Linear regression with a given (non zero) intercept If I have a simple linear regression model with the intercept $\beta_0$ known, would the least squares estimator of $\beta_1$ be 
$\frac{\sum(y_ix_i)}{\sum({x_i}^2)} - \frac{\beta_0*\sum(x_i)}{\sum(x_i^2)}?$
Because I am treating $\beta_0$ as a constant?
I have also got that the variance of $\hat\beta_1$ is variance of $\left(\frac{\sum y_ix_i}{\sum x_i^2}\right)$? But how do I compute this?
Also how do I find a 100(1-$\alpha$)% confidence interval for $\beta_1$? How do I find MSE?
Thanks 
 A: I assume that $y_i=\beta_0+\beta_1 x_i +\varepsilon_i$
Estimation of beta 1
why would 

$$\hat \beta_1=\frac{\sum(y_ix_i)}{\sum({x_i}^2)} - \frac{\beta_0*\sum(x_i)}{\sum(x_i^2)}?$$

the intercept $\beta_0$ known, the least squares estimator of $\beta_1$ is still
\begin{align*}
\hat\beta_1 &=\frac{cov(x,y)}{var(x)}\\
&=\frac{\sum (x_i-\bar x)(y_i-\bar y)}{\sum (x_i-\bar x)}\\
&=\frac{\sum (x_i y_i-\bar xy_i-x_i\bar y+\bar x\bar y)}{\sum (x_i-\bar x)}\\
&=\frac{\sum (x_i y_i)-\bar x\bar y}{\sum (x_i-\bar x)}\\
\end{align*}
even if: $\bar y=\beta_0+\beta_1\bar x+\bar\varepsilon$ I can't see where you can place $\beta_0$
variance of beta 1
I don't remember that 

$var(\hat \beta_1)=\left(\frac{\sum y_ix_i}{\sum x_i^2}\right)$

To my mind it is rather:
$$var(\hat\beta_1)=\frac{\sigma^2}{(\sum x_i-\bar x)^2}$$
If you want to compute it, please make you confortable and use the following set of data.
$$\begin{array}{lll}
\hline
   t & y_i & x_i  \\
\hline
   1 & 100 & 100  \\
   2 & 106 & 104 \\
   3 & 107 & 106  \\
   ... & ... & ... \\
   9 & 137 & 126 \\ 
\hline
\end{array}$$
I have not yet studied confidence tests.
