As part of assumption testing for ordinary least squares (OLS), I am examining the linearity between the variables in my model. I'm using scatterplots between the independent variables (IVs) and dependent variable (DV) to do this. I'm not sure if the graph below violates the linearity assumption or not. It seems that there is no relationship between my IV (dominance) and DV (leadership potential)... yet, there's also a slight quadratic shape (inverted U-shape) present. Is this problematic enough to require a transformation? Or, can I assume linearity between these variables and proceed with OLS? Thanks!

enter image description here

*Updated to add residuals vs. fitted plot.

enter image description here

*Updated to add jitter plot with LOESS imposed. enter image description here

  • 1
    $\begingroup$ It looks fine, usually you would fit the regression and plot the residuals vs fits and then assess linearity. $\endgroup$ – Cameron Chandler Oct 26 at 7:26
  • $\begingroup$ Thanks, Cameron! The residuals vs fits plot (now included above) looks quite similar, so it looks like the linearity assumption has been met. Appreciate your help! $\endgroup$ – Christa Oct 26 at 13:10
  • $\begingroup$ Forgot to tag you, @CameronChandler $\endgroup$ – Christa Oct 26 at 13:19
  • 1
    $\begingroup$ Two issues: (1) the data are obviously discrete, so you cannot tell from the graph how many data points are on each dot, and this matters a great deal in assessing the curvature of the response function. The solution is to show the jittered scatterplot. (2) The simplest way to assess the degree of curvature is to superimpose the LOESS fit (calculated from the raw, non-jittered data) over the jittered scatterplot. Then you will have a better idea of the degree of curvature in your process, and can decide whether you need to do something about it. $\endgroup$ – BigBendRegion Oct 26 at 15:56
  • $\begingroup$ Thanks so much, @BigBendRegion. I've included a new scatterplot above with the LOESS fit line imposed. There is a slight curvature, but I'm not sure if it's enough to violate the linearity assumption? $\endgroup$ – Christa Oct 26 at 17:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.