The ranking procedures described in the present chapter have an advantage over parametric methods in that they do not depend on the underlying distributions of parent populations.

**PERMUTATION TESTS**

The ranking procedures described in the present
chapter have an advantage over parametric methods in that they do not depend on
the underlying distributions of parent populations. As we will discuss in
Section 14.9, ranking procedures are not sensitive to one or a few outlying
observations. However, a disadvantage of ranking procedures is that they are
less informative than corresponding parametric tests. In-formation is lost as a
result of the rank transformations. For the sake of constructing a
distribution-free method, we ignore the numerical values and hence the
magnitude of differences among the observations. Note that if we observed the
values 4, 5, and 6 we would assign them ranks 1, 2, and 3 respectively. On the
other hand, had we observed the values 4, 5, and 10, we would still assign the
ranks 1, 2, and 3, respec-tively. The fact that 10 is much larger than 6 is
lost in the rankings.

Is there a way for us to have our cake and eat it
too? Permutation tests retain the information in the numerical data but do not
depend on parametric assumptions. They are computer-intensive techniques with
many of the same virtues as the boot-strap.

In the late 1940s and early 1950s, research
confirmed that under certain condi-tions, permutation methods can be nearly as
powerful as the most powerful para-metric tests. This observation is true as
sample sizes become large [see, for exam-ple, Lehmann and Stein (1949) and
Hoeffding (1952)]. Although permutation tests have existed for more than 60
years, their common usage has emerged only in the 1980s and 1990s. Late in the
twentieth century, high-speed computing enabled one to determine the exact
distributions of permutations under the null hypothesis. Per-mutation
statistics generally have discrete distributions. Computation of all possible
values of these statistics and their associated probabilities when the null
hypothesis is true allows one to calculate critical values and *p*-values; the resulting tables are much
like normal probability tables used for parametric Gaussian distributions

The concepts underlying permutation tests, also
called randomization tests, go back to Fisher (1935). In the case of two
populations, assume we have data from two distributions denoted as *X*_{1}, *X*_{2}, . . . , *X _{n}*
for the first population, and

The first assumption for this test is that both
samples are independent random samples from their respective parent
populations. The second is that at least an in-terval measurement scale is
being used. Under these conditions, and assuming the null hypothesis to be
true, it makes sense to pool the data because each *X* and *Y* gives information
about the common distribution for the two samples.

As it makes no difference whether we include an *X* or a *Y* in the calculation of T, any arrangement of the *n* + *m*
observations that assigns *n* to group
one and *m* to group two is as probable
as the other. Hence, under the null hypothesis, any assign-ment to the *X*s of *n* out of the *n* + *m* observations constitutes a value for
T. Recall from Chapter 5 that there are exactly *C*(*n* + *m*, *n*)
= (*n* + *m*)!/[*n*! *m*!] ways to select *n* observations from pooled data to serve as the *X*s.

Each arrangement leads to a potentially different
value for *T* (some arrangements may
give the same numerical values if the *X*s
and *Y*s are not all different values).
The test is called a permutation test because we can think of the pooled
observations as *Z*_{1}, *Z*_{2}, . . . , *Z _{n}*,

The other name for a permutation test—randomization
test—comes about be-cause each selection of *n*s
assigned to the *X*s can be viewed as a
random selection of *n *of the samples.
This condition applies when the samples are selected at random out* *of the set of *n* + *m* values. Physically,
we could mark each of the *n* + *m* values on a piece of paper, place and
mix them in a hat and then reach in and randomly draw out *n *of them without replacing any in the hat. Hence, permutation
methods also are said* *to be sampling
without replacement. Contrast this to a bootstrap sample that is se-lected by
sampling a fixed number of times but always with replacement.

Since under the null hypothesis each permutation
has the probability 1/*C*(*n* + *m*,
*n*), in principle we have the null
distribution. On the other hand, if the two popula-tions really are different,
than the observed *T* should be
unusually low if the *X*s tend to be
smaller than the *Y*s and unusually
large if the *X*s tend to be larger
than the *Y*s. The *p*-value for the test is then the sum of the probabilities for all
permutations leading to values of *T*
as extreme or more extreme (equal or larger or smaller) than the observed *T*.

