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I'm trying to understand the definition of Partial R-squared values in the context of a regression model. Does anyone have a layman's definition or an intuitive example that might help me better understand what it means?

Thank You!

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Let's look at the equation.

$$ R^2_{partial} = \dfrac{SSE(reduced) - SSE(full)}{SSE(reduced)} $$

In words, this compares the reduction in SSE from the reduced to the full model to the SSE of the reduced model. (We know the full model will have lower SSE.)

When the reduced model is the intercept-only model, the reduced model always predicts $\bar{y}$. This means that $SSE(reduced) = SSTotal$. Thus...

$$ R^2_{partial} = \dfrac{SSTotal - SSE(full)}{SSTotal} $$

This is the usual $R^2!$

Therefore, much as $R^2$ describes the variance (proportional to $SSE$) of a model, compared to the variance of the intercept-only model (naïvely guessing $\bar{y}$, no matter the predictor values), partial $R^2$ describes the variance explained by a model, compared to a more complex model than the intercept-only model.

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  • $\begingroup$ Thank you for your explanation! This is very helpful and illuminating. $\endgroup$ – Sharif Amlani Jun 29 at 7:28
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See this answer by amoeba.

This site and this site make it clear that the partial R-squared (also known as the coefficient of partial determination) can be used to assess the goodness of fit of a reduced multiple linear regression model as compared to a full(er) model.

The partial R-squared gives the proportion of variation explained by the explanatory variables in the full(er) model that cannot be explained by the explanatory variables in the reduced model. If the reduced model is a good fit compared to the full(er) model, then it will have a low partial R-squared.

EDIT: Please note that I just learned about this topic to answer this question because I had never heard of it before.

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  • $\begingroup$ This is consistent with how $R^2$ compares your model fit to the naïve model that always predicts $\bar{y}$ (intercept-only model). $\endgroup$ – Dave Jun 16 at 19:24
  • $\begingroup$ Thank you so much for your explaination! It is very good, especially considering you just learned it! Thank You! $\endgroup$ – Sharif Amlani Jun 29 at 7:27

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