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smallest value is 0 for Ri = 1, R2 = 2,...,Rn =n), rank sequences, compute d for each one, and then
s0 Ho should be rejected in favor of H, ifd < c. When obtain the null distribution of D. See the Lehmann
Ho is true, any sequence of ranks has probability I/n!. book (in the chapter bibliography), for more infor-
Use this to find c for which the test has a level as close mation.]
The Wilcoxon Rank-Sum Test
When at least one of the sample sizes in a two-sample problem is small, the r test
requires the assumption of normality (at least approximately). There are situations,
though, in which an investigator would want to use a test that is valid even if the
underlying distributions are quite nonnormal. We now describe such a test, called
the Wilcoxon rank-sum test. An alternative name for the procedure is the Mann—
Whitney test, although the Mann-Whitney test statistic is sometimes expressed in a
slightly different form from that of the Wilcoxon test. The Wilcoxon test procedure
is distribution-free because it will have the desired level of significance for a very
large class of underlying distributions.
ASSUMPTIONS X, ... , Xm and Y;, ... , Y, are two independent random samples from
continuous distributions with means jz; and fl, respectively. The X and Y
distributions have the same shape and spread, the only possible difference
between the two being in the values of j4, and fly.
When Ho: ft, — fy = Ao is true, the X distribution is shifted by the amount Ao to the
right of the Y distribution; whereas when Hp is false, the shift is by an amount other
than Ao.
Development of the Test When m = 3, n = 4
Let’s first test Ho: 4; — fy = 0. If yz; is actually much larger than ji, then most of
the observed x’s will fall to the right of the observed y’s. However, if Ho is true, then
the observed values from the two samples should be intermingled. The test statistic
will provide a quantification of how much intermingling there is in the two samples.
Consider the case m = 3, n = 4. Then if all three observed x’s were to the
right of all four observed y’s, this would provide strong evidence for rejecting Hp in
favor of Hy: 4; — fy # 0, with a similar conclusion being appropriate if all three
x’s fall below all four of the y’s. Suppose we pool the x’s and y’s into a combined
sample of size m + n = 7 and rank these observations from smallest to largest,
with the smallest receiving rank 1 and the largest, rank 7. If either most of the
largest ranks or most of the smallest ranks were associated with X observations, we
would begin to doubt Ho. This suggests the test statistic
W = the sum of the ranks in the combined sample (14.1)
associated with X observations ,
For the values of m and n under consideration, the smallest possible value of W is
w = 1+2+3 = 6 (if all three x’s are smaller than all four y’s), and the largest
possible value is w = 5 + 6 + 7 = 18 (if all three x’s are larger than all four y’s).
--- Trang 780 ---
14.2 The Wilcoxon Rank-Sum Test 767
As an example, suppose x, = —3.10, x, = 1.67, x3 = 2.01, y, = 5.27,
y2 = 1.89, y3 = 3.86, and y, = .19. Then the pooled ordered sample is —3.10, .19,
1.67, 1.89, 2.01, 3.86, and 5.27. The X ranks for this sample are 1 (for —3.10), 3 (for
1.67), and 5 (for 2.01), so the computed value of Wisw = 1+3+5=9.
The test procedure based on the statistic (14.1) is to reject Ho if the computed
value w is “too extreme” — that is, > ¢ for an upper-tailed test, < ¢ for a lower-
tailed test, and either > c, or < c> for a two-tailed test. The critical constant(s) c
(ci, €2) should be chosen so that the test has the desired level of significance %. To
see how this should be done, recall that when Hp is true, all seven observations
come from the same population. This means that under Ho, any possible triple of
ranks associated with the three x’s — such as (1, 4, 5), (3, 5, 6), or (5, 6, 7) — has
the same probability as any other possible rank triple. Since there are (G) =35
possible rank triples, under Hy each rank triple has probability 1/35. From a list of
all 35 rank triples and the w value associated with each, the probability distribution
of W can immediately be determined. For example, there are four rank triples that
have w value 11 — (1, 3, 7), (1, 4, 6), (2, 3, 6), and (2, 4, 5) — so PW = 11) =
4/35. The summary of the listing and computations appears in Table 14.3.
Table 14.3 Probability distribution of W(m = 3, n = 4) when Ho is true
PW =w) 1 1 2 3 4 4 5 4 4 3 2 1 1
The distribution of Table 14.3 is symmetric about w = (6 + 18)/2 = 12,
which is the middle value in the ordered list of possible W values. This is because
the two rank triples (r, s, t) (with r < s < t) and (8 — t, 8 — s, 8 — r) have values
of w symmetric about 12, so for each triple with w value below 12, there is a triple
with w value above 12 by the same amount.
If the alternative hypothesis is Hy: ft; — ty > 0, then Ho should be rejected
in favor of H, for large W values. Choosing as the rejection region the set of
W values {17, 18}, « = P(type I error) = P(reject Hy when Ho is true) = P(W =
17 or 18 when Ap is true) = 4 +4 = % = .057; the region {17, 18} therefore
specifies a test with level of significance approximately .05. Similarly, the region
{6, 7}, which is appropriate for H,: 4) — fy < 0, has « = .05S7 = .05. The region
{6, 7, 17, 18}, which is appropriate for the two-sided alternative, has « = * = .114.
The W value for the data given several paragraphs previously was w = 9, which is
rather close to the middle value 12, so Hp would not be rejected at any reasonable
level « for any one of the three H,’s.
General Description of the Rank-Sum Test
The null hypothesis Ho: 1; — fy = Apo is handled by subtracting Ay from each X;
and using the (X; — Ao)’s as the X;’s were previously used. Recalling that for any
positive integer K, the sum of the first K integers is K(K + 1)/2, the smallest
possible value of the statistic W is m(m + 1)/2, which occurs when the (X; — Ao)’s
are all to the left of the Y sample. The largest possible value of W occurs when the