Solved – Finding the the Confidence Interval with Linear Extrapolation

I'm assuming you are talking about an interval for the observations, rather than for the regression line.

Given the x, the outcome y is assumed to be normally distributed like so

$$ yvert x sim mathcal{N}(hat{beta}_0 + hat{beta}_1 x, hat{sigma}^2) $$

Here, $hat{sigma}^2$ has been estimated from the data. This thread seems to discuss the computation of both prediction and confidence intervals quite well.

In R, it is easy to get prediction intervals

library(tidyverse)   x = rnorm(100) xpred = seq(-3,3,0.01) y = 2*x+1+rnorm(length(x), 0, 2) model = lm(y~x) ypred = predict(model, list(x = xpred), interval = 'predict' ) %>%as.data.frame()  d = tibble(xpred=xpred) %>% bind_cols(ypred)   d %>%    ggplot(aes(xpred, fit))+   geom_line()+   geom_ribbon(aes(ymin = lwr, ymax = upr),alpha = 0.5)+   geom_point(data = tibble(x,y), aes(x,y)) 

Yielding

If instead you want a confidence interval for the regression line, then the variance conditional on x is given by

$$operatorname{Var}(y) = operatorname{Var}(hat{beta}_0) + x^2operatorname{Var}(hat{beta}_1) + 2xoperatorname{Cov}(hat{beta}_0, hat{beta}_1) = mathbf{x}^T Sigma mathbf{x}$$

Here, $mathbf{x} = [1,x]$. Using this, we can apply the standard confidence interval formula. Obtaining confidence intervals in R is the same procedure, except now we pass interval="conf" to the predict function. This yields

Note that the the precision is greatest near the sample mean of the x. As you extrapolate more and more, the uncertainty increases as evidenced by the widening of the confidence interval.