统计学分析

2023-12-19 21:44:17 浏览数 (1)

一.t检验

1.单样本t检验

代码语言:text复制
> daily.intake<-c(1,2,3,4,5)
> t.test(daily.intake,mu=10)#mu为已知总体均数

	One Sample t-test

data:  daily.intake
t = -9.8995, df = 4, p-value = 0.0005844#p值<0.05,认为daily.intake与总体均数有差异
alternative hypothesis: true mean is not equal to 10
95 percent confidence interval:#样本均数的置信区间
 1.036757 4.963243
sample estimates:
mean of x 
        3 

2.两独立样本均数比较的t检验

代码语言:text复制
install.packages("ISwR")
library(ISwR)
attach(energy)
energy#瘦子与胖子24小时消耗的能量
> t.test(expend~stature)

	Welch Two Sample t-test

data:  expend by stature
t = -3.8555, df = 15.919, p-value = 0.001411
alternative hypothesis: true difference in means between group lean and group obese is not equal to 0
95 percent confidence interval:#瘦子与胖子消耗能量差值的可信区间
 -3.459167 -1.004081
sample estimates:
 mean in group lean mean in group obese 
           8.066154           10.297778
           
           
 t.test(expend~stature,var.equal=T)#假定两样本方差齐,其结果与上相同         

如何判断方差齐不齐?——方差齐性检验

代码语言:text复制
> var.test(expend~stature)

	F test to compare two variances

data:  expend by stature
F = 0.78445, num df = 12, denom df = 8, p-value = 0.6797#p值大于0.05,认为方差齐,读取以上两种结果均可,第一种结果更保守
alternative hypothesis: true ratio of variances is not equal to 1
95 percent confidence interval:
 0.1867876 2.7547991
sample estimates:
ratio of variances 
          0.784446 

3.配对样本t检验

代码语言:text复制
attach(intake)#载入数据集
intake#同一个人绝经前与绝经后的能量摄入
> t.test(pre,post,paired=T)

	Paired t-test

data:  pre and post
t = 11.941, df = 10, p-value = 3.059e-07
alternative hypothesis: true mean difference is not equal to 0
95 percent confidence interval:
 1074.072 1566.838
sample estimates:
mean difference 
       1320.455

二.非参数秩和检验

1.单样本秩和检验

代码语言:javascript复制
> daily.intake<-c(1,2,3,4,5)
> wilcox.test(daily.intake,mu=10)

	Wilcoxon signed rank exact test

data:  daily.intake
V = 0, p-value = 0.0625
alternative hypothesis: true location is not equal to 10

2.两样本wilcoxon秩和检验

代码语言:text复制
library(ISwR)
attach(energy)
energy#瘦子与胖子24小时消耗的能量
> wilcox.test(expend~stature)

	Wilcoxon rank sum test with continuity correction

data:  expend by stature
W = 12, p-value = 0.002122
alternative hypothesis: true location shift is not equal to 0

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