Data Analysis Statistics An Introduction to Statistical Inference and Data Analysis

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determine k.
If we specify � in
n - k - .5 - .5n �
P (S+ e" n - k) = P Z e" " = z = ,
.5 n 2
194 CHAPTER 9. 1-SAMPLE LOCATION PROBLEMS
then
"
k = .5(n - 1 - z n).
For example, � = .05 entails z = 1.96. If n = 100, then
"
k = .5(100 - 1 - 1.96 100) = 39.7
and the desired confidence interval is approximately
(x(41), x(60)),
which is slightly liberal, or
(x(40), x(61)),
which is slightly conservative.
9.3 The Symmetric 1-Sample Location Problem
" Assume that X1, . . . , Xn
" We assume that the Xi are continuous random variables with symmet-
ric pdf f. Let � denote the center of symmetry. Note, in particular,
that � = M, the population median.
9.3.1 Hypothesis Testing
" As before, we initially consider testing a 2-sided alternative, H0 : � = �0
vs. H1 : � = �0.
" Let Di = Xi - �0. Because the Xi are continuous, P (Di = 0) = 0 and
P (|Di| = |Dj|) = 0 for i = j. Therefore, we can rank the absolute
differences as follows:
|Di1|
Let Ri denote the rank of |Di|.
9.3. THE SYMMETRIC 1-SAMPLE LOCATION PROBLEM 195
" The Wilcoxon Signed Rank Test is the following procedure:
Let x1, . . . , xn denote the observed sample and let di = xi - �0.
Initially, we assume that no di = 0 or |di| = |dj| were observed.
We define two test statistics,
T+ = k,
Dik >0
the sum of the positive ranks, and
T- = k,
Dik
the sum of the negative ranks.
Notice that
n
T+ + T- = k = n(n + 1)/2,
k=1
so that it suffices to consider only T+ (or T-, whichever is more
convenient).
By symmetry, under H0 : � = �0 we have
ET+ = ET- = n(n + 1)/4.
The Wilcoxon signed rank test rejects H0 if and only if we observe
T+ sufficently different from ET+, i.e. if and only if
p = PH0(|T+ - n(n + 1)/4) e" |t+ - n(n + 1)/4|) d" �.
" For n d" 15, we can compute the significance probability p from Table
G in Gibbons.
" Example 3.1 from Gibbons
Suppose that we want to test H0 : M = 10 vs. H1 : M = 10 at
significance level � = .05.
Suppose that we observe the following sample:
196 CHAPTER 9. 1-SAMPLE LOCATION PROBLEMS
xi di ri
9.83 -.17 7
10.09 .09 3
9.72 -.28 10
9.87 -.13 5
10.04 .04 1
9.95 -.05 2
9.82 -.18 6
9.73 -.27 9
9.79 -.21 8
9.90 -.10 4
Then n = 10, ET+ = 10(11)/4 = 27.5, t+ = 3 + 1 = 4, and
p = P (|T+ - 27.5| e" |4 - 27.5| = 23.5)
= P (T+ d" 4 or T+ e" 51)
= 2P (T+ d" 4)
= 2 � .007 = .014,
from Table G in Gibbons (see handout).
Since p = .014
" For n e" 16, we convert T+ to standard units and use the normal
approximation:
Under H0 : � = �0, ET+ = n(n + 1)/4, and
VarT+ = n(n + 1)(2n + 1)/24.
For n sufficiently large,
T+ - ET+
Z = "
�
VarT+
In the above example,
VarT+ = 10(11)(21)/24 = 96.25
and
t+ - ET+ 4 - 27.5
.
z = " = " = -2.40,
VarT+
96.5
9.3. THE SYMMETRIC 1-SAMPLE LOCATION PROBLEM 197
which gives an approximate significance probability of
p = 2P (Z d" z = -2.40)
= 2 [.5 - P (0 d" Z
= 2(.5 - .4918)
= .0164.
" Ties. Now suppose that the |di| > 0, but not necessarily distinct. If
the number of ties is small, then one can perform the test using each
possible ordering of the |di|. Otherwise:
If several |Di| are tied, then each is assigned the average of the
ranks to be assigned to that set of |Di|. These ranks are called
midranks. For example, if we observe |di| = 8, 9, 10, 10, 12, then
the midranks are ri = 1, 2, 3.5, 3.5, 5.
We then proceed as above using the midranks. Since Table G was
calculated on the assumption of no ties, we must use the normal
approximation. The formula for ET+ is identical, but the formula
for VarT+ becomes more complicated.
Suppose that there are J distinct values of |Di|. Let uj denote
the number of |Di| equalling the jth distinct value. Then
J
n(n + 1)(2n + 1) 1
VarT+ = - (u3 - uj).
j
24 48
j=1
Notice that, if uj = 1 (as typically will be the case for most of
the values), then u3 - uj = 0.
j
" If any di = 0, i.e. if any xi = �0, then we can adopt any of the strategies
that we used with the sign test when we observed xi = M0.
9.3.2 Point Estimation
�
" We derive an estimator � of � by determining the value of �0 for which
the Wilcoxon signed rank test is least inclined to reject H0 : � = �0 in [ Pobierz całość w formacie PDF ]

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