Can we calculate the standard error of prediction just based on simple linear regression output? The standard error of prediction in simple linear regression is $\hat\sigma\sqrt{1/n+(x_j-\bar{x})^2/\Sigma{(x_i-\bar{x})^2}}$.
My question is to calculate the standard error of prediction for $pop=1029$ just based on the following regression output. I can get all except $\bar{x}$. And I also know how to calculate the approximate standard error of prediction based on the standard errors of intercept and coefficient of $pop$, ignoring their correlation.

 A: The question is to calculate the following statistic from the above regression output:
$$s.e.(\hat\mu|x_j)=\hat\sigma\sqrt{1/n+(x_j-\bar{x})^2/\Sigma{(x_i-\bar{x})^2}}.$$
The answer is inspired by @whuber:


*

*get $\hat\sigma$ from $\hat\sigma^2=SS_{Residual}/(n-p-1)$,
where $p=1$;

*$n$ and $x_j$ are known;

*obtain $\bar{x}$ from
$\hat{Var}(β_{cons})=\hatσ^2(1/n+\bar{x}^2/\Sigma{(x_i-\bar{x})^2})$;

*$\Sigma{(x_i-\bar{x})^2}=SS_{Model}/\hat{\beta}_{pop}^2$.

A: For simplicity, we are working with the following model:
$$y=\beta_0 + \beta_1x + \varepsilon,$$
where $\varepsilon\sim N(0, \sigma^2)$.
Now suppose that for $x=5$, we would like to predict $E(y|x=5) = \beta_0 + 5\beta_1$, denoted by $pre$.
We note that $pre$ is just a value if we know $\beta_0$ and $\beta_1$. However, they are unknow and each of them has its own sampling distribution (where the sd of this distribution is called the se). Therefore, the standard error associated with $pre$ is computed as:
$$Var(pre) = Var(\beta_0 + 5\beta_1) = Var(\beta_0) + 10 Cov(\beta_0, \beta_1) + 25Var(\beta_1).$$
Now you can see that you cannot compute this variance if you do not know the covariance $Cov(\beta_0, \beta_1)$.
A: See Section 6-4a of Wooldridge (2020), Introductory Econometrics: A Modern Approach, 7ed, Cengage.
Variance of prediction error
It's not $\sigma^2 [ 1/n + (x_j-\bar{x})^2 / \sum_i (x_i - \bar{x})^2 ]$.
For the correct expression, let me use vector notations. The model is $y=x\beta + u$, where $x=(1,pop)$. Let $x_0 = (1,1029)$. The value to predict ($y_0$) is not yet labeled. The predictor is $\hat\theta = x_0 \hat\beta$, where $\hat\beta$ is the OLS estimator, and the label to predict is $x_0 \beta + u_0$. The prediction error is thus $y_0 - \hat\theta = u_0 - x_0 (\hat\beta-\beta)$.
As $\hat\beta$ is a function of the sample and $u_0$ (out of sample) is assumed to be independent of the sample, the variance of the prediction error is $\sigma^2 + x_0 V(\hat\beta) x_0'$, where $'$ stands for transpose. As $V(\hat\beta) = \sigma^2 (X'X)^{-1}$, where $X$ is the feature matrix (including the constant term in the first column), the variance of the prediction error is $\sigma^2 [ 1+ x_0 (X'X)^{-1} x_0' ]$ so the standard error is
$$\sigma \sqrt{1 + x_0 (X'X)^{-1} x_0'}.$$
Calculation of se(prediction error)
If we change the variable $pop$ to $pop-1029$, then $\hat\theta$ is reported as the intercept estimate:
/* Stata */
gen pop2 = pop - 1029
reg fuel pop2

You see that the reparameterizated model $fuel = \theta + \beta_1 (pop-1029) + u$ gives $\theta = \beta_0 + 1029 \beta_1$ so the intercept estimator $\hat\theta$ is your predictor.
The reported standard error is, however, $se(\hat\theta) = \hat\sigma \sqrt{x_0 (X'X)^{-1} x_0'}$, not the desired $\hat\sigma \sqrt{1+x_0 (X'X)^{-1} x_0'}$. For the correct one, you can simply compute $\sqrt{\hat\sigma^2 + se(\hat\theta)^2}$.
