Simple Linear Regression: why do we estimate conditionla mean when we can estimate the parameters? If we can estimate $\beta_0$, $\beta_1$ and $\sigma^2$ in a simple linear regression model, why do we want to estimate the conditional mean $\beta_0+\beta_1x_0$ at a value $x_0$? I mean we have already all the information that we might possibly need. If I want to estimate the conditional mean I can just substitute the estimates $\hat{b_1}$ and $\hat{b_0}$ to obtain $\hat{y}=\hat{b_0}+\hat{b_1}x_0$ which is indeed the fitted value.
 A: You have to distinguish between the prediction of the conditional mean at $x=x_0$ and the prediction of a given point at $x=x_0$. 
Recall that your population conditional mean model is $E[Y|X=x]=g(x) = \beta_0 + \beta_1x$, thus you are estimating these regression coefficient and then just plug in the $x_0$ and the point estimators of the $\beta$-s, i.e., $\widehat{E[Y|X=x_0]}=\hat{\beta_0} + \hat{\beta_1}x_0$. This estimator depends on a random sample, hence it has a distribution. If you assume that your data generating model is $y_i = \beta_0 + \beta_1 x_i + \epsilon_i$, where $\epsilon_i \sim N(0, \sigma^2)$, then it is straight forward to show that
$$
\widehat{E[Y|X=x_0]} \sim N(\beta_0 + \beta_1x_0,  \sigma^2 (1/n + (x_0 - \bar{x})^2/S_{xx}).
$$
But what happens if you are interested in the prediction of $Y(x_0)$ itself and not its (conditional) mean? The point estimator is the same, i.e., 
$$
\hat{Y}(x_0)=\hat{\beta_0} + \hat{\beta_1}x_0,
$$
however now you have to consider also the noise term $\epsilon_0$, as such its distributions should be calculated by 
$$
\hat{Y}(x_0)=\hat{\beta_0} + \hat{\beta_1}x_0 + \epsilon_0,
$$
that is 
$$
\hat{Y}(x_0) \sim N(\beta_0 + \beta_1x_0,  \sigma^2 (1+1/n + (x_0 - \bar{x})^2/S_{xx}).
$$
Intuitively, the point estimator does not change because the best "guess" (in the given model) of $\epsilon_0$ value is $0$, however this consideration introduces much more noise to the estimator.  
