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How can adding a 2nd IV make the 1st IV significant?

For two of my main variables, the bivariate correlation coefficients (with the outcome measure) are non-significant (variable $A$ is $-.15$ with the outcome, and variable $B$ is $.04$ with the outcome, $p>.01$ in both cases). However, when entered in the second step of my regression model $A$ and $B$ add to the model ($R^2$ jumps from $.28$ to $.5$), and both variables are significant predictors ($p<.01$). Squared semi-partial correlation for variable $A$ is $.21$ and for variable $B$ it's $.12$.

I'm not sure what to make of this. Is the effect I'm seeing (in the regression model) genuine? Or is rendered suspect by virtue of the fact that the zero-order correlations are so weak? Any thoughts and comments would be appreciated.

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  • $\begingroup$ first of all what are the p-values actually if some are near the border of 0.01 that matters. would you still get this conflict if you use 0.05? $\endgroup$ – Michael R. Chernick Sep 28 '12 at 13:25
  • $\begingroup$ Have you tried plotting A vs your dependent variable and B vs your dependent variable? That's always a good first step! $\endgroup$ – wcampbell Sep 28 '12 at 13:36
  • $\begingroup$ Yes, both A and B correlate very poorly with the dependent variable. Yet, in the regression, they are significant predictors. In the write-up, it's a bit difficult to justify why I'm even including A and B in the analysis given that they correlate so poorly with the outcome variable. $\endgroup$ – Archaeopteryx Sep 30 '12 at 5:58
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This can happen if the variance that is captured in the first step is high enough to swamp out the bivariate correlation. Here is an example with made up data (using R):

x <- rnorm(100, sd=10)
z <- rnorm(100)
y <- x + z + rnorm(100, sd=0.1)

Obviously, y and z are related, but the correlation between them is going to be small because of all the variance due to x

cor.test(y,z)
Pearson's product-moment correlation

data:  y and z 
t = 0.5533, df = 98, p-value = 0.5813
alternative hypothesis: true correlation is not equal to 0 
95 percent confidence interval:
 -0.1421736  0.2494864 
sample estimates:
       cor 
0.05580281

But in a linear regression it will be a significant predictor:

dat <- data.frame(x=x, z=z, y=y)
m.1 <- lm(y~x, data=dat)
m.2 <- lm(y ~ x+z, data=dat)
anova(m.1, m.2)
Analysis of Variance Table

Model 1: y ~ x
Model 2: y ~ x + z
  Res.Df    RSS Df Sum of Sq     F    Pr(>F)    
1     98 95.280                                 
2     97  0.872  1    94.407 10500 < 2.2e-16 ***
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
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  • $\begingroup$ Does this mean that the effect is still genuine, just not captured by a correlation? $\endgroup$ – Archaeopteryx Sep 28 '12 at 13:22
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    $\begingroup$ That would be my interpretation: the bivariate correlation is obscured by noise that is cleaned up in the first step of the regression model. $\endgroup$ – Dan M. Sep 28 '12 at 16:55

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