完整版非参数统计第二版习题R程序Word格式.docx
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完整版非参数统计第二版习题R程序Word格式.docx
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t1<
for(iin1:
length(ss)){
if(ss[i]<
0)t<
-t+1#求小于80的个数
elset1<
-t1+1求大于80的个数
}
t;
t1
t;
[1]13
[1]15
binom.test(sum(scores<
80),length(scores),0.75)
p-value=0.001436<
0.01
Cox-Staut趋势存在性检验P47
例2.6
year<
-1971:
2002;
year
length(year)
rain<
c(206,223,235,264,229,217,188,204,182,230,223,
227,242,238,207,208,216,233,233,274,234,227,221,214,
226,228,235,237,243,240,231,210)
length(rain)
#
(1)该地区前10年降雨量是否变化?
t1=0
for(iin1:
5){
if(rain[i]<
rain[i+5])t1<
-t1+1
k<
-0:
t1-1
sum(dbinom(k,5,0.5))#=0.1875
y<
-6/(2A5);
y#=0.1875
#
(2)该地区前32年降雨量是否变化?
例2.9
rl<
-1+2*n1*n0/(n1+n0)*(1-1.96/sqrt(n1+n0));
rl
ru<
2*n1*n0/(n1+n0)*(1+1.96/sqrt(n1+n0));
ru#=15.3
3476(课本为ru=17)
t=0
16)(
rain[i+16])t<
-t+1
}t
k1<
min(t,16-t)-1
sum(dbinom(k1,16,0.5))#=0.0002593994pbinom(max(k1),16,0.5)#=0.0002593994y1<
-(1+16)/(2A16);
y1#=0.0002593994plot(year,rain)
abline(v=(1971+2002)/2,col=2)lines(year,rain)
anova(lm(rain~(year)))
随机游程检验(P50)例2.8
client<
-c("
F"
"
M"
"
clientn<
-length(client);
nn1<
-sum(client=="
n1
n0<
-n-n1;
n0t1<
-0for(iin1:
(length(client)-1))(
if(client[i]==client[i+1])t1<
-t1
-t1+1}
R<
-t1+1;
R#=12#findrejectionregion(不写)shuju39<
-data.frame(read.table
("
SHUJU39.txt"
header=TRUE));
shuju39
attach(shuju39)
sum.a=0
sum.b=0
sum.c=0
length(id))(
if(pinzhong[i]=="
A"
)sum.a<
-sum.a+chanliang[i]
elseif(pinzhong[i]=="
B"
)sum.b<
-sum.b+chanliang[i]
elsefuhao<
-sum.c<
-sum.c+chanliang[i]
sum.a;
sum.b;
sum.c
ma<
-sum.a/4
mb<
-sum.b/4
mc<
-sum.c/4
ma;
mb;
mc
fuhao<
-rep("
a"
12);
fuhao
&
((chanliang[i]-ma)>
0))fuhao[i]<
-"
+"
((chanliang[i]-mb)>
C"
((chanliang[i]-mc)>
elsefuhao[i]<
#利用上题编程解决检验的随机性
n<
-length(fuhao);
n
n1<
-sum(fuhao=="
n0
(length(fuhao)-1))(
if(fuhao[i]==fuhao[i+1])t1<
-t1+1;
R
#findrejectionregion
-2*n1*n0/(n1+n0)*(1+1.96/sqrt(n1+n0));
ru
例2.10(P52)library(quadprog)#不存在叫
'
quadprog'
这个名字的程辑包
library(zoo)#不存在叫’zoo'
library(tseries)#不存在叫’tseries'
这个名字的程辑包
run1=factor(c(1,1,1,0,rep(1,7),0,1,1,0,0,rep(1,6),0,rep(1,4),
0,rep(1,5),rep(0,4),rep(1,13)));
run1
y=factor(run1)
runs.test(y)#错误:
没有"
runs.test"
这个函数
Wilcoxon符号秩检验
W+在零假设下的精确分布
#下面的函数dwilxonfun用来计算W+分布密度函数,
即P(W+=x)的一个参考程序!
dwilxonfun=function(N){
a=c(1,1)#whenn=1frequencyofW+=1oro
n=1
pp=NULL#distributeofallsizefrom2toN
aa=NULL#frequencyofallsizefrom2toN
for(iin2:
N){
t=c(rep(0,i),a)
a=c(a,rep(0,i))+t
p=a/(2Ai)#densityofWilcoxdistributwhen
size=N
N=19#samplesizeofexpecteddistributionofW+
-dwilxonfun(N);
y
#计算P(W+=x)中的x取值的R参考程序!
