Let’s return to nonparametrics this week with the **Friedman two-way analysis of variance by ranks**!

**When Would You Use It?
**The Friedman two-way analysis of variance by ranks is a nonparametric test used to determine if, in a set of k (k ≥ 2) independent samples, at least two of the samples represent populations with different median values.

**What Type of Data?
**The Friedman two-way analysis of variance by ranks requires ordinal data.

**Test Assumptions**

- The presentation of the k experimental conditions should be random or counterbalanced.
- If dealing with matched samples, the subjects should be randomly assigned to the k experimental conditions.

**Test Process**

Step 1: Formulate the null and alternative hypotheses. The null hypothesis claims that the k population medians are equal. The alternative hypothesis claims that at least two of the k population medians are different.

Step 2: Compute the test statistic, a chi-square value. It is computed as follows:

The ranks themselves are obtained by ranking each of the k scores of a subject within that subject. That is, an individual’s scores in each of the k conditions are ranked from highest to lowest (or lowest to highest) for that particular individual. See the example below for more explanation.

Step 3: Obtain the p-value associated with the calculated chi-square statistic. The p-value indicates the probability of observing a chi-square value equal to or larger than the observed chis-square value from the sample under the assumption that the null hypothesis is true. The degrees of freedom for this test are k – 1.

Step 4: Determine the conclusion. If the p-value is larger than the prespecified α-level, fail to reject the null hypothesis (that is, retain the claim that the population medians are equal). If the p-value is smaller than the prespecified α-level, reject the null hypothesis in favor of the alternative.

**Example
**The example I want to look at today is similar to last week’s. The data come from a previous semester’s STAT 213 grades. The class had two midterms and I final. Taking a sample of n = 20 from this class, I wanted to see if the average test grades were all similar across all three tests or if there were some statistically significant differences. Let α = 0.05.

H_{0}: θ_{midterm1}= θ_{midterm2} = θ_{final
}H_{a}: at least one pair of medians are different

The following table shows the midterm and final scores as well as the corresponding within-subject ranks.

**Computations:**

Here, our computed p-value is smaller than our α-level, which leads us to reject the null hypothesis, which is the claim that the median grade is equal across the three tests.

**Example in R
**No example in R this week, as this is probably easier to do by hand than using R!