Suppose I have some slightly messy data with an r squared of 0.9, and so fits a line pretty well.

If I were to extrapolate based on the slope and intercept of the fit, I would expect my y values to be pretty close while the x values are close to the range of the data, but as my x values got farther out, I would expect more uncertainty.

What would be the best way to find how big the confidence interval is of y values.

So where 95% of scenarios so that match the same slope, intercept and r squared will pass through?


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

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

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


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_ribbon(aes(ymin = lwr, ymax = upr),alpha = 0.5)+
  geom_point(data = tibble(x,y), aes(x,y))


enter image description here

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

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

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

enter image description here

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.

  • $\begingroup$ Please review your formula, because it's not correct. $\endgroup$ – whuber Dec 6 '19 at 20:03
  • $\begingroup$ @whuber Which? There are 3. $\endgroup$ – Demetri Pananos Dec 6 '19 at 20:04
  • $\begingroup$ There's only one for the interval. See the duplicate for correct formulas. $\endgroup$ – whuber Dec 6 '19 at 20:05
  • 1
    $\begingroup$ @whuber I've removed them and linked to a better answer than my own for reference. $\endgroup$ – Demetri Pananos Dec 6 '19 at 20:15
  • $\begingroup$ That's a good one. Thanks for finding it! $\endgroup$ – whuber Dec 6 '19 at 20:36

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