#1 从相同的正态分布总体N(50,52)中随机抽两个样本,样本含量均为10。
#对两个样本进行成组t检验,检验水准为0.05,以比较其均数。重复上述步骤1000次,
#有多少次拒绝H0? 改变检验水准,例如0.10,结果如何?改变样本含量,结果如何?
#完成表格并下结论总结。
#样本含量为10时的模拟实验
#H1,H2,H3分别为α为0.05,0.10,0.20时拒绝无效假设H0的次数
H1<-0
for(i in 1:1000)
{
n1=rnorm(10,50,5)
n2=rnorm(10,50,5)
t<-t.test(n1,n2,var.equal = T,conf.level = 0.95)
if (t$p.value<0.05)
{
H1=H1+1
}
}
H2<-0
for(i in 1:1000)
{
n1=rnorm(10,50,5)
n2=rnorm(10,50,5)
t<-t.test(n1,n2,var.equal = T,conf.level = 0.95)
if (t$p.value<0.10)
{
H2=H2+1
}
}
H3<-0
for(i in 1:1000)
{
n1=rnorm(10,50,5)
n2=rnorm(10,50,5)
t<-t.test(n1,n2,var.equal = T,conf.level = 0.95)
if (t$p.value<0.20)
{
H3=H3+1
}
}
#样本含量为20时的模拟实验
H1<-0
for(i in 1:1000)
{
n1=rnorm(20,50,5)
n2=rnorm(20,50,5)
t<-t.test(n1,n2,var.equal = T,conf.level = 0.95)
if (t$p.value<0.05)
{
H1=H1+1
}
}
H2<-0
for(i in 1:1000)
{
n1=rnorm(20,50,5)
n2=rnorm(20,50,5)
t<-t.test(n1,n2,var.equal = T,conf.level = 0.95)
if (t$p.value<0.10)
{
H2=H2+1
}
}
H3<-0
for(i in 1:1000)
{
n1=rnorm(20,50,5)
n2=rnorm(20,50,5)
t<-t.test(n1,n2,var.equal = T,conf.level = 0.95)
if (t$p.value<0.20)
{
H3=H3+1
}
}
#样本含量为50时的模拟实验
H1<-0
for(i in 1:1000)
{
n1=rnorm(50,50,5)
n2=rnorm(50,50,5)
t<-t.test(n1,n2,var.equal = T,conf.level = 0.95)
if (t$p.value<0.05)
{
H1=H1+1
}
}
H2<-0
for(i in 1:1000)
{
n1=rnorm(50,50,5)
n2=rnorm(50,50,5)
t<-t.test(n1,n2,var.equal = T,conf.level = 0.95)
if (t$p.value<0.10)
{
H2=H2+1
}
}
H3<-0
for(i in 1:1000)
{
n1=rnorm(50,50,5)
n2=rnorm(50,50,5)
t<-t.test(n1,n2,var.equal = T,conf.level = 0.95)
if (t$p.value<0.20)
{
H3=H3+1
}
}