Why does a paired t-test (when appropriate) result in better variance?

I didn't quite get the Wikipedia explanation here:

http://en.wikipedia.org/wiki/Paired_difference_test#Use_in_reducing_variance

I agree that both the unpaired and paired means are the same...then I see how

$\text{var}(\bar{Y_2} - \bar{Y_1})$

will include a covariance term..but what is the alternative variance? Take the paired differences first, and then the variance of the result? And how would that be any different?

Let's say we have the two conditions in Table 1. Each condition has a variance of 4 yielding a pooled variance of 4 as well and we double that to get the variance of the effect, 8. What if they were actually paired values and qualify for a paired t-test? We take the variance of the differences, the variance of the actual effect, which can be seen from the table to be 0 because they're all equal. This is the kind of thing that can happen when you have a paired test and how it can be more sensitive with a smaller standard error.

Table 1.

A1  A2  A1-A2
11   5   6
13   7   6
15   9   6
var(A1) = 4
var(A2) = 4
var(A1-A2) = 0
• In this case, if we were testing for pair difference of means equal to say $2$ wouldn't the t-stat blow up? Since $t = (\bar{X}_D - \mu_0)/(s_D / \sqrt{n})$ and $s_D = 0$? Sep 20 '11 at 5:17
• Palace Chan... yes. :) But that's not the point of having the variance go to 0 in the example. I'm just showing how the paired t-test denominator can be lower than the independent test. It's because A1 and A2 are correlated. In this case the correlation is 1.0, and therefore the t-test can't be done. But that's a rare incidence just provided as an extreme example with easy math one can do off the top of their head.
– John
Sep 20 '11 at 9:32