If a statistic $T(X)=\Sigma_{i=1}^n X_i$ is sufficient does that imply the mean is also sufficient?

I've been working on some problems, the question asked me if the mean of a sample is a sufficient statistic for poisson distribution. I've already proved that $$T(X)=\Sigma_{i=1}^n X_i$$ is a sufficient statistic. Can I just conclude that the mean is also a sufficient statistic since we are just dividing by $$n$$?

And can we say the same about completeness? If $$T(X)=\Sigma_{i=1}^n X_i$$ is complete then the mean is also complete?

• Yes by factorization theorem and definition of complete statistic Apr 1 '21 at 21:04
• Two statistics $S$ and $T$ are equivalent if there exists a one-to-one function $f$ such that $S = f(T)$. If two statistics are equivalent and one is a sufficient statistic, then so is the other. Apr 1 '21 at 21:13
• @bdeonovic thank you for answering, can you check my other question stats.stackexchange.com/questions/517792/…? Apr 1 '21 at 21:14

Roughly, given a set $$\mathbf {X}$$ of independent identically distributed data conditioned on an unknown parameter $$\theta$$ , a sufficient statistic is a function $$T(\mathbf {X} )$$ whose value contains all the information needed to compute any estimate of the parameter.
If $$T(X)$$ contains all the information, $$T(X)/n$$ contains all the information as well.
$$f(X|mean=y,\theta)=f(X|\sum X_i =ny, \theta)=f(X| \sum X_i=ny) =f(X| mean =y)$$