Normality requirements for One-Way ANOVA, SPSS vs SigmaPlot


New Member

I’m a bit confused about the requirements for normality using a One-Way ANOVA. I had always thought that in order to use a One-Way ANOVA, 2 requirements had to be met:

1 - The data should be normally distributed for each category of the independent variable. That is, say I had 3 groups, each group was drawn randomly from the same population, and each group underwent a different treatment; the data for each group assessed individually (not collectively), should be normally distributed.
2 - There should be approximately equal variances.

The reason I’m a bit confused, is because depending on the package I use to analyse my data, I get different results.

If I use SPSS, I manually check that each set of data is normally distributed and that there are equal variances, which there are, then I run my One-Way ANOVA, no problems. However, if I use SigmaPlot, it automatically tests for normality and equal variances for me, however normality isn’t passed, and it will only allow me to do a Kruskal-Wallis.

Seen as I thought that it’s a bit weird that SPSS passes normality just fine (high P values for the 3 data sets), and the P value SigmaPlot gives me is very low (p = 0.018), and also as there’s just a single P value, and not 3! I manually tested for normality in SigmaPlot. Similar to SPSS, it tells me that the data are normally distributed. If however, I paste all of the data for my 3 groups into a single column, and assess normality it gives me the same P value (0.018) as when I try to run a One-Way ANOVA. So clearly SigmaPlot is checking for normality by collating all the 3 individual data sets into 1, and then checking for normality.

So, I guess my question is which method is right? I’m 99.999% sure that its the individual data sets should pass normality, and not the collective data grouped together, but then it also seems unlikely that SigmaPlot would use this method if it wasn’t a justifiable way of doing it. If it is a justifiable way, are there any benefits for using this methodology, opposed to the one SPSS uses?

Many thanks



TS Contributor
You want each population to be approximately normal and you want the populations to have a common variance (i.e. each group's variance is the same as the others, with respect to the dependent variable, of course).