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Suppose that I have a linear regression: $ y = \beta_{0} + \beta_{1}x_{1} + \beta_{2}x_{2} + \epsilon $

I would like to demonstrate that one regression coefficients is significantly greater than the other:
$H_{0} = \beta_{1} > \beta_{2}$ or $H_{0} = \beta_{1} - \beta_{2} > 0$.

Therefore, I would conduct a t-test:

$t = \mu_{d} / \sigma_{d} $
with $\mu_{d} = \beta_{1} - \beta_{2} $ and $\sigma_{d} = \sqrt{\sigma_{\beta_{1}}^2 + \sigma_{\beta_{2}}^2 - Cov(\beta_{1}, \beta_{2})}$

and $Cov (\beta_{1}, \beta_{2}) = \sigma_{\beta_{1}} * \sigma_{\beta_{2}} * \rho_{\beta_{1}, \beta_{2}} $

As $\beta_{1}$, $\beta_{2}$, $\sigma_{\beta_{1}}$ and $\sigma_{\beta_{2}}$ are known from the regression results, the only mising part is the correlation of $\beta_{1}$ and $\beta_{2}$. However, I don't know how this can be computed and would be really thankful for any help!

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    $\begingroup$ Hi: You can calculate the covariance matrix $\Sigma = \sigma^2 (X^{\prime}X)^{-1}$. The correlation is then the square root of the element in the second row and the third column. $\endgroup$
    – mlofton
    Jun 16 '20 at 9:23
  • $\begingroup$ Thank you very much! $\endgroup$
    – Emil
    Jun 18 '20 at 4:01
  • $\begingroup$ no problem. glad to help. $\endgroup$
    – mlofton
    Jun 19 '20 at 2:22