Meaning of variance term in confidence interval for Multiple Linear Regression I am currently struggling on the meaning of the variance term $\sigma$ in the equation for computing the variance and the confidence interval of the mean reponse for a MLR:
$$
Var[\hat{y}(x_0)]=\sigma^2\cdot x_0'(X'X)^{-1}x_0
$$
and
$$
\hat{y}(x_0) - t_{\alpha /2, df(error)}\sqrt{\hat{\sigma}^2\cdot x_0'(X'X)^{-1}x_0} 
$$
$$
\leq \mu_{y|x_0} \leq \hat{y}(x_0) + t_{\alpha /2, df(error)}\sqrt{\hat{\sigma}^2\cdot x_0'(X'X)^{-1}x_0}  .
$$
(both formulas are taken from Myers, Montgomery, Anderson-Cook, "Response Surface Methodology" fourth edition, page 33-34)
1) Does $\sigma^2$ represents the variance of my data, or does it represent the variance of my Errors ?
Usually one uses the MSE as an estimator for $\sigma^2$, thus using the variance of my errors and not only of the data.
2) If I have already done a Measuring System Analysis (MSA), can I use the value of the calculated variance instead of the MSE of the Regression, because it would be a better estimator ?
 A: MSE measures the variance of the error. To be clear -- that's the variance of the model errors, not the variance of the data. You can see this by looking at $SSE = (y_i - f(x_i))^2$. $SSE$ gives the squared difference between the observed and fitted values. Linear regression models are fit by minimizing $MSE$. From the Gauss-Markov theorem, we know that minimizing $MSE$ (i.e. using the "ordinary least squares" estimator) gives the best linear unbiased estimator of the coefficients, where "best" means the estimator with the lowest variance.
So the algorithm used to calculate an OLS regression model depends on the use of $MSE$, which estimates the variance of the model error (not the variance of the data), and gives the best linear unbiased estimator of the model coefficients (assuming the assumptions of regression, uncorrelated errors with expectation 0 and equal variance, are met). So there isn't a straightforward way to swap in a different estimate of variance, and it might not be what you're looking for anyway (again, variance of data vs. variance of model error). Furthermore, using a different approach will result in estimated coefficients no better than those obtained through OLS regression (based on MSE).
A: The value $\sigma^2 = \mathbb{V}(\varepsilon_i)$ is the error variance in the regression model, and the variance result in your post is a consequence of the underlying variance of the OLS coefficient estimator:
$$\mathbb{V}(\hat{\boldsymbol{\beta}}) = \sigma^2 (\mathbf{x}^\text{T} \mathbf{x})^{-1}.$$
Since $\hat{y}(\mathbf{x}_0) = \mathbf{x}_0^\text{T} \hat{\boldsymbol{\beta}}$ you can use the ordinary rules for variances of random vectors to obtain
$$\mathbb{V}(\hat{y}(\mathbf{x}_0)) = \mathbb{V}(\mathbf{x}_0^\text{T} \hat{\boldsymbol{\beta}}) = \mathbf{x}_0^\text{T}  \mathbb{V}(\hat{\boldsymbol{\beta}}) \mathbf{x}_0 = \sigma^2  \cdot \mathbf{x}_0^\text{T} (\mathbf{x}^\text{T} \mathbf{x})^{-1} \mathbf{x}_0.$$
The formula for the confidence interval then follows from the following pivotal quantity:
$$\frac{\hat{y}(\mathbf{x}_0) - \mu_0}{\hat{\sigma}/df_{Res}} \sim \text{Student's T}(df_{Res}),$$
where $\mu_0 = \mathbb{E}(y(\mathbf{x}_0))$ and $\hat{\sigma}$ is the standard bias-corrected MLE in the linear regression.  Now, there is no particular reason you could not substitute this with a different estimator for $\sigma$ if you want to, but bear in mind that it could change the distribution of the pivotal quantity you are using to form your confidence interval.  So the thing you would need to do if you want to substitute a different estimator is to see how this would affect the distribution of the newly created quantity.  Many variance estimators have an asymptotic chi-squared distribution (via the CLT) so you might end up with the same distribution, and hence the same form for your confidence interval, but you should still check this.
