What is the main motivation behind using linear regression to model linear data as opposed to finding the means at 2 values and inferring a line?

After doing some tests I know the answer to this. If the standard deviation is high enough, the means will not converge fast enough to give an acceptable error. Assume the distribution of residuals is uniform and normal. If you take $n$ samples from each of the 2 points, you can take the "realm of possibilities" as some confidence interval with bounds: $$ME = z^*\frac{\sigma}{\sqrt{n}}$$ For whatever z-score is acceptable. The maximum error is going to be 2 times this (the bottom of one interval to the top of the other). If the relationship is already very small, then you must have large $n$ anyways, but there also is the fact that $n$ is proportional to $\sigma^2$ so if the standard deviation is high, then n must be extremely high to get a good answer.

Is there a more official way to say this, or a resource I could look up that gives a formal reason/proof/explanation?

Also, could this possibly invalidate the results of matching if the underlying relationship is linear and the standard deviation is high? Or is matching merely looking for any answer, as opposed to a precise one?

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The Gauss-Markov Theorem is one major motivation to prefer least-squares regression over some other arbitrary linear estimator :) –  Macro Aug 1 '12 at 14:58
Are you referring to a question of experimental design, where (for example) you could choose $2n$ $x$ values and you might elect to take $n$ at one value of $x$ and another $n$ at a second value of $x$, as opposed to spreading all $2n$ $x$ values out? Or are you talking about regression with $(x,y)$ data? And what do you mean that "$n$ is proportional to $\sigma^2$? How is that possible when either you control $n$ or the data are already given to you? –  whuber Aug 1 '12 at 15:05
Hi @whuber you got the first one right. 2n x values, with n at an $x_1$ and n at an $x_2$. And when I say "n is proportional to $\sigma^2$ I mean that in general, if you have an experiment with high standard deviation in the results, you are going to have to set $n$ to be very high to get a good estimate. In my tests I could change $\sigma$ because I was running a simulation. –  Mike Flynn Aug 1 '12 at 15:15
And thank you @Macro I will definitely check that out. –  Mike Flynn Aug 1 '12 at 15:15
Mike OK, but exactly what do you mean by the "maximum error"? Error of what? Residuals? Coefficients? Predictions for $y$ at given values of $x$? Where you refer to a "good answer," what exactly constitutes an answer? –  whuber Aug 1 '12 at 15:23