Manuals/calci/MOODSMEDIANTEST

MOODSMEDIANTEST (XRange1,XRange2,ConfidenceLevel,NewTableFlag)

• $XRange1$ is the array of values.
• $XRange2$ is the array of values.
• $ConfidenceLevel$ is the value between 0 and 1.
• $NewTableFlag$ 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.

• where $n=\sum _{i=1}^{k}n_{i}$.
• $x_{(1)},x_{(2)}....x_{(n)}$,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)

MOODSMEDIANTEST STATISTICS

MEAN1	25
MEDIAN1	22
MEAN2	27.6
MEDIAN2	32
OVERALLMEDIAN	28
GREATERMEDIAN1	2
GREATERMEDIAN2	3
LESSEQUALMEDIAN1	3
LESSEQUALMEDIAN2	2
OBSERVED FREQUENCY	2 3 3 2
EXPECTED FREQUENCY	2.5 2.5 2.5 2.5
PVALUE	0.5270892568655381

RESULT AS PVALUE > 0.05, MEDIANS OF THE POPULATIONS FROM WHICH THE TWO SAMPLES ARE DERIVED ARE EQUAL

Related Videos

Moods Median Test

See Also

• LEVENESTEST
• FRIEDMANTEST
• RIEMANNZETA
• MANNWHITNEYUTEST

References

• Mood's Median Test

• List of Main Z Functions

• Z3 home