!
p=a/(2Ai)#densityofWilcoxdistributwhensize=N
a
length(y)-1
hist(y,freq=FALSE)
lines(density(y),col="
)
例2.12(P59)
ceo<
-c(310,350,370,377,389,400,415,425,440,295,
325,296,250,340,298,365,375,360,385);
length(ceo)
#方法一
wilcox.test(ceo-320)
#方法二
ceo.num<
-sum(ceo>
320);
ceo.num
n=length(ceo)
binom.test(ceo.num,n,0.5)
例2.13(P61)
a<
-c(62,70,74,75,77,80,83,85,88)
walsh<
-NULL
(length(a)-1)){
for(jin(i+1):
length(a)){
-c(walsh,(a[i]+a[j])/2)
walsh=c(walsh,a)
NW=length(walsh);
NW
median(walsh)
2.5单组数据的位置参数置信区间估计(P61)
例2.14'
stu<
-c(82,53,70,73,103,71,69,
80,54,38,87,91,62,75,65,77);
stu
alpha=0.05
rstu<
-sort(stu);
rstu
conff<
-NULL;
conff
n=length(stu);
(n-1)){
n){
conf=pbinom(j,n,0.5)-pbinom(i,n,0.5)
if(conf>
1-alpha){conff<
-c(conff,i,j,conf)}
length(conff)
min<
-103-38;
min
c<
-seq(1,(length(conff)-1),3);
c
for(iinc){
col<
-c(rstu[conff[i]],rstu[conff[i+1]],conff[i+2])
min1<
-rstu[conff[i+1]]-rstu[conff[i]]
if(min1<
min){min<
-min1;
l<
-i}
print(col)
col1<
c(rstu[conff[l]],rstu[conff[l+1]],conff[l+2]);
col1
例2.14“
conf=pbinom(n,n,0.5)-pbinom(0,n,0.5);
conf
for(kin1:
conf=pbinom(n-k,n,0.5)-pbinom(k,n,0.5)
if(conf<
1-alpha){loc=k-1;
break}
print(loc)
(剩余的例题参考程序在课本)
3.6正态记分检验
例2.18
baby1<
-c(4,6,9,15,31,33,36,65,77,88)
baby=(baby1-34);
baby
baby.mean=mean(baby);
baby.mean
qiuzhi<
-function(x){
n=length(x)
a=rep(2,n)
a[i]=sum(x<
=x[i])
-function(x,y){
sgn=rep(2,n)
n)(
if(x[i]>
y)
sgn[i]=1
elseif(x[i]==y)
sgn[i]=0
else
sgn[i]=-1
sgn
-length(baby)
babyzhi=qiuzhi(baby)
q=(n1+1+babyzhi)/(2*n1+2)
babysgn<
-fuhao(baby,34)
babysgn=sign(baby1-34);
babysgn
s=qnorm(q,0,1)
W<
-t(s)%*%babysgn;
W
sd<
-sum((s*babysgn)A2);
sd
T=W/sd;
T
2.7分布的一致性检验
例2.19
shuju1<
-data.frame(month=c(1:
6),
customers=c(27,18,15,24,36,30));
shuju1
attach(shuju1)
-sum(customers);
expect<
-rep(1,6)*(1/6)*n;
expect
x.squ=sum((customers-expect)A2)/25;
x.squ
value<
-qchisq(1-0.05,length(customers)-1);
value
pvalue<
-1-pchisq(x.squ,length(customers)-1);
pvalue
例2.20
shuju2<
-data.frame(chongshu=c(0:
zhushu=c(10,24,10,4,1,0,1));
shuju2
attach(shuju2)
n=sum(zhushu);
lamda<
-sum(chongshu*zhushu)/n;
lamda
-dpois(chongshu,lamda);
n*p
x.squ=sum((zhushuA2)/(n*p))-n;
-qchisq(1-0.05,length(zhushu)-1);
-1-pchisq(x.squ,length(zhushu)-1);
例2.21
shuju3<
-c(36,36,37,38,40,42,43,43,44,45,48,48,
50,50,51,52,53,54,54,56,57,57,57,58,58,58,58,
58,59,60,61,61,61,62,62,63,63,65,66,68,68,70,
73,73,75);
shuju3
n=length(shuju3)
n0=sum(shuju3<
30);
n1=sum(shuju3>
30&
shuju3<
=40);
n2=sum(shuju3>
40&
=50);
n2
n3=sum(shuju3>
50&
=60);
n3
n4=sum(shuju3>
60&
=70);
n4
n5=sum(shuju3>
70&
=80);
n5
n6=sum(shuju3>
80);
n6
nn<
-c(n0,n1,n2,n3,n4,n5,n6);
nn#计算45位学生
体重分类的频数!
