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I do not fully understand how to interpret the difference between two statistical models where they only differ based on whether a certain variable is included on the right hand side.

If the results do not change much, then do we say that that this omitted variable has little effect on the result?

If the coefficient on the variable of interest changes significantly in magnitude and is no longer significant, we say that the variable of interest on the right hand side is not robust to that omitted variable. That omitted variable effects the relationship between the variable of interest on the right hand side and the dependent variable.

What is the estimate becomes more statistically significant when this variable is included? It appears that including these variables makes the results more precisely estimated. How can this be? What does that mean about the relationship between the dependent variable and the main variable of interest on the right hand side if the relationship is more precisely estimated only with the additional control variable?

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What you're asking about centers on the concept of partial correlation, which is dealt with, among other places, at stats.stackexchange.com/questions/17336/… . –  rolando2 Mar 19 '12 at 1:34

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I think this hinges on what you means by "not changes much". The estimated parameters could change and so could their standard errors. That's two separate effects. Let's focus on the variance part for now.

Suppose the true DGP is really $y=x\beta+z\alpha+v$, but you leave out the relevant variable $z$ from your model, so you actually estimate $y=x\beta+(z\alpha+v)=x\beta+\epsilon$ using OLS. Your error term is now effectively larger since it includes an extra term. The estimator of the variance-covariance matrix of $\beta$ will be biased upward, because the estimator of $\sigma^{2}$, the variance of the error term, will be biased upward. This causes the inference about $\beta$ to be inaccurate. This is the case even if $z$ is orthogonal to the other explanatory variables. When you add $z$ back in, the variance should go down.

Does that make sense?

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