Considering following linear multiple regression model: \begin{equation} y=X\beta + e, \end{equation} where observations $y\in\Re^n$, coefficents $\beta\in\Re^p$ and $e\sim N(0,\sigma I)$ is a white Gaussian noise term.
By the definition of the confidence interval: \begin{equation} \hat{y} \pm t_{n-p}^{\alpha/2} \times \mathrm{SE}_\bar{y} \end{equation} where $t_{n-p}^{\alpha/2}$ is the critical value, and standard error \begin{equation} \mathrm{SE}_\bar{y} = \frac{\sigma_y}{\sqrt{n}} \end{equation} I understand that \begin{equation} \mathrm{Var}(\hat{\beta})=\hat{\sigma}^2 (X^\top X)^{-1} \qquad\text{and} \qquad \mathrm{Var}(\hat{y}_{n+1})=\mathrm{Var}(x_{n+1}\hat{\beta})=\hat{\sigma}^2x_{n+1}^\top(X^\top X)^{-1}x_{n+1} \end{equation} My question is that why in post:Difference between confidence intervals and prediction intervals and Obtaining a formula for prediction limits in a linear model (i.e.: prediction intervals) they use square root of Variance \begin{equation} \hat{y}_{n+1} \pm t_{n-p}^{\alpha/2} \times \sqrt{\mathrm{Var}(\hat{y}_{n+1})} \end{equation} rather than the standard error?? Where is the $\frac{1}{\sqrt{n}}$?
