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13.9 THE FRIEDMAN TWO-WAY ANALYSIS OF VARIANCE BY RANKS 727
4. Test statistic. By means of the Friedman test we will be able to determine
if it is reasonable to assume that the columns of ranks have been
drawn from the same population. If the null hypothesis is true we would
expect the observed distribution of ranks within any column to be the
result of chance factors and, hence, we would expect the numbers 1, 2,
and 3 to occur with approximately the same frequency in each column.
If, on the other hand, the null hypothesis is false (that is, the models are
not equally preferred), we would expect a preponderance of relatively
high (or low) ranks in at least one column. This condition would be
reflected in the sums of the ranks. The Friedman test will tell us whether
or not the observed sums of ranks are so discrepant that it is not likely
they are a result of chance when H 0 is true.
Since the data already consist of rankings within blocks (rows), our
first step is to sum the ranks within each column (treatment). These sums
are the R j shown in Table 13.9.1. A test statistic, denoted by Friedman
as x 2 r, is computed as follows:
x 2 r =
k
12
nk1k + 12 a
j =1
1R j 2 2 - 3n1k + 12
(13.9.1)
where n = the number of rows (blocks) and k = the number of columns
(treatments).
5. Distribution of test statistic. Critical values for various values of n and
k are given in Appendix Table O.
6. Decision rule. Reject H 0 if the probability of obtaining (when H 0 is
true) a value of x 2 r as large as or larger than actually computed is less
than or equal to a.
7. Calculation of test statistic. Using the data in Table 13.9.1 and Equations
13.9.1, we compute
x 2 r =
12
913213 + 12 311522 + 1252 2 + 1142 2 4 - 319213 + 12 = 8.222
8. Statistical decision. When we consult Appendix Table Oa, we find that
the probability of obtaining a value of x 2 r as large as 8.222 due to chance
alone, when the null hypothesis is true, is .016. We are able, therefore,
to reject the null hypothesis.
9. Conclusion. We conclude that the three models of low-volt electrical
stimulator are not equally preferred.
10. p value. For this test, p = .016.
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Ties When the original data consist of measurements on an interval or a ratio scale
instead of ranks, the measurements are assigned ranks based on their relative magnitudes
within blocks. If ties occur, each value is assigned the mean of the ranks for which
it is tied.