0
$\begingroup$

enter image description here

I think this graph shows homoscedasticity. Is this true?

$\endgroup$
1
  • 2
    $\begingroup$ Hello, and welcome to the community. While most of us might be able to infer that this is probably a predicted vs residual plot, it would still help to have a little bit of context (so, maybe tell us what we're looking at). $\endgroup$ – John Madden Dec 7 '15 at 22:11
-1
$\begingroup$

Honestly, the linear association between those two variables is too low to know for sure. I'd also go with homoskedasticity, but in this case it won't matter much since there is practically no correlation between the variables...

$\endgroup$
13
  • 1
    $\begingroup$ How did you get the strength of linear association between the variables out of that graph? (I'm genuinely curious, it can be hard to convey that on the internet) $\endgroup$ – John Madden Dec 7 '15 at 22:12
  • $\begingroup$ If you try to wrap the points within an ellipse, you'll get a very fat ellipse which almost resembles a circle. Also, since the scale of the variable on the y axis is smaller, that means the points are even more spread in the up-down direction than they look in the graph, which in this case will make the ellipse even fatter. You get high associations for instance (although there are other ways) when the scatterplot looks like a thin tilted ellipse. $\endgroup$ – Felipe Gerard Dec 7 '15 at 22:17
  • $\begingroup$ oh, I thought this was a predicted vs residual plot (the mean on the y axis seems to be zero). As i stated in my comment to the OP, the vague context of this graph makes it difficult to interpret. $\endgroup$ – John Madden Dec 7 '15 at 22:19
  • $\begingroup$ You are right. He's probably looking for random errors in order to have a "valid" regression. So zero association would be his best case scenario right? $\endgroup$ – Felipe Gerard Dec 7 '15 at 22:22
  • $\begingroup$ yes, i believe so. His graph seems to indicate constant error (if that is indeed predicted vs residual), and the regression I assume hes doing has that particular assumption met. $\endgroup$ – John Madden Dec 7 '15 at 22:24

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.