Difference between revisions of "Manuals/calci/MOODSMEDIANTEST"

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*=MOODSMEDIANTEST(A1:A5,B1:B5,0.05,TRUE)
 
*=MOODSMEDIANTEST(A1:A5,B1:B5,0.05,TRUE)
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==See Also==
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*[[Manuals/calci/LEVENESTEST| LEVENESTEST]]
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*[[Manuals/calci/FRIEDMANTEST| FRIEDMANTEST]]
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*[[Manuals/calci/RIEMANNZETA| RIEMANNZETA]]
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*[[Manuals/calci/MANNWHITNEYUTEST| MANNWHITNEYUTEST]]
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==References==
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*[http://www2.hawaii.edu/~taylor/z631/moods.pdf Mood's Median Test]

Revision as of 11:33, 12 May 2015

MOODSMEDIANTEST(xRange,yRange,Confidencelevel,Logicalvalue)


  • is the array of x values.
  • is the array of y values.
  • is the value between 0 and 1.
  • 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:The population medians all are equal.Alternative hypothesis: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:
,if n is even.
,if n is odd.
  • where .
  • ,is the ordered data of all observations from small to large.
  • 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.

Example

Spreadsheet
A B
1 30 32
2 10 13
3 22 33
4 20 26
5 43 34
  • =MOODSMEDIANTEST(A1:A5,B1:B5,0.05,TRUE)

See Also

References