In a survey of 1000 people, 60% say they vote for Candidate A for president. How can you estimate a margin of error for the 60% estimate?
Confusion:
I am very confuse about the estimation method. I have learnt about the unparametric estimate (plug-in estimation), and to estimate the mean, you simply do $1/N*\sum x_i$.
On the other hand, you can also use the MLE which is simply max $f(x_1..x_n;\theta )$.
In particular, My confusion is that:
1) If we already assume that $X_i$ they are bernoulli, how come we have to go through that long derivation. $E(Y/N) = .... = p$, why can't we say that $E[X_i]=p$, thus, $1/N*\sum x_i=p$? If I do this, am I using the MLE or the plug-in estimation?
It seems like you confused some concepts. I'll try to make it clearer.
The sample mean does not equal to the parameter $p$, rather it converges (in probability, mean, a.s., etc.) to $p$. I.e., $1/N\sum_{i=1}^N X_i \xrightarrow{p} EX_i =p$ as $n \to \infty$. (Of course, assuming that $X_1, X_2,...$ are iid from Bernoulli r.v with $p$).
It is possible to get the same estimator with different estimating methods. E.g., for this case you can check that $\bar{X}_n$ is the MLE, it is clearly also the method of moments estimator and an UMVUE. Actually, it is the most intuitive estimator that once can come with for $EX=\mu$.