The variance of linear regression estimator $\beta_1$ Can we say 
$$\text{Var}(\beta_1) = \text{Var}\left(\frac{\sum (x_i-\bar x)y_i}{\sum (x_i- \bar x)^2}\right) = \left(\frac{\sum (x_i-\bar x)}{\sum (x_i- \bar x)^2}\right)^2 \text{Var}(y_i) \;\;??$$
I am not sure if I can separate the $x$'s i from $\sum (x_i-\bar x)y_i$. This expression seems like a linear combination of $y_i$'s. Is this legit because every $y_i$ follows the same distribution?
 A: This appears to be simple linear regression. If the $x_i$'s are treated as deterministic, then things like "variance" are not associated with them, and so the expression holds, under the additional assumption that the the error term (and hence $y$ also) has identical distribution for all $i$, and also, that the error terms (and hence $y$ also) are independent for all $j\neq i$.  
For compactness, denote 
$$z_i = \frac{x_i-\bar x}{\sum (x_i- \bar x)^2}$$
Then
$$\text{Var}(\beta_1) = \text{Var}\left(\sum z_iy_i\right)$$
The assumption of deterministic $x$'s permits us to treat them as constants. The assumption of independence permits us to set the covariances between $y_i$ and $y_j$ equal to zero. These two give
$$\text{Var}(\beta_1) = \sum z_i^2\text{Var}(y_i)$$
Finally, the assumption of identically distributed $y$'s implies that $\text{Var}(y_i)= \text{Var}(y_j) \;\; \forall i,j$ and so permits us to write
$$\text{Var}(\beta_1) = \text{Var}(y_i)\sum z_i^2$$
A: There are a few way to approach this. Alecos' method is simple and clear. Here's another:
$\hat{\beta}_1=\frac{\sum (x_i-\bar x)y_i}{\sum (x_i- \bar x)^2}=\frac{\sum (x_i-\bar x)y_i}{S_{xx}}=\frac{\sum d_iy_i}{S_{xx}}$
$\text{Var}(\hat{\beta}_1) = \text{Var}\left(\frac{\sum (x_i-\bar x)y_i}{\sum (x_i- \bar x)^2}\right)=\frac{1}{S_{xx}^2}\text{Var}\left(\sum d_iy_i\right)=\frac{1}{S_{xx}^2}(\sum d_i^2)\text{Var}\left(y_i\right)$
You can then inspect $\sum d_i^2$ in order to do some cancelling out in the formula, and you can substitute a value for the variance of the observations.
Compare with the sample estimate of the standard error, $s_{\hat\beta}$ here, which should be similar up to the two obvious differences.
