Sufficient statistic for poisson Possion have mean and variance of the same value, and obviously the mean of samples is a sufficient statistic
Is the variance of the sample a sufficient statistic as well?
1) If not, how do I prove its not? Not able to factorize it? Its numerical value same as mean creates some uncertainty for me.
2) If yes, then it should have a as low variance as the mean by Rao Blackwell, which is obviously not true. Variance is a worse estimator than mean for Poisson.
Any clarification is appreciated.
 A: Answer question (1)
Lets $X_1,\cdots , X_n \sim Poisson(\lambda)$, we want to prove  $U=\frac{1}{n-1}\sum (X_i-\bar{X})^2=\frac{\sum X_i^2 -n \bar{X}^2}{n-1}$ is not a sufficient statistic.
If you want to show a statistic  is not a sufficient statistic  , you can compare it with minimal sufficient statistic. Use the fact that 
a minimal sufficient statistic
 is a  function of any  sufficient statistic. 
It is obvious that $T=\frac{\sum X_i}{n}$ is a minimal sufficient statistic for $\lambda$. Since $T$ is minimal sufficient statistic, so it is a function of any sufficient statistic. It is enough to show that $T$ is not a function of $U$.
$T$ is a function of $U$ if $U(a_1)=U(a_2
)$ $\Rightarrow T(a_1)=T(a_2)$. So it is enough to find two points that $U(a_1)= U(a_2)$  but $T(a_1)\neq T(a_2)$ , and hence $T$ is not a function of $U$  and hence $U$ is not a sufficient  statistic.
$a_1=(x_1=1,x_2=1, \cdots ,x_n=1)$
$a_2=(x_1=0,x_2=0, \cdots ,x_n=0)$
So $0=U(a_1)=U(a_2
)$ but $1=T(a_1)\neq 0=T(a_2
)$
