# Counting samples seem to not be Poisson distributed, need sanity check

I have an exercise where I have to use Poisson one-way classification / Regression of some data. The data I have is a set of 120 samples grouped the following labels A, B, C, D, E, and F.
For each group, there are 20 samples (or 20 repetitions) with a count value.
Now that is all good and from what I can tell is well suited for the assumption that it may fit a Poisson distribution.

However, as I understand it one of the properties of a random variable $$YY$$ that follows $$Y∼Po(λ)Y \sim Po(\lambda)$$
Then it follows that $$E(Y)=V(Y)E(Y) = V(Y)$$
But when I calculate the means (expected) and the variance for the data according to the grouping: Then I get

|          |         A |         B |         C |         D |          E |         F |
|----------+-----------+-----------+-----------+-----------+------------+-----------|
|----------+-----------+-----------+-----------+-----------+------------+-----------|
| Mean     |      4.90 |      9.45 |      8.65 |      1.45 |      18.35 |      0.80 |
| Variance | 9.8842105 | 6.4710526 | 7.3973684 | 1.3131579 | 15.6078947 | 0.5894737 |
|----------+-----------+-----------+-----------+-----------+------------+-----------|


So from what I can tell $$E(Y)≠V(Y)E(Y) \neq V(Y)$$ and just to show how I computed this using R:

( Means <- tapply(D$NumberPGrains, D$Era, mean) )
( Variances <- tapply(D$NumberPGrains, D$Era, var) )


This means, from my understanding that the data is not Poisson distributed.
So my question is: Am I wrong can this still be Poisson distributed, is there something I have missed?

And just to clarify, the exercise literally states to follow Poisson one-way classification (the title of the exercise: “Question 3 -Poisson one-way classification model”), but right now I have hard time seeing the purpose of that.