两组数（mean、median）差异性检验

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想要得到如下表格：

	Group 1	Group 2	Difference  in mean (p-value)	Difference  in median (p-value)
Capex	Mean(median)	Mean(median)	diff (p)	diff (p)

用ttest，很容易得到mean的差异值。
* difference in mean;
proc ttest data=temp1;
   class group;
   var capex;
run;


但是用npar1way，结果里面没有输出median 和 difference in median，只有p-value。如下。请问，如何得到上面的表格呢？想到一个土办法，把两组的median分别算出来，然后自己减一下。。。这个，这个，是不是有点傻？



*difference in median;
proc npar1way data = temp1 wilcoxon;
        class group;
        var capex;
run;

Wilcoxon Scores (Rank Sums) for Variable capex Classified by Variable gr
gr	N	Sum of Scores	Expected Under H0	Std Dev Under H0	Mean Score
top	4284	18500092.0	18389070.0	114799.062	4318.41550
bottom	4300	18346728.0	18457750.0	114799.062	4266.68093
Average scores were used for ties.

Wilcoxon Two-Sample Test
Statistic	18500092.0000
Normal Approximation
Z	0.9671
One-Sided Pr > Z	0.1667
Two-Sided Pr > |Z|	0.3335
t Approximation
One-Sided Pr > Z	0.1668
Two-Sided Pr > |Z|	0.3335
Z includes a continuity correction of 0.5.

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关键词：median 差异性检验 Media mean Med median

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？

intheangel 发表于 2014-5-20 10:04
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.

jy1st 发表于 2014-5-20 10:33
我在网上查的，就是说用proc npar1way data = temp1 wilcoxon;
说这个就是检验median的。

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

intheangel 发表于 2014-5-20 16:33
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

hkhl237 发表于 2014-5-20 16:40
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， ...

额~~~秩和检验我还是学过的。。。thx

intheangel 发表于 2014-5-20 21:20
额~~~秩和检验我还是学过的。。。thx

楼上两位，所以，我应该怎么做呢？还是困惑ing

You can get medians from other procedures, for examples, proc means.
It is not strange why NPAR1WAY does not provide medians, etc. The reason is that this procedure does the computation based all on  the ranks or scores, but not the data values themselves, with an exception of anova scores. it gives group means.
JingJu

jingju11 发表于 2014-5-21 09:43
You can get medians from other procedures, for examples, proc means.
It is not strange why NPAR1WA ...

Thanks a lot.
Although I can get median from proc means, the p-value of difference-in-median is still calculated by NPAR1WA. Right?