# Inferencing from Pearson correlation and multiple linear regression

My hypothesis is: Company size (CSIZE) is positively correlated with Return on Assets (ROA).

I create a Pearson correlation matrix in Stata using pwcorr csize roa and it returns a correlation of 0.2 with a 5% significance level.

Now I also want create a multiple regression to test Company Age (CAGE) and Company Size (CSIZE) on Return on Assets (ROA).

I create a linear regression model in Stata using regress roa cage csize

I returns low R^2=0,007 and low F-statistic 1,08. Neither independent variable had significant T-statistic either.

My question is how to interpret such a result: The low R^2 means my model does not explain much of the variation. Plus it is not significant. But is that only for the combined effect of both independent variables (CAGE+CSIZE) on the dependent variable (ROA)? The individual t-test was also insigificant, so what about the result from my Pearson correlation?

I am having a hard time seeing how linear regression and Pearson correlation relates to each other, and what implications the results of both have on my hypothesis.

• It isn't possible for a Pearson correlation to be 1.2, because Pearson correlations are limited to [-1, 1]. Did you make a typo? Likewise, should "Plus it is significant" be "Plus it is not significant"? – Kodiologist May 20 '16 at 23:55
• Ah yes you are right. I just took a random number to illustrate. It's just an example. I guess that backfired :-) it has been corrected now. – Esben May 21 '16 at 7:51
• You will probably get better answers if you use the real numbers you obtained. – Kodiologist May 21 '16 at 11:26

First, I think regress is the multiple linear regression in stata, not correlate. Second, "theoretically," the relationship (or model) between your response and covariates cannot weaken (lower $R^2$) when you add variables (for more reading on this, I recommend google searching "degrees of freedom"). Third, if in the first case you mentioned, if correlation is .2, then $$R^2 = (.2)^2 = .04 < .07$$ meaning the response was explained somewhat better when you added the aditional covariate.