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+ +2 +3 —4 Ss 6 +1 +2 +3 +4 —5 10
-l 2 -3 -4 45 5 -l -2 -3 44 45 9
+1 -2 -3 -4 45 6 +1 -2 -3 44 45 10
ew | +2 =3: —4 +5 7 wa] +2 <3, +4 +5. if |
-l -2 43 -4 45 8 -l -2 43 44 45 12
+1 42 -3 -4 45 8 +1 42 -3 44 #45 #12
+1 -2 43 -4 459 +1 -2 43 44 #45 #13
-1 42 43 -4 45 10 =1 42443 #44 #45 #14
ht +2 +3 —4 +5 11 a +2 +3 +4 +5 15
Table 14.2 Null distribution of S, when n = 5
P(S4) 1/32 1/32 1/32 2/32 2/32 3/32 3/32 3/32
Sy 8 9 10 11 12 13 14 15
P(S4) 3/32 3/32 3/32 2/32 2/32 1/32 1/32 1/32
--- Trang 774 ---
14.1 The Wilcoxon Signed-Rank Test 761
is symmetric about 7.5 [more generally, symmetrically distributed over the possible
values 0, 1, 2,...,2(m + 1)/2]. This symmetry is important in relating the rejection
region of lower-tailed and two-tailed tests to that of an upper-tailed test.
For n = 10 there are 2'° = 1024 possible signed rank sequences, so a listing
would involve much effort. Each sequence, though, would have probability 1/1024
when Hp is true, from which the distribution of S,. when Hp is true can be easily obtained.
We are now in a position to determine a rejection region for testing Ho: i = 0
versus H,: ji>0 that has a suitably small significance level x. Consider the
rejection region R = {s, :s, > 13} = {13, 14,15}. Then
a = P(reject Ho when Hp is true)
= P(S, = 13,14, or 15 when Hp is true)
= 1/32 + 1/32 + 1/32 = 3/32
= .094
so that R = {13, 14, 15} specifies a test with approximate level .1. For the rejec-
tion region {14, 15}, « = 2/32 = .063. For the sample x; = .58, x2 = 2.50,
x3 = —.21, x4 = 1.23, xs = .97, the signed rank sequence is —1, +2, +3, +4, +5,
sos, = 14 and at level .063 Ho would be rejected.
A General Description of the Wilcoxon Signed-Rank Test
Because the underlying distribution is assumed symmetric, j = fi, so we will state
the hypotheses of interest in terms of j rather than ji.'
ASSUMPTION X,, Xo, ... , X, is a random sample from a continuous and symmetric
probability distribution with mean (and median) ju.
When the hypothesized value of jm is fo, the absolute differences
[x1 — Hol, ---[tn — Mol, must be ranked from smallest to largest.
Null hypothesis: Ho: 1 = Uo
Test statistic value: s, = the sum of the ranks associated with positive (x; — [Uo)’s
Alternative Hypothesis Rejection Region for Level « Test
Hy: > ly S20
Ay: h< bo 54 <cy [where cy = n(n + 1)/2—c4]
Hy: RF tly either s, >cors, <n(n+1)/2—e
where the critical values c, and c obtained from Appendix Table A.12 satisfy
P(S, > 1) © xand P(S, > c) © a/2 when Hy is true.
'If the tails of the distribution are “too heavy,” as was the case with the Cauchy distribution of Chapter 7,
then j1 will not exist. In such cases, the Wilcoxon test will still be valid for tests concerning jt
--- Trang 775 ---
762 = cuarrer 14 Alternative Approaches to Inference
A producer of breakfast cereals wants to verify that a filler machine is operating
correctly. The machine is supposed to fill one-pound boxes with 460 g, on the
average. This is a little above the 453.6 g needed for one pound. When the contents
are weighed, it is found that 15 boxes yield the following measurements:
454.4 470.8 447.5 453.2 462.6 445.0 455.9 458.2
461.6 457.3 452.0 464.3 459.2 453.5 465.8
It is believed that deviations of any magnitude from 460 g are just as likely to be
positive as negative (in accord with the symmetry assumption) but the distribution
may not be normal. Therefore, the Wilcoxon signed-rank test will be used to see if
the filler machine is calibrated correctly.
The hypotheses are Ho: 4 = 460 versus H,: ¢ # 460, where y is the true
average weight. Subtracting 460 from each measurement gives
—5.6 10.8 -125 -68 26 -150 -41 -I18 16 —2.7
—8.0 43 -8 -65 5.8
The ranks are obtained by ordering these from smallest to largest without regard to sign.
Absolute
Magnitude] .8 1.6 1.8 2.6 2.7 4.1 4.3 5.6 5.8 65 68 8.0 10.8 12.5 15.0
Rank 12 3 4 5 6 7 8 9 10 11 12 13) 14° «15
Sign eee eo ee te Be Se ee OS
Thus s, =2+4+7+9+ 13 = 35. From Appendix Table A.12, P(S, > 95) =
P(S, < 25) = .024 when Hy is true, so the two-tailed test with approximate level
.05 rejects Ho when either s, > 95 or < 25 [the exact « is 2(.024) = .048]. Since
5; = 35 is not in the rejection region, it cannot be concluded at level .05 that
differs from 460. Even at level .094 (approximately .1), Ho cannot be rejected, since
P(S, < 30) = P(S, > 90) = .047 implies that s, values between 30 and 90 are not
significant at that level. The P-value of the data is thus >.1. a
Although a theoretical implication of the continuity of the underlying distri-
bution is that ties will not occur, in practice they often do because of the discrete-
ness of measuring instruments. If there are several data values with the same
absolute magnitude, then they would be assigned the average of the ranks they
would receive if they differed very slightly from one another. For example, if in
Example 14.1 xg = 458.2 is changed to 458.4, then two different values of
(x; — 460) would have absolute magnitude 1.6. The ranks to be averaged would
be 2 and 3, so each would be assigned rank 2.5.
Paired Observations