# Slope uncertainty of linear regression with negative $R^2$ value

When I have a linear regression and I want to determine uncertainty in the slope from the quality of the fit (ignoring any uncertainty from error bars for now), I generally use

$$\sigma_m = m \sqrt{\frac{1/R^2 - 1}{n-2}}$$

where $$R^2$$ is the coefficient of determination, $$n$$ is the number of data points, $$m$$ is the slope, and $$\sigma_m$$ is the uncertainty in the slope.

For a set of data that is highly non-linear, and thus has a very low-quality fit, $$R^2$$ may become negative. However, when $$R^2 \leq 0$$, the argument of the square root becomes negative, and thus the uncertainties become imaginary. Is there a method for determining uncertainty due to the quality of a fit under these circumstances?

• Can you provide a reference for the formula you are using? Typically, standard errors, which seem to be what you are trying to calculate are always positive. Jun 3, 2020 at 22:30

When there is an intercept, in-sample $$R^2 \ge 0$$, so there is no risk of an imaginary root. Even $$R^2=0$$ is so unlikely in real (or even simulated) data that I would consider it to be practically impossible. You can use your equation without fear of an imaginary root or dividing by zero.
If you substitute above function to the numerator of your $$\sigma_m$$, we get $$\frac{SS_{total}}{SS_{total}-SS_{res}}$$. The uncertainty seems to depend on which direction SS of residual differs from total SS and n-2 is just a standardized term. I think you can add an absolute calculation outside the numerator to make it non-negative without changing the meaning this formula wants to express.