Manuals/calci/MOODSMEDIANTEST

MOODSMEDIANTEST(xRange,yRange,Confidencelevel,Logicalvalue)

• $xRange$ is the array of x values.
• $yRange$ is the array of y values.
• $Confidencelevel$ is the value between 0 and 1.
• $Logicalvalue$ is either TRUE or FALSE.

Description

• This function gives the test statistic of the Mood's median test.
• It is one of the Non parametric test.
• This function is used to test the equality of medians from two or more populations.
• So it provides a nonparametric alternative to the one way ANOVA.
• It is a special case of Pearson's chi-squared test.
• This function works when the Y variable is continuous,discrete-ordinal or discrete -count,and the X variable is discrete with two or more attributes.
• This test does not require normally distributed data,which is does not mean that it is assumption free.
• The following assumptions are required to test this function:
• 1.Sample data drawn from the populations of interest are unbiased and representative.
• 2.Data of k populations are continuous or ordinal when the spacing between adjacent values is not constant.
• 3.k populations are independent from each other.
• 4.The distributions of the populations the samples were drawn from all have the same shape.
• The test interpretation is:
• Null hypothesis$(H_{0})$:The population medians all are equal.Alternative hypothesis$(H_{a})$:Atleast one of the medians is different from another.
• If the null hypothesis is true, any given observation will have probability 0.5 of being greater than the shared median. *For each sample,the number of observations greater than the shared median would have a binomial distribution with p=0.5
• The procedure of the test is:
• 1. Determine the overall median.
• The combined data from all groups are sorted and the median is calculated:

$md={\frac {(x_{(}{{\frac {n}{2}})}+x_{({\frac {n}{2}}+1)})}{2}}$,if n is even.

$md=x_{(}{\frac {(n+1)}{2}})$,if n is odd.

• 2. For each sample, count how many observations are greater than the overall median, and how many are equal to or less than it.
• 3. Put the counts from step 2 into a 2xk contingency table:
• 4. Perform a chi-square test on this table, testing the hypothesis that the probability of an observation being greater than the overall median is the same for all populations.