Mood’s Median Test is used to compare the medians of two or more independent groups.
The median_test function from the coin library can be used to perform this test in R, which uses the following syntax:
- response: a vector of response values
- group: a vector of grouping values
- data: a data frame containing the response and group vectors
The following example illustrates how to use this function to perform Mood’s Median Test in R.
Example: Mood’s Median Test in R
Suppose a teacher wants to know whether or not two different studying methods produce different exam scores among her class of students. To test this, she randomly assigns 10 students to use one studying method and another 10 students to use another. After two weeks, each student takes the same exam.
She decides to use Mood’s Median Test to determine if the median exam score differs between the two groups.
Step 1: Create the data frame.
#create data method = rep(c('method1', 'method2'), each=10) score = c(75, 77, 78, 83, 83, 85, 89, 90, 91, 97, 77, 80, 84, 84, 85, 90, 92, 92, 94, 95) examData = data.frame(method, score) #view data examData method score 1 method1 75 2 method1 77 3 method1 78 4 method1 83 5 method1 83 6 method1 85 7 method1 89 8 method1 90 9 method1 91 10 method1 97 11 method2 77 12 method2 80 13 method2 84 14 method2 84 15 method2 85 16 method2 90 17 method2 92 18 method2 92 19 method2 94 20 method2 95
Step 2: Perform Mood’s Median Test.
#load the coin library library(coin) #perform Mood's Median Test median_test(score~method, data = examData) #output Asymptotic Two-Sample Brown-Mood Median Test data: score by method (method1, method2) Z = -0.43809, p-value = 0.6613 alternative hypothesis: true mu is not equal to 0
The p-value of the test is 0.6613. Since this value is not less than 0.05, we fail to reject the null hypothesis. We do not have sufficient evidence to say that there is a statistically significant difference in the median exam scores between the two groups.
By default, this function assigns a score of 0 to observations that are exactly equal to the median. However, you can specify this value to instead be 0.5 or 1 by using the mid.score argument.
For example, the following code performs the exact same Mood’s Median Test but it assigns a value of 0.5 to observations that are equal to the median:
#perform Mood's Median Test median_test(score~method, mid.score="0.5", data = examData) #output Asymptotic Two-Sample Brown-Mood Median Test data: score by method (method1, method2) Z = -0.45947, p-value = 0.6459 alternative hypothesis: true mu is not equal to 00
The p-value of the test turns out to be 0.6459, which is slightly smaller than the previous p-value of 0.6613. However, the conclusion of the test is still the same: we do not have sufficient evidence to say that the median exam scores between the two groups is significantly different.