Manuals/calci/KRUSKALWALLISTEST

KRUSKALWALLISTEST(Array,Confidencelevel,Logicalvalue)

• $Array$ is the set of values to find the test statistic.
• $Confidencelevel$ is the value between 0 and 1.
• $Logicalvalue$ is either TRUE or FALSE.

Description

• This function gives the test statistic value of the Kruskal Wallis test.
• It is one type of Non parametric test.
• It is a logical extension of the Wilcoxon-Mann-Whitney Test.
• The parametric equivalent of the Kruskal-Wallis test is the one-way analysis of variance (ANOVA).
• This test is used for comparing more than two sample that are independent or not related.
• It is used to test the null hypothesis that all populations have identical distribution functions against the alternative hypothesis that at least two of the samples differ only with respect to Median.
• Kruskal–Wallis is also used when the examined groups are of unequal size.
• When the Kruskal-Wallis test leads to significant results, then at least one of the samples is different from the other samples.
• The test does not identify where the differences occur or how many differences actually occur.
• Since it is a non-parametric method, the Kruskal–Wallis test does not assume a normal distribution of the residuals, unlike the analogous one-way analysis of variance.
• However, the test does assume an identically shaped and scaled distribution for each group, except for any difference in medians.
• The Kruskal Wallis test data are having the following properties:
• 1.The data points must be independent from each other.
• 2.The distributions do not have to be normal and the variances do not have to be equal.
• 3.The data points must be more than five per sample.
• 4.All individuals must be selected at random from the population.
• 5.All individuals must have equal chance of being selected.
• 6.Sample sizes should be as equal as possible but some differences are allowed.
• Steps for Kruskal Wallis Test:
  • 1. Define Null and Alternative Hypotheses:
• Null Hypotheses:There is no difference between the conditions.
• Alternative Hypotheses:There is a difference between the conditions.
  • 2.State Alpha:Alpha=0.05.
  • 3.Calculate degrees of freedom:df = k – 1, where k = number of groups.
  • 4.State Decision Rule:From the Chi squared table calculate the critical value.
• Suppose the $\chi$ is greater than the critical value then reject the null hypothesis
  • 5.Calculate the Test Statistic:$H={\frac {12}{N(N+1)}}\sum _{i=1}^{k}{\frac {T_{i}^{2}}{n_{i}}}-3(N+1)$
  • 6.State Results:In this step we have to take a decision of null hypothesis either accept or reject depending on the critical value table.
  • 7.State Conclusion:To be significant, our obtained H has to be equal to or LESS than this critical value.