So if *k*
is the number of values as extreme as or more extreme than the observed *T*, the *p*-value is *k*/*C*(*n*
+ *m*, *n*). Such a *p*-value can be
one-sided or two-sided depending on how we define “more extreme.” The process
of determining the distribution of the test statistic (*T*) is in principle a very simple procedure. The problem is that we
must enumerate all of these permutations and calculate *T* for each one to construct the cor-responding permutation
distribution. As *n* and *m* become large, the process of
gener-ating all of these permutations is a very computer-intensive procedure.

The basic idea of enumerating a multitude of
permutations has been generalized to many other statistical problems. The
problems are more complicated but the idea remains the same, namely, that a
permutation distribution for the test can be calcu-lated under the null
hypothesis. The null distribution will not depend on the shape of the
population distributions for the original observations or their scores.

Several excellent texts specialize in permutation
tests. See, for example, Good (2000), Edgington (1995), Mielke and Berry
(2001), or Manly (1997). Some books with the word “resampling” in the title
include permutation methods and comparethem with the bootstrap. These include
Westfall and Young (1993), Lunneborg (2000), and Good (2001).

Another name for permutation tests is exact tests.
The latter term is used be-cause, conditioned on the observed data, the
significance levels that are determined for the hypothesis test have a special
characteristic: The significance levels satisfy the exactness property
regardless of the population distribution of the pooled data.

In the 2 × 2 contingency table in Section 11.6, we
considered an approximate chi-square test for independence. The next section
will introduce an exact permuta-tion test known as Fisher’s exact test. This
test can be used in a 2 × 2 table when the chi-square approximation is not very
good.

In a 2 × 2 contingency table, the elements are
sometimes all random, but there are occasions when the row totals and the
column totals are restricted in advance. In such cases, a permutation test for
independence (or differences in group propor-tions), known as Fisher’s exact
test, is appropriate. The test is attributed to R. A. Fisher, who describes it
in his design of experiments text (Fisher, 1935). However, as Conover (1999)
points out, it was also discovered and presented in the literature almost
simultaneously in Irwin (1935) and Yates (1934).

Fisher and others have argued for its more general
use based on conditioning ar-guments. As Conover (1999) points out, it is very
popular for all types of 2 × 2 ta-bles because its exact *p*-values can be determined easily (by enumerating all the more extreme
tables and their probabilities under the null hypothesis). As in the chapter on
contingency tables, the null hypothesis is that if the rows represent two
groups, then the proportions in the first column should be the same for each
group (and, consequently, so should the proportions in the second column).

Consider *N*
observations summarized in a 2 × 2 table. The row totals *r* and *N – r* and the
column totals *c* and *N – c* are fixed in advance (or
conditioned on after-wards). Refer to Table 14.13.

Because the values of *r*, *c*, and *N* are fixed in advance, the only
quantity that is ran-dom is *x*, the
entry in the cell corresponding to the intersection of Row 1 and Column 1. Now,
*x* can vary from 0 up to the minimum
of *c* and *r*. This limit on the value is due to the requirement that the row
and column totals must always be *r*
for the first row and *c* for the first
column. Each different value of *x*
determines a new distinct contin-gency table. Let us specify the null
hypothesis that the probability *p*_{1}
of an observa-tion in row 1, column 1 is the same as the probability *p*_{2} of an observation in row
2, *T*, defined to be equal to *x*, is the hypergeometric distribution. Equation 14.9 defines the test statistic *T.*

**TABLE 14.13. Basic 2 × 2 Contingency Table for Fisher’s Exact Test **

While not covered explicitly in previous chapters,
the hypergeometric distribu-tion is similar to discrete distributions that were
discussed in Chapter 5. Remember that a discrete distribution is defined on a
finite set of numbers. The hypergeometric distribution used for calculating
test statistic for Fisher’s exact test is given in Equa-tion 14.9. Let *T* be the cell value for column 1, row 1
in a 2 × 2 contingency table with the constraints that the row one total is *r* and the column one total is *c*, with *r* and *c* less than or
equal to the grand total *N*. Then for *x* = 0, 1, . . . , min(*r*, *c*),

and *P*(*T* = *x*)
= 0 for all other values of *x*.

