Home Â» How to Perform a Kruskal-Wallis Test in SPSS

How to Perform a Kruskal-Wallis Test in SPSS

AÂ Kruskal-Wallis TestÂ is used to determine whether or not there is a statistically significant difference between the medians of three or more independent groups. It is considered to be the non-parametric equivalent of theÂ One-Way ANOVA.

This tutorial explains how to conduct a Kruskal-Wallis Test in SPSS.

Example: Kruskal-Wallis Test in SPSS

A researcher wants to know whether or not three drugs have different effects on knee pain, so he recruits 30 individuals who all experience similar knee pain and randomly splits them up into three groups to receive either Drug 1, Drug 2, or Drug 3.

After one month of taking the drug, the researcher asks each individual to rate their knee pain on a scale of 1 to 100, with 100 indicating the most severe pain. The ratings for all 30 individuals are shown below:

Use the following steps to perform a Kruskal-Wallis Test to determine whether or not there is a difference between the reported levels of knee pain between the three groups:

Step 1: Perform a Kruskal-Wallis Test.

Click theÂ AnalyzeÂ tab, thenÂ Nonparametric Tests, thenÂ Legacy Dialogs, thenÂ K Independent Samples:

In the window that pops up, drag the variableÂ painÂ into the box labelled Test Variable List andÂ drugÂ into the box labelled Grouping Variable. Then clickÂ Define RangeÂ and set the Minimum value to 1 and the Maximum value to 3. Then click Continue. Make sure the box is checked next toÂ Kruskal-Wallis HÂ and then clickÂ OK.

Step 2: Interpret the results.

Once you clickÂ OK, the results of the Kruskal-Wallis test will appear:

The second table in the output displays the results of the test:

• Kruskal-Wallis H:Â This is the X2 test statistic.
• df:Â This is the degrees of freedom, calculated as #groups-1 = 3-1 = 2.
• Asymp. Sig:Â This is the p-value associated with aÂ X2 test statistic of 3.097 with 2 degrees of freedom. This can also be found by using the Chi-Square Score to P Value Calculator.

Since the p-value (.213) is not less than .05, we fail to reject the null hypothesis. We do not have sufficient evidence to say that there is a statistically significant difference between the knee pain ratings across these three groups.