Tag Archives: test 31

Week 31: The Cochran Q Test

Let’s do some more nonparametric testing today with the Cochran Q test!

When Would You Use It?
The Cochran Q test is a nonparametric test used to determine if, in a set of k dependent samples (k ≥ 2), at least two of the samples represent different populations.

What Type of Data?
The Cochran Q test requires categorical or nominal data.

Test Assumptions

  • The presentation of the k experimental conditions is random or counterbalanced.
  • With matched samples, within each set of matched subjects, each of the subjects should be randomly assigned to one of the k experimental conditions.

Test Process
Step 1: Formulate the null and alternative hypotheses. For the Cochran Q test, we are interested in variables that are dichotomous (let’s say that they have a “yes” and a “no” response). The null hypothesis claims that the proportions of one of the responses is the same across all j experimental conditions. The alternative hypothesis claims otherwise (at least two population proportions are not equal).

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


Step 3: Obtain the p-value associated with the calculated chi-square. The p-value indicates the probability of observing a Q value equal to or larger than the one calculated for the test, under the assumption that the null hypothesis is true.

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 proportion of “yes” responses is equal across the k experimental conditions). If the p-value is smaller than the prespecified α-level, reject the null hypothesis in favor of the alternative.

The example for this test comes from a previous semester’s STAT 213 grades. Students took two midterms and a final. I wanted to determine if there was a difference in the proportion of students who passed midterm 1, midterm 2, or the final, in a sample of n = 30. Let α = 0.05.

H0: πmid1 = πmid2 = πfinal
Ha: At least two of the underlying population proportions are not equal.

The following table shows the data for this example. Here, a passing grade is coded as “1” and a failing grade is coded as “0”.




Since our p-value is larger than our alpha-level, we fail to reject H0 and claim that the proportions for each of the three tests are equal.

Example in R

Since the calculations for this week’s test are quite easy, it’s probably faster to do them by hand than use R!