平均值的差是可以拟合到一个分布的（方差相同正态分布，不同t分布），所以SAS是有办法进行差异性检验，但是我不知道原来中位数的差值也可以拟合到一个分布么？SAS有程序可以检验median的差异性？

我在网上查的，就是说用proc npar1way data = temp1 wilcoxon;
说这个就是检验median的。

Using Wilcoxon scores in the linear rank statistic for two-sample data produces the rank sum statistic of the Mann-Whitney-Wilcoxon test. Using Wilcoxon scores in the one-way ANOVA statistic produces the Kruskal-Wallis test. Wilcoxon scores are locally most powerful for location shifts of a logistic distribution

Using median scores in the linear rank statistic for two-sample data produces the two-sample median test. The one-way ANOVA statistic with median scores is equivalent to the Brown-Mood test. Median scores are particularly powerful for distributions that are symmetric and heavy-tailed.

也是，wilcox秩和检验，非参倒是可以，果然理论已经忘得差不多了

恩，wilcox是给所有的观测数排个序，从大到小记个分，然后检验，median scores是观测数和中位数做个比较，按照谁大谁小来记个分，然后检验