1. pairing data values (before-after, method1 vs
pairing data values (before-after, method1 vs. method2 on same subjects, subject is own control, etc.) often reduces variability and the analysis reduces to single sample methods.
but be careful about giving the same subject both treatments - if the treatments are drugs for example, what about residual effects of the first drug? how long does it take to completely leave the subject's system?
see the example data in Table on page both methods of estimating calorie intake are given to 5 subjects (though not mentioned, I would randomize the order of the method - i.e., flip a coin to see which of the 2 methods is asked first.)
if there were no differences between the two methods, then the 2 calorie values could have come from either method, so the difference between the two could be either + or - … see Table on page this lists all the possible +/- differences in calories as estimated by the two methods (note there are 25 = 32 ( ) such permutations).
now use this table to perform a permutation test: compare our observed mean difference with the distribution of all possible mean differences…

2. When there are more than a few subjects, 2n becomes too large, so we'll do random sampling of signs of differences by randomly picking the sign (+1 or -1) and multiplying it by the absolute value of the differences, then get the mean difference. Do this 5000 times and compare our observed mean difference to this permutation distribution.
HW (due next Tuesday): Do a permutation test (randomly selected permutations, not all 217=131,072 possible differences in signs!) for the data in Table on page Use R as we've been doing. HINT: See section to guide you through the process. Look at the function sample(c(-1,1),1,prob=c(.5,.5))
Instead of using the mean difference as the test statistic, we may use either S+ or S- , defined as the sum of the positive and negative differences, respectively.
Since
we see that either one of the sums may be used as the test statistic in place of the mean of the differences - you may also use one of the sums in your permutation test if you'd like to …
Go over the formulas in section 4.1.3

3. Assuming no difference in treatments in this paired-comparison situation, let’s find the means and variances of the permutation distributions of the important statistics in this context (they will tend to normality):
Let Ui = +/- 1 with probability = .5 First show that E(Ui ) = 0 and Var(Ui ) = 1
Now so
Now do similar computations of mean and variance for
Thus for large samples we may use the normal approximations to the permutation distributions as in Example 4.1.2
HW: Complete reading section 4.1 – we’ll begin section 4.2 next time...

4. One last note about this paired data procedure:
We may use it to test hypotheses about the median of a symmetric population of measurements as follows... If Xi = the ith observation in the sample from this population and if M is the hypothesized median, then if the null hypothesis is true, Di = Xi – M is equally likely to be positive or negative and so the Di ‘s are symmetrically distributed around 0 so our paired data procedure may be used instead of the parametric one-sample t-test for example...