Datasets:
vnpthack
/
vnpt_data

Dataset card Data Studio Files
xet
 Community
1
text
stringlengths
0
6.73k
Section 14.4 introduces the Bayesian approach to inference. The standard
frequentist view of inference is that the parameter of interest, @, has a fixed but
unknown value. Bayesians, however, regard @ as a random variable having a prior
probability distribution that incorporates whatever is known about its value. Then
to learn more about 6, a sample from the conditional distribution f (x|@) is
obtained, and Bayes’ theorem is used to produce the posterior distribution of 0
given the data x, ... , 1). All Bayesian methods are based on this posterior
distribution.
JL. Devore and K.N. Berk, Modern Mathematical Statistics with Applications, Springer Texts in Statistics, 758
DOI 10.1007/978-1-4614-0391-3_14, © Springer Science+Business Media, LLC 2012
--- Trang 772 ---
14.1 The Wilcoxon Signed-Rank Test 759
The Wilcoxon Signed-Rank Test
A research chemist replicated a particular experiment a total of 10 times and obtained
the following values of reaction temperature, ordered from smallest to largest:
—57 —19 -05 76 1.30 2.02 2.17 246 2.68 3.02
The distribution of reaction temperature is of course continuous. Suppose the
investigator is willing to assume that this distribution is symmetric, so that the pdf
satisfies f (ji + t) =f (fi — 1) for any t >0, where 71 is the median of the distribution
(and also the mean 1 provided that the mean exists). This condition on f(x) simply
says that the height of the density curve above a value any particular distance to the
right of the median is the same as the height that same distance to the left of the
median. The assumption of symmetry may at first thought seem quite bold, but
remember that we have frequently assumed a normal distribution. Since a normal
distribution is symmetric, the assumption of symmetry without any additional
distributional specification is actually a weaker assumption than normality.
Let’s now consider testing the null hypothesis that ji = 0. This amounts to
saying that a temperature of any particular magnitude, say 1.50, is no more likely
to be positive (+1.50) than to be negative (—1.50). A glance at the data casts doubt
on this hypothesis; for example, the sample median is 1.66, which is far larger in
magnitude than any of the three negative observations.
Figure 14.1 shows graphs of two symmetric pdf’s, one for which Ho is
true and the other for which the median of the distribution considerably exceeds 0.
In the first case we expect the magnitudes of the negative observations in the
sample to be comparable to those of the positive sample observations. However,
in the second case observations of large absolute magnitude will tend to be positive
rather than negative.
iT. ir.
0 0. @F
Figure 14.1 Distributions for which (a) 72 = 0; (b) ji > 0
For the sample of ten reaction temperatures, let’s for the moment disregard
the signs of the observations and rank the absolute magnitudes from 1 to 10, with
the smallest getting rank 1, the second smallest rank 2, and so on. Then apply the
sign of each observation to the corresponding rank (so some signed ranks will be
negative, e.g. —3, whereas others will be positive, e.g. 8). The test statistic will be
S, = the sum of the positively signed ranks.
Absolute Magnitude 05 19 57.76 «1.30 2.02 2.17 2.46 2.68 3.02
Rank 1 2 3 4 5 6 i, 8 9 10
Signed Rank “t -2 =3 4 5 6 7 8 9 10
54 =44546474+849410=49
--- Trang 773 ---
760 charter 14 Alternative Approaches to Inference
When the median of the distribution is much greater than 0, most of the observa-
tions with large absolute magnitudes should be positive, resulting in positively
signed ranks and a large value of s,. On the other hand, if the median is 0,
magnitudes of positively signed observations should be intermingled with those
of negatively signed observations, in which case s, will not be very large. Thus
we should reject Ho: ji = 0 when s, is “quite large”— the rejection region should
have the form s, > c.
The critical value c should be chosen so that the test has a desired significance
level (type I error probability), such as .05 or .01. This necessitates finding the
distribution of the test statistic S, when the null hypothesis is true. Let’s consider
n = 5, in which case there are 2° = 32 ways of applying signs to the five ranks 1, 2, 3,
4, and 5 (each rank could have a — sign or a + sign). The key point is that when Ho is
true, any collection of five signed ranks has the same chance as does any other
collection. That is, the smallest observation in absolute magnitude is equally likely to
be positive or negative, the same is true of the second smallest observation in absolute
magnitude, and so on. Thus the collection — 1,2, 3, —4, 5 of signed ranks is just as likely
as the collection 1, 2, 3,4, —5, and just as likely as any one of the other 30 possibilities.
Table 14.1 lists the 32 possible signed-rank sequences when n = 5 along
with the value s, for each sequence. This immediately gives the “null distribution”
of S, displayed in Table 14.2. For example, Table 14.1 shows that three of the
32 possible sequences have s, = 8, so P(S, = 8 when Hp is true) = 1/32 +
1/32 + 1/32 = 3/32. This null distribution appears in Table 14.2. Notice that it
Table 14.1 Possible signed-rank sequences for n = 5
Sequence s Sequence Ss
-l 2 -3 -4 -5 0 -l -20 -3 44-54
410-20 --30 4-5 1 +1 -2 -3 44-5 5
-1 42 -3 -4 -5 2 -l 42 -3 44 -5 6
-l -2 43 -4 -5 3 -1 -2 43 44 -5 #7
+1 +2 3 —4 a 3 +1 42 3 +4 = 7
+1 -2 43 -4 -5 4 +1 -2 43 44 -5 8
ok +2 +3 —4 Ss 5 ok +2 +3 +4 ac} 9