Standard Error - Statistics A box contains twelve tickets labeled with numbers. The number on the tickets are:
-10,-6,-4,-3,-2,1,3,4,4,5,8,8
a) The standard error of the sample sum of the ticket labels in 5 independent random draws with replacement from the box is? 
b) The standard error of the sample mean of the ticket labels in 5 independent random draws with replacement from the box is?
I've calculated using the method --> n½ ×SD(box) and ((N−n)/(N−1) )½ × n½ ×SD(box) for the sample sum however neither of them work. The SD value was 5.67824129836.
I appreciate the help! Thanks
 A: Let $X_1$ be the result of the first draw, $X_2$ the result of the second, and so on up to $X_5$. 
Then the sum $Y$ is $X_1+\cdots+X_5$, and the sample is $\frac{Y}{5}$.
We want to find (i) the standard error of $Y$, and (ii) the standard error of $\frac{Y}{5}$.
First we calculate the variance of $X_1$, say. The mean $E(X_1)$ of $X_1$ is not hard to compute. It is 
$$\frac{1}{12}((-10)+(-6)+\cdots + 8+8).$$
We also need to compute $E(X_1^2)$. This is
$$\frac{1}{12}((-10)^2+(-6)^2_\cdots +(8^2)+(8)^2).$$
Now we know the variance of $X_1$, for $\text{Var}(X_1)=E(X_1^2)-(E(X_1))^2$.
(i) The $X_i$ are independent, so the variance of the sum is the sum of the variances, and therefore $Y$ has variance $5\text{Var}(X_1)$. For the standard deviation of $Y$, take the square root.
(ii) For the standard deviation of $\frac{Y}{5}$, take the standard deviation of $Y$, and divide by $5$. Or the standard deviation of $X_1$, and divide by $\sqrt{5}$. 
A: a) Since the draws are with replacement,  the trials are independent and so we use the usual formula
$$\text{SE}_{\text{sum}} = \sqrt{n}\times\text{SD}_{\text{box}} = \sqrt 5\,(5.44) = 12.2$$
where $a = \text{avg}_\text{box}\approx  .667$ and 
$$\text{SD}_{\text{box}} = \sqrt{\frac{(-10-a)^2+(-6-a)^2+\dotsb+(8-a)^2+(8-a)^2}{12}} \approx 5.44.$$
b) We simply recall the usual formula
$$\text{SE}_\text{avg} = \frac{\text{SE}_\text{sum}}{n} = \frac{12.2}{5} \approx 2.44$$
