Mann-Whitney Test

1
Mann-Whitney Test

• If X1, X2, Xm is a sample of size m from a
  population, and Y1, Y2, Yn is a sample of size
  n from another population, then the Mann-Whitney
  statistic, U, is the number of (Xi,Yj) pairs for
  which XiltYj. U is used to test the same null
  hypothesis of equal population distributions and
  p-values are given in Table A4.
• Go over Example 2.6.1 - use R
• First note that W (sum of ranks, the Wilcoxon
  statistic) and U (the Mann-Whitney statistic) are
  equivalent
• Let W2sum of ranks of the Y's and note that for
  any Yj, R(Yj) of X's lt Yj of Y's lt Yj
• So W2 Sum over all j'sR(Yj)
• sum over all j's of X's lt Yj of Y's lt Yj
  
• U (123n) , since we may assume the Y's
  are ordered
• Un(n1)/2
• NOTE In R, the W given in wilcox.test is U
• (or mn - U )
• If there are ties, then Upairs of XltY
  1/2(ties)

2

• Review the relationship between a confidence
  interval and hypothesis testing in the parametric
  case
• If the null hypothesis of F1F2 is rejected then
  one possible alternative is the so-called shift
  alternative F1(x) F2(x-D) where D can be
  thought of as the difference in means (or
  medians) do a sketch.
• A consequence of the truth of the shift
  alternative is that X and Y D have the same
  distribution. So think of D as the difference of
  the medians - we'll use the Hodges-Lehmann
  estimate of D H-Lmedian(all pwds)
• The 95 confidence interval for D is found by
• get all the pwds of X-Y
• arrange them from smallest to largest
• find 2 numbers ka, and kb s.t.
• P(ka lt U lt kb ) P(pwds(ka ) lt D lt pwds(kb
  )) as close to the level of confidence as
  possible
• for large samples, we'll use the normal
  approximation (more on this later)
• Now go over Example 2.6.2 on page 46ff - use the
  R code in R4.doc
• Redo problem 4 on page 73 - get a 95 CI for D
  and its H-L estimate. Use R.