When comparing the average of two or more groups with the help of hypothesis tests, the assumption is that the data is a sample from a normally distributed population. That is why hypothesis tests such as the t-test, paired t-test and analysis of variance (ANOVA) are also called parametric tests.
Nonparametric tests do not make assumptions about a specific distribution. If assumptions do not hold, nonparametric tests are a better safeguard against drawing wrong conclusions.
The Mood’s median test is a nonparametric test that is used to test the equality of medians from two or more populations. Therefore, it provides a nonparametric alternative to the one-way ANOVA. The Mood’s median test works when the Y variable is continuous, discrete-ordinal or discrete-count, and the X variable is discrete with two or more attributes.
Examples for the usage of the Mood’s median test include:
A project team wants to determine what drives the lead times of quality control (QC) analyses. One potential X they analyze is the products (A, B, C).Thus, they collect the data of all analysis times over the last three months. A dot plot (Figure 1) of the data shows a lot of overlap between the lead times of the three product groups, but it is hard to tell whether there are significant differences.
The team decides to use a hypothesis test to determine if there are “true differences” between the three product types or simply random differences due to the samples taken.
The team now has the choice between the nonparametric Kruskal-Wallis and the Mood’s median test. Because the latter is more robust against outliers and some extreme values are observed in the QC data, the team decides to use the Mood’s median test.
The null hypothesis, H0, is: The samples come from the same distribution, or there is no difference between the medians of the three products’ analysis times.
The alternative hypothesis, Ha, states: The samples come from different distribution (i.e., at least one median is different).
Although the Mood’s median test does not require normally distributed data, that does not mean that it is assumption free. The assumptions of Mood’s median test are that the data from each population is an independent random sample and the population distributions have the same shape.
Testing for same shape can ideally be done with the probability plot. A practitioner would now look for a distribution that is the same for all three product groups.
In this case, the probability plot (Figure 3) shows that all data follows a lognormal distribution (p>0.05), which is also typical for cycle time data. If the probability plot does not provide distribution that matches all groups under comparison, a visual check of the data may help. Do the distributions look similarly (e.g., are they all left- or right-skewed, with only some extreme values)?
If the assumptions are met, the Mood’s median test can be conducted. If the p-value is less than the agreed Alpha risk of 5 percent (0.05), the null hypothesis is rejected and at least one significant difference can be assumed. For the QC analysis time, the p-value is 0.016 – in other words, less than 0.05.
The 95 percent confidence intervals of the individual group medians now help to find where the significant difference is. The rule is: If there is no overlap between the confidence intervals, a significant difference can be assumed. In this example, at least product A and C have significantly different analysis times (Figure 4).
The test statistic of the Mood’s median test is actually based on another well known hypothesis test: the chi-square test. This test is usually used to find differences between proportions of two or more groups. But how can it be used to compare medians?
First, practitioners should aggregate the original data into a two-way table following this procedure:
|Table 1: Two-way Contingency Table|
|Overall median = 1.66||Product Type|
|Number of Observations…||A||B||C|
|Less than or equal to overall median||20||16||9|
|Greater than overall median||10||14||21|
The assumption (or null hypothesis) is that if there were no median difference between the groups, the percentage of values below and above the overall median should be equal for each group. The chi-square test can now be used to test this assumption. Low values of chi-square would prove this assumption true; large values would indicate that the null hypothesis is false.
In this project example, the chi-square value is 8.27. The p-value of 0.016 indicates that the probability that such a chi-square value occurs if there are actually no differences between the product type groups is only 1.6 percent. Therefore, the practitioners can conclude that there is at least one significant difference between the groups, with just a 1.6 percent risk of being wrong.