Mood’s median test is a nonparametric test to compare the medians of two independent samples. It is also used to estimate whether the median of any two independent samples are equal. Therefore, Mood’s median non parametric hypothesis test is an alternative to the one-way ANOVA. This test works when dependent variable is continuous or discrete-count, and the independent variables are discrete with two or more attributes.

Mood’s median test is a primitive two sample version of sign test. This test can be applied for more than two samples, but it is not as powerful as Kruskal-Wallis Test.

While Mood’s median test is more useful for smaller sample sizes, when the data contains few outliers, because this test is only focuses on median value instead of ranks.

Usually the researchers prefers Wilcoxon Rank Sum test or Mann-Whitney U test as they provides more robust results when compared to Mood’s Median Test.

The Mood’s median test compares whether k independent samples have either drawn from the same population or from populations with equal medians.

## Assumptions of the Mood’s Median Test

• The Observations are independent both within and between samples.
• The Observations come from population with continuous distribution function
• The distributions of the populations the samples were drawn from all have the same shape.

## Uses of Mood’s Median Test

Following are the few examples where Mood’s median test can be used:

• Moods median test is to compare the medians of customer satisfaction level (dependent variable) of different account holders (independent variables: current account and savings account customers).
• To compare the medians of machining times of different production lines in a manufacturing plant (Line 1, 2, and 3 )

## Procedure to execute Mood’s Median Non Parametric Hypothesis Test

• Compute the overall median (M) of the combination of k samples—list all the data into single set and sort the data into ascending order, finally find the middle number (median M)
• Count the number of observation that are greater than the overall median (M), and then count the observations that are equal to or less than the median (M).
• Create 2xk contingency table with the above data
• Compute chi-square test for the completed contingency table
• Compare the calculated value with chi-square critical value
• Finally, formulate decision and conclusion

## Calculation of Mood’s Median Test

The Mood’s Median Test is used to determine whether there is sufficient evidence to conclude that samples are drawn from populations with different medians. The test statistic used is the chi-square test statistic

Where O is Observed Frequency and E is Expected Frequency

• Null Hypothesis H0: The medians of the populations all are equal
• Alternative Hypothesis H1: The medians of the population are not all equal (at least one population median is different)

DF= (rows-1)*(columns-1)

## Example of Mood’s Median Non Parametric Hypothesis Test

A major wheat supplier from Texas analyzing the yields of various crop methods. He randomly assigned two different wheat crop methods to a very high number of different acres of farm land and recorded the production rate (yield per acre) for each plot.

Step 1: Calculate total median from two methods, first sort the data into ascending order and find the middle number

1,3,5,6,8,10,11,11,12,12,14,14,15,15,15,16,17,19,19,20,21,22,24,26,27,28,31,32

Overall Median = 15+15/2=15

Step 2: Create 2X2 contingency table from the above values

Step 3: Perform chi-square (χ2) test

Observed Frequencies :

Expected Frequencies :

Expected value for each cell= row total * column total/grand total

(A) Cell (A1XB1) = 14*13/28=6.5
(B)  Cell (A2XB1) = 14*15/28=7.5
(C) Cell (A1XB2) = 14*13/28=6.5
(D) Cell (A2XB2) = 14*15/28=7.5

Step 4: Compute the degrees of freedom = (2-1)*(2-1) =1

Critical value of χ2 (0.05, 1) = 3.841

Step 5: Calculated χ2 value is less than the critical value of χ2 for a 0.05 significance level, hence we have no convincing evidence to reject the null hypothesis

So, the data are consistent with the null hypothesis that there is no difference between the two wheat crop methods.

## Six Sigma Black Belt Certification Mood’s Median Test Questions:

Question: What statistical test would be used to compare medians of two independent samples?

A) Wlcoxon Ranked Sum test
(B) Mann-Whitney U test
(C) Mood’s Median test
(D) All the above

OR

This section requires you to be logged in. Mohamed Hany says:

sorry sir,
as you mentioned before ” Mann-Whitney test will helps to compare the medians of the two populations.”

so why Answer C is the right answer? also Mann-whitney is alternative to two sample t test and mood median is alternative to one way anova Ted Hessing says:

Hi Mohamed,

By definition Moods is for MEDIANs of independent samples.

Mann-Whitney is for MEANS.

Mean and Median are different concepts.

Hope this helps.

Best, Ted. Youssef Boudoudouh says:

Hi Ted
I think in non parametric test you compare Median not mean
the difference between mood median and kruwals is that mood median is with outliers and kriswlas is without outliers Ramana PV says:

Hi Youssef Boudoudouh,

Mood’s median test and Kruskal-wallis compares the medians of two independent samples, but all non parametric tests are not compare the medians of two populations. In the above explanation Ted talking about Mann-Whitney U Test.

Mann-Whitney test is a non-parametric test that is to compare two sample means that may come from the same population, and used to test whether two sample means are equal or not. Greg Chesterton says:

You said:

Step 5: Calculated χ2 value is less than the critical value of χ2 for a 0.05 significance level, hence failed to reject the null hypotheses.
So, there is enough evidence to conclude that all population medians are equal.

This is not correct. Failure to reject the null should not lead you to conclude that the null is true. It simply says that there is not strong evidence against the null. If you want to assert that the population means/medians are the same, you use an equivalence test (e.g., TOST). Ramana PV says:

Thanks Greg,
For better clarity i have updated the verbiage in the step #5

Thanks Francesco De Angelis says:

Hello Ramana,
Shouldn’t the value of cell A1B1 be 5, as there are 5 values (namely, 20,21, 21, 24, 29) greater than 15?

Best regards. Ramana PV says:

Hello Francesco De Angelis

Somehow old data file existed in the system. I have uploaded the correct “Production rate” picture.