The 3rd homework

1.10

(1)计算对数经验生存函数的标准差

Sd=Variance−−−−−−−−√

Variance=var[lnSn(t)]≈1nF(t)1−F(t)

因为F(t)=1−e−t 且n=100

所以Sd=110et−1−−−−−√

(2)生成4个类似的容量为100的样本，画出它们的对数经验生存函数，可以发现随着t的增加，对数经验生存函数会变得不稳定，因为标准差变大了。

t1<-log(1-sort(rexp(100))/100)
t2<-log(1-sort(rexp(100))/100)
t3<-log(1-sort(rexp(100))/100)
t4<-log(1-sort(rexp(100))/100)

par(mfrow=c(2,2))
plot.ecdf(t1)
plot.ecdf(t2)
plot.ecdf(t3)
plot.ecdf(t4)

1.11

(1)画出beenswax数据的经验累积分布、直方图和Q-Q图

y=read.table("R\\beenswax.txt",header = TRUE)
library(lattice)

par(mfrow=c(1,2))
plot.ecdf(y$MeltingPoint,main="Meltingpoint")
plot.ecdf(y$Hydrocarbon,main="Hydrocarbon")

histogram(y$MeltingPoint,main="直方图1")
histogram(y$Hydrocarbon,main="直方图2")

qqnorm(y$MeltingPoint)
qqnorm(y$Hydrocarbon)

(2)找出0.90,0.75,0.50,0.25和0.10的分位数

“`{r}

c<-c(0.9,0.75,0.5,0.25,0.1)

算出meltingpoint的分位数

y1<-quantile(ecdf(y$MeltingPoint),c);y1

算出hydrocarbon的分位数

y2<-quantile(ecdf(y$Hydrocarbon),c);y2

“`

k	0	1	2	21	22	23
B(30,0.5)	9.31E-10	2.79E-08	4.05E-07	1.33E-02	5.45E-03	1.90E-03
N(15,7.5)	7.20E-03	9.32E-03	1.18E-02	3.86E-02	3.44E-02	3.01E-02
N(15−1/2,7.5)	NA	NA	NA	3.65E-02	3.23E-02	2.80E-02
N(15+1/2,7.5)	6.29E-03	8.21E-03	1.05E-02	NA	NA	NA