A one-sided *p*-value
for Fisher’s exact test is calculated as follows:

1. Find all 2 × 2 tables with the
row and column totals of the observed table and with row 1, column 1 cell
values equal to or smaller than the observed *x*.

2. Use the hypergeometric
distribution from Equation 14.9 to calculate the probability of occurrence of
these tables under the null hypothesis.

3. Sum the probabilities over all
such tables.

The result at step (3) is the one-sided *p*-value. Two-sided and opposite
one-sided *p*-values can be obtained
according to a similar procedure. One needs to define the re-jection region
such that it is the area on one tail of the distribution that is comprised of
probabilities that are as extreme as or more extreme than the significance
level of the test. The second side or the opposite side would be the
corresponding area on the opposite end of the distribution. The next example
will illustrate how to carry out the procedure described above.

Fisher (1935) gave a now famous example of a lady
who claims that she can tell sim-ply by tasting tea whether milk or tea was
poured into a cup first. Fisher used this ex-ample to demonstrate the
principles of experimental design and hypothesis testing.

Let us suppose, as is described in Agresti (1990),
page 61, that an experiment was conducted to test whether the lady simply is
taking guesses versus the alterna-tive that she has the skill to determine the
order of pouring the two liquids. The lady is given eight cups of tea, four
with milk poured first and four with tea poured first. The cups are numbered 1
to 8. The experimenter has recorded on a piece of paper which cup numbers had
the tea poured first and which had the milk poured first.

The lady is told that four cups had milk poured
first and four had tea poured first. Given this information, she will designate
four of them for each group. This design is important because it forces each
row and column total to be fixed (see Table

**TABLE 14.14. Lady Tasting Tea Experiment: 2 × 2 Contingency Table for
Fisher’s Exact Test**

In this experiment, the use of Fisher’s
exact test is appropriate and uncon-troversial. For other designs, the
application of Fisher’s exact test may be debatable even when there are some
similarities to the foregoing example.

For this problem, there are only five contingency
tables: (1) Correctly labeling all four cups with milk poured first and, hence,
all with tea poured first; (2) incorrectly labeling one with milk poured first
and, hence, one with tea poured first; (3) incor-rectly labeling two with milk
poured first (also two with tea poured first); (4) incor-rectly labeling three
with milk poured first (also three with tea poured first); and (5) incorrectly
labeling all four with milk poured first (also all four with tea poured first).

Case (3) is the most likely under the null
hypothesis, as it would be expected from random guessing. Cases (1) and (2)
favor some ability to discriminate, and (4) and (5) indicate good
discrimination but in the wrong direction. However, the sam-ple size is too
small for the test to provide very strong evidence for the lady’s abili-ties,
even in the most extreme cases in this example when she guesses three or four
outcomes correctly.

Let us first compute the *p*-value when *x* is 3. In
this case, it is appropriate to per-form a one-sided test, as a significant
test statistic would support the claim that she can distinguish the order of
pouring milk and tea. We are testing the alternative hy-pothesis that the lady
can determine that the milk was poured before the tea versus the null
hypothesis that she cannot tell the difference in the order of pouring. Thus,
we must evaluate two contingency tables, one for *x* = 3 and one for *x* = 4.
The ob-served data are given in Table 14.15.

The probability associated with the observed table
under the null hypothesis is *C*(4, 3)* C*(4, 1)/*C*(8, 4) = (4 4 4!)/(8 7 6 5) = 8/35 = 0.229. The only table more
ex-treme that favors the alternative hypothesis is the perfect table, Table
14.16.

**TABLE 14.15. Lady Tasting Tea Experiment: Observed 2 × 2 Contingency
Table for Fisher’s Exact Test**

**TABLE 14.16. Lady Tasting Tea Experiment: More Extreme 2 × 2 Contingency
Table for Fisher’s Exact Test**

The probability of this table under the null
hypothesis is 1/*C*(8, 4) = 1/ 70 =
0.0142. So the *p*-value for the
combined tables is 0.229 + 0.014 = 0.243. If we ran the tea drinking experiment
and observed an *x* of 3, we would have
an observed *p*-value of 0.243; this
outcome would suggest that we cannot reject the null hypothe-sis that the lady
is unable to discriminate between milk or tea poured first.

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