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: true mean is not equal to 0 95 percent confidence interval:  63.1585 71.6415 sample estimates: mean of x      67.4 平均脉搏点估计为 67.4 ，95%区间估计为 63.1585 71.6415 。 > 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: true mean is less than 72 95 percent confidence interval:      -Inf 70.83705 sample estimates: 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] 140 137 136 140 145 148 140 135 144 141 > y<-c(135,118,115,140,128,131,130,115,131,125);y  [1] 135 118 115 140 128 131 130 115 131 125 > t.test(x,y,var.equal=TRUE)         Two Sample t-test data:  x and y t = 4.6287, df = 18, p-value = 0.0002087 alternative hypothesis: true difference in means is not equal to 0 95 percent confidence interval:   7.53626 20.06374 sample estimates: mean of x mean of y     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 and y t = 1.198, df = 7, p-value = 0.2699 alternative hypothesis: true difference in means is not equal to 0 95 percent confidence interval:  -0.001996351  0.006096351 sample estimates: mean of x mean of y   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 and y F = 0.2353, num df = 9, denom df = 9, p-value = 0.04229 alternative hypothesis: true ratio of variances is not equal to 1 95 percent confidence interval:  0.05845276 0.94743902 sample estimates: ratio of variances          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 and y t = 4.6287, df = 13.014, p-value = 0.0004712 alternative hypothesis: true difference in means is not equal to 0 95 percent confidence interval:   7.359713 20.240287 sample estimates: mean of x mean of y     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: true mean is greater than 0 95 percent confidence interval:  920.8443      Inf sample estimates: 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