统计建模与R软件第四章习题…

原文地址：统计建模与R软件第四章习题答案(参数估计)作者：蘓木柒

Ex4.1
只会极大似然法，不会矩法...


Ex4.2
指数分布，λ的极大似然估计是n/sum(Xi)
>x<-c(rep(5,365),rep(15,245),rep(25,150),rep(35,100),rep(45,70),rep(55,45),rep(65,25))
>lamda<-length(x)/sum(x);lamda
[1] 0.05

Ex4.3
Poisson分布P(x=k)=λ^k/k!*e^(-λ)
其均数和方差相等，均为λ，其含义为平均每升水中大肠杆菌个数。
取均值即可。
>x<-c(rep(0,17),rep(1,20),rep(2,10),rep(3,2),rep(4,1))
>mean(x)
[1] 1
平均为1个。

Ex4.4
>obj<-function(x){f<-c(-13+x[1]+((5-x[2])*x[2]-2)*x[2],-29+x[1]+((x[2]+1)*x[2]-14)*x[2]);sum(f^2)} #其实我也不知道这是在干什么。所谓的无约束优化问题。
>x0<-c(0.5,-2)
>nlm(obj,x0)
$minimum
[1] 48.98425

$estimate
[1] 11.4127791-0.8968052

$gradient
[1] 1.411401e-08 -1.493206e-07

$code
[1] 1

$iterations
[1] 16

Ex4.5
>x<-c(54,67,68,78,70,66,67,70,65,69)
>t.test(x)       #t.test()做单样本正态分布区间估计

       One Sample t-test

data: x
t = 35.947, df = 9, p-value =4.938e-11
alternative hypothesis: truemean is not equal to 0
95 percent confidenceinterval:
 63.158571.6415
sampleestimates:
mean of x
    67.4
平均脉搏点估计为 67.4 ，95%区间估计为 63.158571.6415 。
>t.test(x,alternative="less",mu=72) #t.test()做单样本正态分布单侧区间估计

       One Sample t-test
data: x
t = -2.4534, df = 9, p-value =0.01828
alternative hypothesis: truemean is less than 72
95 percent confidenceinterval:
    -Inf 70.83705
sampleestimates:
mean of x
    67.4
p值小于0.05，拒绝原假设，平均脉搏低于常人。
要点：t.test()函数的用法。本例为单样本；可做双边和单侧检验。

Ex4.6
>x<-c(140,137,136,140,145,148,140,135,144,141);x
 [1] 140 137136 140 145 148 140 135 144 141
>y<-c(135,118,115,140,128,131,130,115,131,125);y
 [1] 135 118115 140 128 131 130 115 131 125
>t.test(x,y,var.equal=TRUE)

       Two Sample t-test

data:  x andy
t = 4.6287, df = 18, p-value =0.0002087
alternative hypothesis: truedifference in means is not equal to 0
95 percent confidenceinterval:
  7.5362620.06374
sampleestimates:
mean of x mean ofy
   140.6    126.8
期望差的95%置信区间为 7.53626 20.06374 。
要点：t.test()可做两正态样本均值差估计。此例认为两样本方差相等。
ps：我怎么觉得这题应该用配对t检验？

Ex4.7
>x<-c(0.143,0.142,0.143,0.137)
>y<-c(0.140,0.142,0.136,0.138,0.140)
>t.test(x,y,var.equal=TRUE)

       Two Sample t-test

data:  x andy
t = 1.198, df = 7, p-value =0.2699
alternative hypothesis: truedifference in means is not equal to 0
95 percent confidenceinterval:
 -0.001996351 0.006096351
sampleestimates:
mean of x mean ofy
 0.14125  0.13920
 期望差的95%的区间估计为-0.001996351 0.006096351

Ex4.8
接Ex4.6
>var.test(x,y)

       F test to compare two variances

data:  x andy
F = 0.2353, num df = 9, denomdf = 9, p-value = 0.04229
alternative hypothesis: trueratio of variances is not equal to 1
95 percent confidenceinterval:
 0.058452760.94743902
sampleestimates:
ratio ofvariances
        0.2353305
要点：var.test可做两样本方差比的估计。基于此结果可认为方差不等。
因此，在Ex4.6中，计算期望差时应该采取方差不等的参数。
>t.test(x,y)

       Welch Two Sample t-test

data:  x andy
t = 4.6287, df = 13.014,p-value = 0.0004712
alternative hypothesis: truedifference in means is not equal to 0
95 percent confidenceinterval:
  7.35971320.240287
sampleestimates:
mean of x mean ofy
   140.6    126.8
期望差的95%置信区间为 7.359713 20.240287。
要点：t.test(x,y，var.equal=TRUE)做方差相等的两正态样本的均值差估计
     t.test(x,y)做方差不等的两正态样本的均值差估计

Ex4.9
>x<-c(rep(0,7),rep(1,10),rep(2,12),rep(3,8),rep(4,3),rep(5,2))
>n<-length(x)
>tmp<-sd(x)/sqrt(n)*qnorm(1-0.05/2)
>mean(x)
[1] 1.904762
>mean(x)-tmp;mean(x)+tmp
[1] 1.494041
[1] 2.315483
平均呼唤次数为1.9
0.95的置信区间为1.49,2,32


Ex4.10
>x<-c(1067,919,1196,785,1126,936,918,1156,920,948)
>t.test(x,alternative="greater")

       One Sample t-test

data: x
t = 23.9693, df = 9, p-value =9.148e-10
alternative hypothesis: truemean is greater than 0
95 percent confidenceinterval:
 920.8443     Inf
sampleestimates:
mean of x
   997.1
灯泡平均寿命置信度95%的单侧置信下限为920.8443 
要点：t.test()做单侧置信区间估计