After fitting the regression model, $$y=\beta_0 + \beta_1 x_1 + \beta_2 x_2 + \beta_3 x_3 + \epsilon$$ on 15 cases, it is found that the mean square errors $$s^2 = 3$$ and
$$(X^T X)^{-1}= \begin{bmatrix} 0.5 & 0.3 & 0.2 & 0.6 \\ 0.3 & 6.0 & 0.5 & 0.4 \\ 0.2 & 0.5 & 0.2 & 0.7 \\ 0.6 & 0.4 & 0.7 & 3.0 \\ \end{bmatrix}$$
This is a practice problem for my midterm... I cannot figure out how I would figure out the estimated correlation between $$\hat\beta_1 \mbox{ and } \hat\beta_3.$$
Write $\hat{\mathbf\beta} = (\hat\beta_0,\hat\beta_1,\hat\beta_2,\hat\beta_3,)'$. You have
$$ \hat{\mathbf\beta} = (X'X)^{-1}X'y $$
You're interested in the variance-covariance matrix of $\hat{\mathbf\beta}$, which I will denote by $\text{cov}(\hat{\mathbf\beta} )$:
$$ \text{cov}(\hat{\mathbf\beta} ) = (X'X)^{-1}X'\text{var(y)}X(X'X)^{-1} = \text{var(y)}(X'X)^{-1}X'X(X'X)^{-1} = \text{var(y)}(X'X)^{-1} $$
You'll get an estimate for $\text{var(y)}$ by using $s^2$ appropriately, and you'll get the covariance between $\hat\beta_1$ and $\hat\beta_3$ by looking at the element $(2,4)$ of the covariance matrix $\text{cov}(\hat{\mathbf\beta} )$ (this is not element $(1,3)$ because the index $i$ in $\hat\beta_i$ starts at 0 for the intercept...).
From there, you can compute the correlation between $\hat\beta_1$ and $\hat\beta_3$ by standardizing appropriately, using the variances of $\hat\beta_1$ and $\hat\beta_3$, which you get from the elements $(2,2)$ and $(4,4)$ of $\text{cov}(\hat{\mathbf\beta} )$.
