Mood's Median Test, often used in Lean Six Sigma projects, is a non-parametric method for assessing whether two or more groups come from populations with the same median. This test is particularly useful when the assumptions necessary for parametric tests, such as the ANOVA, cannot be met. These assumptions might include data normality or homogeneity of variances across groups, which are not required for Mood's Median Test. It's an alternative that provides a robust means of comparison when dealing with ordinal data or non-normally distributed interval data.

How Mood's Median Test Works

The procedure for Mood's Median Test involves several steps:

1. Combine and Rank the Data: All data from the groups involved in the test are combined into a single dataset. This pooled dataset is then arranged in ascending order, irrespective of the original groupings.

2. Find the Median: The median of the combined dataset is identified. This median serves as a common benchmark against which the distributions of the groups are compared.

3. Create a Contingency Table: The data is divided into categories based on being above or below the combined median. A contingency table is then created, with the groups forming the columns and the categories (above or below the median) forming the rows.

4. Apply the Chi-Square Test: The contingency table is analyzed using the Chi-Square test to assess whether the observed frequencies of cases above and below the median significantly differ from the expected frequencies if there were no difference between the groups.
   

Assumptions and Limitations

• Assumptions: The primary assumption of Mood's Median Test is that the samples are independent random samples from their respective populations. It does not assume normal distribution or equal variances.

• Limitations: While Mood's Median Test is less sensitive to outliers and non-normal distributions, it may be less powerful than parametric tests like the ANOVA when those tests' assumptions are met. Essentially, it may require a larger sample size to detect the same effect size.
  

Application in Lean Six Sigma

In Lean Six Sigma projects, Mood's Median Test can be particularly useful during the Analyze phase, where understanding differences between groups is crucial to identifying root causes of variation. For example, if a project team wants to compare processing times across several shifts in a manufacturing process without assuming those times are normally distributed, Mood's Median Test offers a straightforward way to assess whether median processing times significantly differ by shift.

Conclusion

Mood's Median Test provides a valuable tool for Lean Six Sigma practitioners to compare group medians without the stringent assumptions required by parametric tests. By allowing for the analysis of non-parametric data, it broadens the range of statistical tools available for driving data-based decision-making in process improvement projects. Understanding when and how to apply Mood's Median Test, alongside other statistical methods, enhances the analytical capabilities of Lean Six Sigma teams, enabling more robust and inclusive analysis of process data.

Mood's Median Test: A Real-Life Example

Mood's Median Test is a non-parametric test that is used to assess whether two or more groups come from populations with the same median. This test is particularly useful when the assumptions for a parametric test, like the ANOVA, cannot be met due to the data not being normally distributed or when ordinal data is involved.

Imagine a company is testing two different training programs to improve the productivity of new employees. The company wants to know if there is a significant difference in the median productivity increase between the two groups of employees who underwent these different training programs. Let's say the productivity increase is measured by the percentage increase in tasks completed correctly within a standard time.

Data Set

• Group A (Training Program 1): 10%, 20%, 15%, 5%, 25%

• Group B (Training Program 2): 30%, 35%, 25%, 40%, 20%

  
  

Step by Step Calculation

1. Combine and Rank the Data:

   First, we combine the two groups and then rank them together, ignoring the group distinction.

   • Combined Data: 10%, 20%, 15%, 5%, 25%, 30%, 35%, 25%, 40%, 20%

   • Ordered Data: 5%, 10%, 15%, 20%, 20%, 25%, 25%, 30%, 35%, 40%

     
     

2. Find the Overall Median:

   The overall median is the middle value of the combined and ordered data. Since we have an even number of observations (10), the median is the average of the 5th and 6th values.

   • Median: (20% + 25%) / 2 = 22.5%

     
     

3. Count Values Above and Below the Median:

   We count how many values from each group are above and below the overall median (22.5%).

   • Group A:

     • Below Median: 4 (10%, 15%, 5%, 20%)

     • Above Median: 1 (25%)

   • Group B:

     • Below Median: 1 (20%)

     • Above Median: 4 (30%, 35%, 25%, 40%)

Construct a Contingency Table:

5. Calculate Expected Frequencies:

1. The expected frequency for each cell in a contingency table is calculated by multiplying the row total by the column total and then dividing by the grand total.

   For Below Median in Group A:

   • Expected Frequency (Below Median, A): (5 * 5) / 10 = 2.5

   • Similarly, all cells will have an expected frequency of 2.5 due to the symmetry in this example.
     

   6. Calculate Chi-Square Statistic:

   The Chi-Square statistic is calculated using the formula:

,where O is the observed frequency, and E is the expected frequency.

7. Determine the p-Value:

The p-value can be found using the Chi-Square distribution table with 1 degree of freedom (since df = (rows - 1) (columns - 1) = (2-1)(2-1) = 1).

A Chi-Square value of 3.841 corresponds to a p-value that is greater. See table below.

Conclusion

Given the Chi-Square statistic of 3.6 and a critical Chi-Square value of 3.841 for the Mood's Median Test, we observe that the calculated statistic is slightly below the critical value needed to reject the null hypothesis at the 5% significance level. This suggests that while there is some evidence pointing towards a difference in the medians of productivity increases between the two groups, it is not statistically significant enough to confidently assert that a difference exists. Therefore, based on this analysis, we cannot reject the null hypothesis that there is no difference in the medians. The margin by which the test statistic falls short of the critical value highlights the need for cautious interpretation of these results. It may be beneficial for the company to gather more data or apply additional statistical methods to achieve a more conclusive outcome regarding the productivity increases between the groups.