# Why does a paired t-test (when appropiate) 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?

Thanks.

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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

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 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 $s_D = 0$? – Palace Chan 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 Ah. Yes Gotcha. Thanks a lot! – Palace Chan Sep 21 '11 at 4:30