I have some continuous variables from my experiment with 6 different treatments, and I would like to ANOVA to compare the means. But the data are not normally distributed even after data transformation. I have tried log, square root, and Box-Cox transformations, and they did not improve the homoscedasticity of variance. Are there any other data transformation to normalize the data? Can I still use ANOVA for non-normally distributed data, or I should use non-parametric analysis?
If data fails normal distribution assumption, then ANOVA is invalid. The simple alternative is the Kruskal Wallis test, available in SPSS, Minitab. It uses the median values to conduct the test. Therefore, if your variables do not have wide variation, then you are unlikely to get very different results from ANOVA versus Kruskal Wallis.
It think it will be okay to employ non parametric tools since the the data did not pass Normality test. There is non parametric tools that is similar to ANOVA.
You mentioned that various transformations you tried "did not improve the homoscedasticity of variance". If heterogeneity of variance is the problem, you could use a form of ANOVA that allows for heterogeneous variances--e.g., the Brown-Forsythe method, or the Welch method. Both are available as part of the ONEWAY procedure in SPSS; and they are no doubt available in other major stats packages too. HTH.
If data fails normal distribution assumption, then ANOVA is invalid. The simple alternative is the Kruskal Wallis test, available in SPSS, Minitab. It uses the median values to conduct the test. Therefore, if your variables do not have wide variation, then you are unlikely to get very different results from ANOVA versus Kruskal Wallis.
Regarding the widely held belief that rank-based tests (like Mann-Whitney U and Kruskal-Wallis) test null hypotheses stating equivalence of the medians, see the links below.
The following is an excerpt from the last link below.
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Do non-parametric tests compare medians?
It is a commonly held belief that a Mann-Whitney U test is in fact a test for differences in medians. However, two groups could have the same median and yet have a significant Mann-Whitney U test. Consider the following data for two groups, each with 100 observations. Group 1: 98 (0), 1, 2; Group 2: 51 (0), 1, 48 (2). The median in both cases is 0, but from the Mann-Whitney test P
Hi, I will go back a bit in your test. Were you, are you, expecting your response to be normally distributed? If yes before to do an ANOVA which is pretty robust against non normal data, you should review why this is not normally distributed. Did you perform a Measurement System Analysis before? Are you sure your resolution is enough? Measurement problems are often responsible for having non normal data where normality is expected.
Yiju, I have four comments.
HTH.
You can still use not-normally distributed data, but with appropriate distribution.
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