# Manuals/calci/MOODSMEDIANTEST

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**MOODSMEDIANTEST (XRange1,XRange2,ConfidenceLevel,NewTableFlag)**

- is the array of values.
- is the array of 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

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

## See Also

## References