This is what I would do which I believe it is what @Dave is hinting at in comment.
Fit GLM with Poisson familiy distribution and see the effect of "condition" on the mean of each group.
Prepare the dataset:
Cond1 <- c(0, 0, 0, 1, 1)
Cond2 <- c(1, 2, 3, 3, 4)
dat <- data.frame(
cond= c(rep('C1', length(Cond1)), rep('C2', length(Cond2))),
count= c(Cond1, Cond2)
)
Fit the model and assess the significance of "condition":
fit <- glm(count ~ cond, data= dat, family= 'poisson')
summary(fit)
Call:
glm(formula = count ~ cond, family = "poisson", data = dat)
Deviance Residuals:
Min 1Q Median 3Q Max
-1.13533 -0.89443 -0.07296 0.65703 0.80391
Coefficients:
Estimate Std. Error z value Pr(>|z|)
(Intercept) -0.9163 0.7071 -1.296 0.1950
condC2 1.8718 0.7595 2.464 0.0137 *
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
(Dispersion parameter for poisson family taken to be 1)
Null deviance: 14.8823 on 9 degrees of freedom
Residual deviance: 5.8682 on 8 degrees of freedom
AIC: 27.731
Number of Fisher Scoring iterations: 5
So the mean of Cond 1 is estimated as exp(-0.9163) = 0.4, the mean of Cond 2 is exp(-0.9163 + 1.8718) = 2.6 and the difference has p = 0.0137 for the null hypothesis of being zero.
Best Answer
Note that Poisson distributions are entirely determined by their parameter, so a test of equality of their mean parameter is a test for whether the distributions are the same.
Some possible tests:
If you have two samples which you treat as iid Poisson each with its own parameter, which you want to test for equality of that parameter; in that case you can simply combine all the observations in each group into a single Poisson count.
a. You could condition on the total count and do a test of proportions (a binomial test in exact form, or via normal approximation, or equivalently a chi-squared test). For example, this binomial test is what you get if you do
poisson.teston two samples in R.b. You could do a likelihood ratio test.
(There are a number of other possibilities under this option.)
If you don't necessarily want to treat them as Poisson except as a rough approximation (but do treat them as iid), you would keep all the individual values.
a. You could then do a permutation test of the means.
b. You could do a Wilcoxon-Mann-Whitney or even a goodness of fit test (e.g. a Kolmogorov-Smirnov test) but you will have to deal with the discreteness of the distributions.
c. If you expect that the means won't be very small, you could perform (say) a t-test (under the null the samples should have equal variance, so it's not important whether you do the equal-variance version).
If instead of being identically distributed, they are of known but different exposures, you could combine into single counts as in option 1, but also combine the exposures into a single exposure for each. You could then follow the approaches in 1.
If they have unknown exposure but the exposures of pairs of observations will be the same, you effectively have pairing. You could perform a paired permutation test -- permuting the group labels within each pair (which corresponds to putting + and - signs on each absolute pair difference of counts). You could also do a sign test, or since (under the null) the differences would be symmetric you could consider a signed rank test (again properly accounting for ties).