Question 62823This question is from textbook
: How could nonparametric data be transformed into parametric data that could be used in an ANOVA? How do the fe and fo in a nonparametric test correspond to the SSE and the SSA in an ANOVA?
This question is from textbook
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! I don't know the answer to your question. This is what Wikipedia
says relating a nonparametric test to ANOVA.
Cheers,
Stan H.
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Kruskal-Wallis one-way analysis of variance
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In statistics, the Kruskal-Wallis one-way analysis of variance by ranks (named after William Kruskal and Allen Wallis) is a non-parametric method for testing equality of population medians among groups. Intuitively, it is identical to a one-way analysis of variance with the data replaced by their ranks. It is an extension of the Mann-Whitney U test to 3 or more groups.
Since it is a non-parametric method, the Kruskal-Wallis test does not assume a normal population, unlike the analogous one-way analysis of variance. However, it still assumes that population variabilities among groups are equal. To get around this limitation of the Kruskal-Wallis test, some statisticians suggest using a robust test for equal locations among groups instead.
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[edit] Method
Rank all data from all groups together; i.e., rank the data from 1 to N ignoring group membership. Assign any tied values the average of the ranks they would have received had they not been tied.
The test statistic is given by: , where:
ng is the number of observations in group g
rij is the rank (among all observations) of observation j from group i
N is the total number of observations across all groups
,
is the average of all the rij, equal to (N + 1) / 2.
Notice that the denominator of the expression for K is exactly (N − 1)N(N + 1) / 12. Thus .
A correction for ties can be made by dividing K by , where G is the number of groupings of different tied ranks, and ti is the number of tied values within group i that are tied at a particular value. This correction usually makes little difference in the value of K unless there are a large number of ties.
Finally, the p-value is approximated by . If some ni's are small (i.e., less than 5) the probability distribution of K can be quite different from this chi-square distribution. If a table of the chi-square probability distribution is available, the critical value of chi-square, , can be found by entering the table at g − 1 degrees of freedom and looking under the desired significance or alpha level. The null hypothesis of equal population medians would then be rejected if . Appropriate multiple comparisons would then be performed on the group medians.
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