How to studentize residuals 
The lecture slide (from PennState Eberly College of Science, STAT 501, Lesson 11.3) says "an ordinary residual divided by an estimate of its standard deviation", but the standard deviation for (-0.2,0.6,-0.6,0.2) is square root of 2, which doesn't match the square of MSE*(1-h)?
 A: In the linear regression the variance of the residuals is given by $$\sigma^2 \left( \mathbf{I} - \mathbf{H} \right) $$
where $H$ is the hat matrix,
$$H = \mathbf{X} \left(\mathbf{X}^{\prime} \mathbf{X} \right)^{-1} \mathbf{X}^{\prime} $$.
You may verify this formula by writing the fitted values and the residuals as afunction of this matrix and using the basic sandwitch theorem along with the symmetry and idempotence of the hat matrix (recall that all projection matrices are symmetric and idempotent). We then have 
$$\mathbf{e} = \left(\mathbf{I} - \mathbf{H} \right) \mathbf{y}$$
and so 
$$\text{Var} \left\{\mathbf{e} \right\} = \left(\mathbf{I} - \mathbf{H} \right) \sigma^2 \mathbf{I} \left(\mathbf{I} - \mathbf{H} \right) = \sigma^2 \left(\mathbf{I} - \mathbf{H} \right)  $$
Thus we studentize the residuals by dividing with an estimate of this quantity. It's only $\sigma^2$ that is unknown here so we use the consistent MSE to esimate it. The MSE however is not exactly the empirical variance of the residuals. We have to adjust for the number of parameters in order to obtain an unbiased estimator, much like we do when we compute the sample variance. Therefore the formula we use is
$$MSE = \frac{\mathbf{e}^{\prime} \mathbf{e}}{n-k}$$
where $k$ is the number of estimated parameters, which is two for the two-variable regression model (slope and intercept). Thus, here we divide the sum of squared residuals by $2$ rather than $4$ and this yields the $0.4$ you asked about. It's worth noting that, contrary to the hypothesized error term, the residuals are correlated in the regression model.
