
关于R语言wilcox.test()函数输出结果的问题
```k1<-rweibull(100,shape=3,scale=4)k2<-rt(200,df=1)wilcox.test(k1,k2)Wilcoxonranksum...
```
k1 <- rweibull(100,shape=3,scale=4)
k2 <- rt(200,df=1)
wilcox.test(k1,k2)
Wilcoxon rank sum test with continuity correction
data: k1 and k2
W = 17512, p-value < 2.2e-16
alternative hypothesis: true location shift is not equal to 0
```
请问,W=17512,是代表什么意思??? 展开
k1 <- rweibull(100,shape=3,scale=4)
k2 <- rt(200,df=1)
wilcox.test(k1,k2)
Wilcoxon rank sum test with continuity correction
data: k1 and k2
W = 17512, p-value < 2.2e-16
alternative hypothesis: true location shift is not equal to 0
```
请问,W=17512,是代表什么意思??? 展开
2个回答
展开全部
你用的这是非参数的秩检验,当然W就是Wilcoxon的非参数统计量了。然后计算机把W=17512找本书一查(类似正态分布表之类的),发现pvalue小到不行 p-value < 2.2e-16
所以拒绝原假设,说明true location shift is not equal to 0
所以拒绝原假设,说明true location shift is not equal to 0
推荐律师服务:
若未解决您的问题,请您详细描述您的问题,通过百度律临进行免费专业咨询
广告 您可能关注的内容 |