shuju3.mean=mean(shuju3);
shuju3.mean
shuju3.var=var(shuju3);
shuju3.var
shuju3.sd=sd(shuju3);
shuju3.sd
e0=pnorm(30,shuju3.mean,shuju3.sd)
e1=pnorm(40,shuju3.mean,shuju3.sd)-
pnorm(30,shuju3.mean,shuju3.sd)
e2=pnorm(50,shuju3.mean,shuju3.sd)-
pnorm(40,shuju3.mean,shuju3.sd)
e3=pnorm(60,shuju3.mean,shuju3.sd)-
pnorm(50,shuju3.mean,shuju3.sd)
e4=pnorm(70,shuju3.mean,shuju3.sd)-
pnorm(60,shuju3.mean,shuju3.sd)
e5=pnorm(80,shuju3.mean,shuju3.sd)-
pnorm(70,shuju3.mean,shuju3.sd)
e6=1-pnorm(80,shuju3.mean,shuju3.sd)
e=c(e0,e1,e2,e3,e4,e5,e6);
e
ee=n*c(e0,e1,e2,e3,e4,e5,e6);
ee
x.squ=sum((nnA2)/(ee))-n;
-qchisq(1-0.05,length(ee)-1);
-1-pchisq(x.squ,length(ee)-1);
例2.22
healthy<
c(87,77,92,68,80,78,84,77,81,80,80,77,92,86,
76,80,81,75,77,72,81,90,84,86,80,68,77,87,76,77,7
8,92,
75,80,78);
healthy
ks.test(healthy,pnorm,80,6)
第三章
#Brown_Mood中位数
#Brown-Mood中位数检验程序
BM.test<
-function(x,y,alt)(
xy<
-c(x,y)
#md.xy<
-quantile(xy,0.25)#利用p分位数的检
验
-sum(xy>
md.xy)
lx<
-length(x)
ly<
-length(y)
lxy<
-lx+ly
A<
-sum(x>
if(alt=="
greater"
(w<
-1-phyper(A,lx,ly,t)}
elseif(alt=="
less"
-phyper(A,lx,ly,t)}
conting.table=matrix(c(A,lx-A,lx,t-A,ly-(t-A),ly,t,lxy-t,lxy),3,3)
col.name<
X”,"
Y”,"
X+Y"
row.name<
MXY"
<
TOTAL"
dimnames(conting.table)<
-list(row.name,col.name)
list(contingency.table=conting.table,p.vlue=w)
例3.2
X<
-c(698,688,675,656,655,648,640,639,620)
Y<
-c(780,754,740,712,693,680,621)
BM.test(X,Y,"
XY<
-c(X,Y)
md.xy<
-median(XY)
-sum(XY>
-length(X)
-length(Y)
-sum(X>
#没有修正时的情形
pvalue1<
-pnorm(A,lx*t/(lx+ly),
-median(xy)#利用中位数的检验
Mx-My的R参考程序:
sqrt(lx*ly*t*(lx+ly-t)/(lx+ly)A3));
pvalue1
#修正时的情形
pvalue2<
-pnorm(A,lx*t/(lx+ly)-0.5,
pvalue2
3.2、Wilcoxon-Mann-Whitney秩和检验
#求两样本分别的秩和的程序.
Qiuzhi<
-function(x,y)(
yy<
wm=0
n1)(
wm=wm+sum(y[i]>
yy,1)
wm
例3.3
weight.low=c(134,146,104,119,124,161,
107,83,113,129,97,123)
m=length(weight.low)
weight.high=c(70,118,101,85,112,132,94)
n=length(weight.high)
wy<
-Qiuzhi(weight.low,weight.high)##wy=50
wxy<
-wy-n*(n+1)/2;
wxy#=22
mean<
-m*n/2
var<
-m*n*(m+n+1)/12
-1-2*pnorm(wxy,mean-0.5,var);
wilcox.test(weight.high,weight.low)
例3.4
x1<
-c(140,147,153,160,165,170,171,193)
x2<
-c(130,135,138,144,148,155,168)
-length(x1)
n2<
-length(x2)
th.hat<
-median(x2)-median(x1)
B=10000
Tboot=c(rep(0,1000))#vectoroflengthBootstrap
B)
(
xx1=sample(x1,5,T)#sampleofsizen1withreplacementfromx1
xx2=sample(x2,5,T)#sampleofsizen2withreplacementfromx2
Tboot[i]=median(xx2)-median(xx1)
th<
-median(Tboot);
th
se=sd(Tboot)
Normal.conf=c(th+qnorm(0.025
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