Dear List! I have a unverse (basic population) which is not normally distributed. Now from this universe I take some subsets. Each subset is normally distributed within itself. I now want to compare the subsets and see if they differ significantly. So what is my assumption - normal distributed data and therefore comparison of means, or non normal distributed data and therefore comparison of medians? Cheers, Anna -- Der Inhalt dieser E-Mail ist vertraulich. Sollte Ihnen die E-Mail irrt?mlich zugesandt worden sein, bitte ich Sie, mich unverz?glich zu benachrichtigen und die E-Mail zu l?schen. This e-mail is confidential. If you have received it in error, please notify me immediately and delete it from your system.
Am 31.03.2011 15:46, schrieb Anna Lee:> Dear List! > > I have a unverse (basic population) which is not normally distributed. > Now from this universe I take some subsets. Each subset is normally > distributed within itself. I now want to compare the subsets and see > if they differ significantly. So what is my assumption - normal > distributed data and therefore comparison of means, or non normal > distributed data and therefore comparison of medians?If you want to do an ANOVA, the assumption is normality /within/ the groups. That is, Y given X=x is normal distributed. Also, you want the same variance within each group (group = your subset = factor value). Short answer: Means. (I am only 95% certain)
On 2011-03-31 07:44, Alexander Engelhardt wrote:> Am 31.03.2011 15:46, schrieb Anna Lee: >> Dear List! >> >> I have a unverse (basic population) which is not normally distributed. >> Now from this universe I take some subsets. Each subset is normally >> distributed within itself. I now want to compare the subsets and see >> if they differ significantly. So what is my assumption - normal >> distributed data and therefore comparison of means, or non normal >> distributed data and therefore comparison of medians? > > If you want to do an ANOVA, the assumption is normality /within/ the > groups. That is, Y given X=x is normal distributed. Also, you want the > same variance within each group (group = your subset = factor value).The equal-variances assumption is not strictly required. R has the multiple-groups equivalent of the two-sample t-test. See ?oneway.test. Peter Ehlers> > Short answer: Means. > (I am only 95% certain) > > ______________________________________________ > R-help at r-project.org mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code.