Suppose we have a linear regression model of the following format : $$ y(x) = \beta_0 + \beta_1 x_1+ \beta_2x_2+\beta_3x_3+\epsilon$$
We know that the prediction interval associated with a level $\alpha$, for a new observation $x_0$, is :
$$ IC_{1-\alpha}=\left[\hat{y}_i \pm t_{\alpha/2, n-p-1}\times \sqrt{\hat{\sigma}^2(1+x_0^T(X^TX)^{-1}x_0)}\right]$$
The asked question is :
Without calculation and using approximations, give a prediction interval at $95\%$ for a given $(x_1,x_2,x_3)$.
Available information :
- Output of the summary of this regression model given by R (so we have the $\beta s$, $\hat{\sigma}^2$ and $std(\beta)$)
- The means and std errors of our variables.
We can approximate $t_{\alpha/2, n-p-1}$ by 2, how can we approximate $x_0^T(X^TX)^{-1}x_0$ ?
The covariance matrix of $\hat{\beta}$ you can obtain in R by
vcov(model). If you are unwilling to do that then can crudely approximate $\hat{\sigma}^2(X'X)^{-1}$ by a diagonal matrix where its diagonal consists out of the variances of each $\hat{\beta}_j$, respectively. These values (standard deviation) you can find in the summary output.