R 如何计算两个50个数据点向量之间的逐点t测试？

我有一个包含3个变量和50个实例（ID、pre和post）的数据框架

ID<- c("1","2","3","4","5","6","7","8","9","10")
pre<- c("2.56802","2.6686","1.0145","0.2568","2.369","1.2365","0.6989","0.98745","1.09878","2.454658")
post<-c("3.3323","2.66989","1.565656","2.58989","5.96987","3.12145","1.23565","2.74741","2.54101","0.23568")

dfw1<-data.frame(ID,pre,post)

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ID该代码表明，在传统的5%置信水平下，您不能拒绝0差异的零假设：

ID<- c("1","2","3","4","5","6","7","8","9","10")
pre<- as.numeric(c("2.56802","2.6686","1.0145","0.2568","2.369","1.2365","0.6989","0.98745","1.09878","2.454658"))
post<-as.numeric(c("3.3323","2.66989","1.565656","2.58989","5.96987","3.12145","1.23565","2.74741","2.54101","0.23568"))
dfw1<-data.frame(ID,pre,post)
t.test(dfw1$pre,dfw1$post,alternative = c("two.sided"), conf.level = 0.95, paired=TRUE)

这背后的数学推理是什么？差异是多少？这可能很有用：。您需要使用原始数据集，而不仅仅是平均值，因此
t.test
可以计算std-dev，并查看您在每组中的观察值。另外，我可以看到pre和post，因此可能需要使用成对的
t.test
？Stack Overflow是一个问答网站，不是代码编写服务，请学习如何编写有效的问题。您将
pre
和
post
数值作为字符（引号）进行介绍，这是返回错误的可能性质。另外，你不应该在两个观测值之间计算t-ttest，而是在两个样本之间计算t-ttest。谢谢你的回答。我之前已经编写了这个代码行（dfw[，c（“pre”）]，dfw[，c（“post”）]），这给了我整个列的结果。我想得到每一行数据的结果。因为这些点是前置和后置的，所以我假设原始问题可以考虑这是配对测试。可能想考虑添加< <代码>成对=真/<代码>到T.测试函数。@ Dave2e：正确，解决。shanky：你的意思是向量被放置为行而不是列吗？考虑到你不能测试人口的点对点的差异，正如你所说的。一个点自身值的平均值和方差为0：即，它在统计上总是与另一个点不同。@000andy8484--Dave我在一列中有50个值，我的前值只是前6个值的平均值，我的后值是第6到第10个值的平均值。我计划进行t检验，以了解在哪里有重大变化。为此，我编写了两个循环，并生成了两个向量，一个包含所有pre值，另一个包含所有post值。我计划用前后值进行t检验。
ID<- c("1","2","3","4","5","6","7","8","9","10")
pre<- as.numeric(c("2.56802","2.6686","1.0145","0.2568","2.369","1.2365","0.6989","0.98745","1.09878","2.454658"))
post<-as.numeric(c("3.3323","2.66989","1.565656","2.58989","5.96987","3.12145","1.23565","2.74741","2.54101","0.23568"))
dfw1<-data.frame(ID,pre,post)
t.test(dfw1$pre,dfw1$post,alternative = c("two.sided"), conf.level = 0.95, paired=TRUE)

Paired t-test

data:  dfw1$pre and dfw1$post
t = -2.1608, df = 9, p-value = 0.05899
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
 -2.18109315  0.04997355
sample estimates:
mean of the differences 
               -1.06556