Say I am given a data-set generated from $Binomial(n, p)$, and these values are: $$5, 6, 2, 6, 3, 6, 8$$I am not given any information about the parameters $n$ or $p$.
First off, I need to set up an estimator $\hat{n}$ for $n$, for which I used: $$\hat{n}=\max{x_i}$$ So in this case $\hat{n}$ would be $8$. Is this considered to be a good estimator for n? Are there any others I could consider?
Also, since I only have an estimated value for $n$, am I correct to say that $$\hat{p}=\dfrac{X}{\hat{n}k}$$ where k is the number of tests (in this case 7), is a biased estimator? This estimator value for the above data-set would be: $$\hat{p}=\frac{5+6+2+6+3+6+8}{8\cdot7}\approx0.643$$
Why not use maximum likelihood estimators? For $N$ i.i.d. observations $x=(x_1,\ldots,x_N)$ with mean $$\bar x=\frac1N\sum_{k=1}^Nx_k$$ the likelihood of the sample $x$ with respect to the binomial $(n,p)$ distribution is $$L(n,p)=c_n(x)p^{N\bar x}(1-p)^{nN-N\bar x}$$ where $$c_n(x)=\prod_{k=1}^N{n\choose x_k}$$ Optimizing $L(n,p)$ with respect to $p$ yields $$\hat p=\frac{\bar x}n$$ hence one is left with optimizing $L(n,\hat p)={\bar x}^{N\bar x}\ell(n)$ over $n$, with $$\ell(n)=c_n(x)n^{-nN}(n-\bar x)^{nN-N\bar x}$$ which can be maximized numerically.
For $x=(5,6,2,6,3,6,8)$, one gets $$N=7\qquad\bar x=\frac{36}7$$ and the plots of $\ell(n)$ shown below indicate a maximum located at $n=14$.
This points at selecting the binomial distribution $(n,p)$ with $$n=14\qquad p=\frac{18}{49}\approx0.367$$ rather than your suggestion $n=8$ and $p\approx0.643$, for a reduced likelihood $\ell(14)=1.392\cdot10^{-32}$ rather than $\ell(8)\approx6.77\cdot10^{-33}$.