Show that $\sigma_\bar Y^2$ , the variance of the sample mean, $\bar Y$ , is given by $\sigma_\bar Y^2 = \frac {N-n}{N-1}\frac {\sigma^2}{n}$

Question

This is really frustrating me,

So I know that $\bar Y$ =$ \frac {1}{n} \sum_{i=1}^n Y_i$

And I have that $Cov(Y_i, Y_j) = \frac {-\sigma^2}{(N-1)}$

I know that population mean $\mu = \frac {1}{N} \sum_{i=1}^N x_i$ and population variance $\frac {\sum_{i=1}^N (x_i-\mu)^2}{N}$

But from this point I am clueless

Answer 1

$Var(\overline{Y})=\dfrac{1}{n^2}Var(\sum_{i=1}^n Y_i)=\dfrac{1}{n^2}[\sum_{i=1}^n Var(Y_i)+\sum_{i\neq j}Cov(Y_i,Y_j)]$

Can you take up from here?