If p > 0.05, we usually conclude that our differences are not statistically significant. 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.

The value of 0.145 basically means thereвЂ™s a 14.5% chance of finding our sample results if creatine doesnвЂ™t have any effect in the population at large. Sig.В is the p-value based on our chi-square approximation. If we compare k groups, we have k вЂ“ 1 degrees of freedom, denoted by dfВ in our output.ДЄsymp.

We therefore usually approximate the significance levelВ with a chi-square distribution. 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 need to know its sampling distributionВ for evaluating whether this is unusually large.Д®xact Sig.В uses the exact (but very complex) sampling distribution of H. A larger value indicates larger differences between the groups weвЂ™re comparing. Our test statistic -incorrectly labeled as вЂњ Chi-SquareвЂќ by SPSS- is known as Kruskal-Wallis H. WeвЂ™ll skip the вЂњRANKSвЂќ table and head over to the вЂњTest StatisticsвЂќ shown below. ItвЂ™s fine to skip this step otherwise.ДЇollowing the previous screenshots results in the syntax below. They are вЂњindependentвЂќ because our groups donвЂ™t overlap (each case belongs to only one creatine condition).Дepending on your license, your SPSSВ version may or may have the E xactВ option shown below. We use KВ Independent SamplesВ if we compare 3 or more groups of cases. WeвЂ™ll run it by following the screenshots below. It basically replaces the weight gain scores with their rank numbers and tests whether these are equal over groups. Well, a test that was designed for precisely this situation is the Kruskal-Wallis test which doesnвЂ™t require these assumptions. So what should we do now? WeвЂ™d like to use an ANOVA but our data seriously violates its assumptions. *В This is a second violation of the ANOVA assumptions. This suggests that creatine does make a real difference.Д«ut donвЂ™t overlook the standard deviationsВ for our groups: they are very different but ANOVA requires them to be equal. The fastest way for doing so is by running the syntax below.ДЇirst, note that our evening creatine group (4 participants) gained an average of 961 grams as opposed to 120 grams for вЂњno creatineвЂќ. Data Check 1 вЂ“ HistogramДЄ very efficient data checkВ is to run histogramsВ on all metric variables. Some basic checks will tell us that these assumptions arenвЂ™t satisfied by our data at hand. The most likely test for this scenario is a one-way ANOVA but using it requires some assumptions.

The basic research question is does the average weight gain depend on the creatine condition to which people were assigned?That is, weвЂ™ll test if three means -each calculated on a different group of people- are equal. After doing so for a month, their weight gains were measured. These were divided into 3 groups: some didnвЂ™t take any creatine, others took it in the morning and still others took it in the evening. Our data contain the result of a small experiment regarding creatine, a supplement thatвЂ™s popular among body builders. But letвЂ™s first take a quick look at whatвЂ™s in the data anyway. WeвЂ™ll show in a minute why thatвЂ™s the case with creatine.sav, the data weвЂ™ll use in this tutorial. The Kruskal-Wallis testВ is an alternative for a one-way ANOVAВ if the assumptions of the latter are violated.

Kruskal wallis test spss