Consider linear regression model $$y = X\beta + e$$
I want to construct confidence interval for fitted values, i.e. $\hat{y} = X\hat{\beta}$. Define linear fucntion of $\beta$ as $$f(\beta) = X\beta$$ and $$f'(\beta) = X$$
Now by Delta Method $$\sqrt{n}(f(\hat{\beta}) - f(\beta)) \xrightarrow{d} f'(\beta)z$$ where $z \sim N(0, V_{\beta})$
$$Var(\sqrt{n}(f(\hat{\beta}) - f(\beta))) = XV_{\beta}X^T$$ Hence $$\sqrt{n}(f(\hat{\beta}) - f(\beta)) \xrightarrow{d} N(0, XV_{\beta}X^T) $$
How do I proceed from now? I can construct 95% confidence interval for the $\sqrt{n}(f(\hat{\beta}) - f(\beta))$ but it does not give much information since I don't know $f(\beta)$. It is stated in the book by Hansen (freely available online, section 6.14) that this interval should be $$CI = x_i^T\hat{\beta} \pm 1.96\sqrt{x_i^T\hat{V}_{\hat{\beta}}x_i}$$ where $\hat{V}_{\hat{\beta}}$ is estimated var-covar matrix for $\hat{\beta}$. Probably $\sqrt{X\hat{V}_{\hat{\beta}}X^T}$ denotes standard error term.
So far I have two questions: (1) how do you come up with Hansen's formula and (2) why dimensions don't match in this formula. $X\hat{V}X^T$ should have dimension $n \times n$ since $X$ is $n\times k$ and $V$ is $k\times k$. However, $X\beta$ is $n\times 1$ matrix.
EDIT When I read my question I realized that if $f(\hat{\beta_i}) - f(\beta_i) \sim N(0, \sigma_i^2)$, then CI for $f(\beta_i)$ is given as $$CI = f(\hat{\beta}_i) \pm 1.96\sqrt{\sigma_i^2}$$
So, assuming that $n$ is big enough one could claim that $$\sqrt{n}(f(\hat{\beta}) - f(\beta)) \sim N(0, XV_{\beta}X^T)$$ But then $$(f(\hat{\beta}) - f(\beta)) \sim N(0, \frac{1}{n}XV_{\beta}X^T)$$ So, I would reformulate my first question: why there is no division by $n$ in Hansen's formula?