# SPSS tutorials

BASICS DATA ANALYSIS T-TEST ANOVA CHI-SQUARE TEST

# SPSS Kruskal-Wallis Test – Simple Tutorial with Example

The Kruskal-Wallis test is an alternative for a one-way ANOVA if the assumptions of the latter are violated. We'll show in a minute why that's the case with creatine.sav, the data we'll use in this tutorial. But let's first take a quick look at what's in the data anyway.

## Quick Data Description

Our data contain the result of a small experiment regarding creatine, a supplement that's popular among body builders. These were divided into 3 groups: some didn't take any creatine, others took it in the morning and still others took it in the evening. After doing so for a month, their weight gains were measured. The basic research question is does the average weight gain depend on the creatine condition to which people were assigned? That is, we'll test if three means -each calculated on a different group of people- are equal. The most likely test for this scenario is a one-way ANOVA but using it requires some assumptions. Some basic checks will tell us that these assumptions aren't satisfied by our data at hand.

## Data Check 1 - Histogram

A very efficient data check is to run histograms on all metric variables. The fastest way for doing so is by running the syntax below.

*Create histogram of weight gain.

frequencies gain
/formats notable
/histogram.

## Histogram Result

First, our histogram looks plausible with all weight gains between -1 and +5 kilos, which are reasonable outcomes over one month. However, our outcome variable is not normally distributed as required for ANOVA. This isn't an issue for larger sample sizes of, say, at least 30 people in each group. The reason for this is the central limit theorem. It basically states that for reasonable sample sizes the sampling distribution for means and sums are always normally distributed regardless of a variable’s original distribution. However, for our tiny sample at hand, this does pose a real problem.

## Data Check 2 - Descriptives per Group

Right, now after making sure the results for weight gain look credible, let's see if our 3 groups actually have different means. The fastest way to do so is a simple MEANS command as shown below.

*Inspect means and standard deviations.

means gain by group.

## SPSS MEANS Output

First, note that our evening creatine group (4 participants) gained an average of 961 grams as opposed to 120 grams for “no creatine”. This suggests that creatine does make a real difference.
But don't overlook the standard deviations for our groups: they are very different but ANOVA requires them to be equal.The assumption of equal population standard deviations for all groups is known as homoscedasticity. This is a second violation of the ANOVA assumptions.

## Kruskal-Wallis Test

So what should we do now? We'd like to use an ANOVA but our data seriously violates its assumptions. Well, a test that was designed for precisely this situation is the Kruskal-Wallis test which doesn't require these assumptions. It basically replaces the weight gain scores with their rank numbers and tests whether these are equal over groups. We'll run it by following the screenshots below.

## Running a Kruskal-Wallis Test in SPSS

We use K Independent Samples if we compare 3 or more groups of cases. They are “independent” because our groups don't overlap (each case belongs to only one creatine condition).

Depending on your license, your SPSS version may or may have the option shown below. It's fine to skip this step otherwise.

## SPSS Kruskal-Wallis Test Syntax

Following the previous screenshots results in the syntax below. We'll run it and explain the output.

*Basic Kruskal-Wallis test syntax.

NPAR TESTS
/K-W=gain BY group(1 3)
/MISSING ANALYSIS.

## SPSS Kruskal-Wallis Test Output

We'll skip the “RANKS” table and head over to the “Test Statistics” shown below.

Our test statistic -incorrectly labeled as “Chi-Square” by SPSS- is known as Kruskal-Wallis H. A larger value indicates larger differences between the groups we're comparing. For our data it's roughly 3.87. We need to know its sampling distribution for evaluating whether this is unusually large.
Exact Sig. uses the exact (but very complex) sampling distribution of H. However, it turns out that if each group contains 4 or more cases, this exact sampling distribution is almost identical to the (much simpler) chi-square distribution.
We therefore usually approximate the significance level with a chi-square distribution. If we compare k groups, we have k - 1 degrees of freedom, denoted by df in our output.
Asymp. Sig. is the p-value based on our chi-square approximation. The value of 0.145 basically means there's a 14.5% chance of finding our sample results if creatine doesn't have any effect in the population at large. So if creatine does nothing whatsoever, we have a fair (14.5%) chance of finding such minor weight gain differences just because of random sampling. If p > 0.05, we usually conclude that our differences are not statistically significant.
Note that our exact p-value is 0.146 whereas the approximate p-value is 0.145. This supports the claim that H is almost perfectly chi-square distributed.

## Kruskal-Wallis Test - Reporting

The official way for reporting our test results includes our chi-square value, df and p as in this study did not demonstrate any effect from creatine, χ2(2) = 3.87, p = 0.15.

So that's it for now. I hope you found this tutorial helpful. Please let me know by leaving a comment below. Thanks!

# Let me know what you think!

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# This tutorial has 58 comments

• ### By Ruben Geert van den Berg on May 30th, 2018

Hi Ludovico!

If your sample size are reasonable (say each n > 25) you don't need any normality assumption. If the sample sizes are roughly equal too, you don't need homogeneity (equal population variances) either. In this case, just go for ANOVA and enter age as the covariate.

If the sample sizes are reasonably large but highly unequal and homogeneity is badly violated (use Levene's test for finding out) run ANOVA but interpret the Welch statistic instead of the normal F test statistic. Again, this allows you to enter age a the covariate.

If sample sizes are small and you don't meet assumptions, then you really should consider Kruskal-Wallis. However, this test doesn't take a covariate (not in SPSS or any other software). If that's really important, perhaps consider going for ANOVA despite possible bias from not meeting assumptions. Neither option is great.

• ### By Ludovico Piccolo on May 30th, 2018

Hi,
So I have 4 independent samples, homogeneity of distribution and homogeneity of variances are not met.
I have to compare these 4 groups on several dependent variables. I found an effect of Age, the problem is that SPSS doesn't allow me to enter Age as a covariate. How can I do it ?

• ### By Ruben Geert van den Berg on May 26th, 2018

Hi!

You don't need any kind of normality check. The Kruskal-Wallist test does not require normality.

Thanks!

• ### By geberehiwot on May 25th, 2018

Hi thank you for your tutorial
question to proced kruskal what are the procedure I mean how many times to transform for normality check

• ### By Ruben Geert van den Berg on May 15th, 2018

Hi Erik!

Nope. Your question refers to post hoc tests, which I'll cover in a new ANOVA tutorial that'll be up perhaps next week.

First off, the KW-test doesn't have any real post hoc tests. However, you could run pairwise Mann-Whitney tests and apply some sort of a Bonferroni correction.

But mind you: the overall effect being statistically significant does not mean that any of the post hoc test will be statistically significant. The reverse neither holds.

So perhaps just look at the means or mean ranks of the KW test and don't further overanalyze.

But that's just my dumb opinion ;-)