poisson, so I estimate it using X, which loses 1df. If I were to test against normality, then I lose 2df from mu = mean(X), sigma = sd(X).

I tried using chisq.test() multiple times and it seems that it always uses df = k-1, where k is number of bins/groups of data, so implying (correct me if I am wrong) that the function assumes the p comes from a fully-specified distribution, which in practice is highly unlikely. It would be ok if I could input the df for chisquare, but the chisquare is programmed to use k-1 df.

My question is how do I do the test properly in R (using built-in functions)?

```
say I have some data that I want to test against Poisson
data = c(0, 0, 0, 1, 0, 1, 2, 2);
lambda = mean(data); #0.75
bins = c(0, 1, 2, 3); #bins for grouping data
x = c(4, 2, 2, 0); #number of observations for bins
p = dpois(bins, lambda);
chisq.test(x, p=p, rescale.p=TRUE); #rescale.p=TRUE because p
doesn't sum to 1
#the df should be number of bins - 1 - number of estimates, so 2, but
R always uses df = k-1?
```