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When the data consisted of pairs (X1,¥,),-..,(Xn,¥,) and the differences
D, =X, —¥1,...,Dn =Xn —Yn were normally distributed, in Chapter 10 we
used a paired f test for hypotheses about the expected difference up. If normality
is not assumed, hypotheses about jp can be tested by using the Wilcoxon signed-
rank test on the D;’s provided that the distribution of the differences is continuous
and symmetric. [f X; and Y; both have continuous distributions that differ only with
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14.1 The Wilcoxon Signed-Rank Test 763
respect to their means (so the Y distribution is the X distribution shifted by
1, — fy = Mp), then D; will have a continuous symmetric distribution (it is not
necessary for the X and Y distributions to be symmetric individually). The null
hypothesis is Ho: lp = Ao, and the test statistic S, is the sum of the ranks
associated with the positive (D; — Ao)’s.
About 100 years ago an experiment was done to see if drugs could help people with
severe insomnia (“The Action of Optical Isomers, II: Hyoscines,” J. Physiol., 1905:
501-510). There were 10 patients who had trouble sleeping, and each patient
tried several medications. Here we compare just the control (no medication) and
levo-hyoscine. Does the drug offer an improvement in average sleep time? The
relevant hypotheses are Ho: [fp = 0 versus H,: [py < 0. Here are the sleep times,
differences, and signed ranks.
Patient 1 2 3 4 5 6 7 8 9 10
Control 0.6 if 25 28 29 30 3.2 47 55 62
Drug 25: 5.7 80 44 6.3 3.8 76 58 56 61
Difference -19 —46 -5.5 -16 -34 —8 -44 -ll -1 1
Signed rank —6 -9 -10 —-5 -7 -3 —8 —4 -15 15
Notice that there is a tie for the lowest rank, so the two lowest ranks are split
between observations 9 and 10, and each receives rank 1.5. Appendix Table A.12
shows that for a test with significance level approximately .05, the null hypothesis
should be rejected if s, < (10)(11)/2 — 44 = 11. The test statistic value is 1.5,
which falls in the rejection region. We therefore reject Hy at significance level .05
in favor of the conclusion that the drug gives greater mean sleep time. The
accompanying MINITAB output shows the test statistic value and also the
corresponding P-value, which is P(S, <1.5 when Hp is true).
Test of median = 0.000000 versus median < 0.000000
N
for Wilcoxon Estimated
N Test Statistic P Median
aiff 10 10 1.5 0.005 ~2.250 |
Efficiency of the Wilcoxon Signed-Rank Test
When the underlying distribution being sampled is normal, either the ¢ test or the
signed-rank test can be used to test a hypothesis about ju. The f test is the best test in
such a situation because among all level « tests it is the one having minimum f. It is
generally agreed that there are many experimental situations in which normality
can be reasonably assumed, as well as some in which it should not be. These two
questions must be addressed in an attempt to compare the tests:
1. When the underlying distribution is normal (the “home ground” of the t test),
how much is lost by using the signed-rank test?
2. When the underlying distribution is not normal, can a significant improvement
be achieved by using the signed-rank test?
If the Wilcoxon test does not suffer much with respect to the f test on the “home
ground” of the latter, and performs significantly better than the f test for a large number
of other distributions, then there will be a strong case for using the Wilcoxon test.
--- Trang 777 ---
764 —cuarrer 14 Alternative Approaches to Inference
Unfortunately, there are no simple answers to the two questions. Upon
reflection, it is not surprising that the ¢ test can perform poorly when the underlying
distribution has “heavy tails” (i.e., when observed values lying far from ju are
relatively more likely than they are when the distribution is normal). This is because
the behavior of the f test depends on the sample mean and variance, which are both
unstable in the presence of heavy tails. The difficulty in producing answers to the
two questions is that f for the Wilcoxon test is very difficult to obtain and study for
any underlying distribution, and the same can be said for the ft test when the
distribution is not normal. Even if f were easily obtained, any measure of efficiency
would clearly depend on which underlying distribution was assumed. A number of
different efficiency measures have been proposed by statisticians; one that many
statisticians regard as credible is called asymptotic relative efficiency (ARE).
The ARE of one test with respect to another is essentially the limiting ratio of
sample sizes necessary to obtain identical error probabilities for the two tests. Thus
if the ARE of one test with respect to a second equals .5, then when sample sizes are
large, twice as large a sample size will be required of the first test to perform as well
as the second test. Although the ARE does not characterize test performance for
small sample sizes, the following results can be shown to hold:
1. When the underlying distribution is normal, the ARE of the Wilcoxon test with
respect to the f test is approximately .95.
2. For any distribution, the ARE will be at least .86 and for many distributions will
be much greater than 1.
We can summarize these results by saying that, in large-sample problems, the
Wilcoxon test is never very much less efficient than the f test and may be much
more efficient if the underlying distribution is far from normal. Although the issue
is far from resolved in the case of sample sizes obtained in most practical problems,
studies have shown that the Wilcoxon test performs reasonably and is thus a viable
alternative to the ¢ test.
Exercises | Section 14.1 (1-8)
1 Reconsider the situation described in Exercise 32 of 7.02 735 734 7.17 728 7.77. 7.09
Section 9.2, and use the Wilcoxon test with x = .05 722 745 695 740 7.10 732 7.14
to test the relevant hypotheses.
. 4. A random sample of 15 automobile mechanics