用R软件进行一元线性回归 实验报告

数理统计上机报告
上机实验题目：用R软件进行一元线性回归
上机实验目的：
1、进一步理解假设实验的基本思想，学会使用实验检验和进行统计推断。

2、学会利用R软件进行假设实验的方法。

一元线性回归基本理论、方法：
基本理论：假设预测目标因变量为Y，影响它变化的一个自变量为X，因变量随自变量的增（减）方向的变化。

一元线性回归分析就是要依据一定数量的观察样本（Xi, Yi），i=1，2…，n，找出回归直线方程Y=a+b*X
方法：对应于每一个Xi，根据回归直线方程可以计算出一个因变量估计值Yi。

回归方程估计值Yi 与实际观察值Yj之间的误差记作e-i=Yi-Yi。

显然，n个误差的总和越小，说明回归拟合的直线越能反映两变量间的平均变化线性关系。

据此，回归分析要使拟合所得直线的平均平方离差达到最小，据此，回归分析要使拟合所得直线的平均平方离差达到最小，简称最小二乘法将求出的a和b 代入式（1）就得到回归直线Yi=a+bXi 。

那么，只要给定Xi值，就可以用作因变量Yi的预测值。

（一）
实验实例和数据资料：
有甲、乙两个实验员，对同一实验的同一指标进行测定，两人测定的结果如
试问：甲、乙两人的测定有无显著差异？取显著水平α=0.05.
上机实验步骤：
1
（1）设置假设：H0：u1-u-2=0：H1：u1-u-2<0
（2）确定自由度为n1+n2-2=14；显著性水平a=0.05 （3）计算样本均值样本标准差和合并方差统计量的观测值alpha<-0.05;
n1<-8;
n2<-8;
x<-c(4.3,3.2,3.8,3.5,3.5,4.8,3.3,3.9);
y<-c(3.7,4.1,3.8,3.8,4.6,3.9,2.8,4.4);
var1<-var(x);
xbar<-mean(x);
var2<-var(y);
ybar<-mean(y);
Sw2<-((n1-1)*var1+(n2-1)*var2)/(n1+n2-2)
t<-(xbar-ybar)/(sqrt(Sw2)*sqrt(1/n1+1/n2));
tvalue<-qt(alpha,n1+n2-2)；
（4）计算临界值：tvalue<-qt(alpha,n1+n2-2)
（5）比较临界值和统计量的观测值，并作出统计推断
实例计算结果及分析：
alpha<-0.05;
> n1<-8;
> n2<-8;
> x<-c(4.3,3.2,3.8,3.5,3.5,4.8,3.3,3.9);
> y<-c(3.7,4.1,3.8,3.8,4.6,3.9,2.8,4.4);
> var1<-var(x);
> xbar<-mean(x);
> var2<-var(y);
> ybar<-mean(y);
> Sw2<-((n1-1)*var1+(n2-1)*var2)/(n1+n2-2)
> t<-(xbar-ybar)/(sqrt(Sw2)*sqrt(1/n1+1/n2));
> var1
[1] 0.2926786
> xbar
[1] 3.7875
> var2
[1] 0.2926786
2
> ybar
[1] 3.8875
Sw2
[1] 0.2926786
> t
[1] -0.3696873
tvalue
[1] -1.76131
分析：t=-0.3696873>tvalue=-1.76131，所以接受假设H1即甲乙两人的测定无显著性差异。

(二)
实验实例和数据资料：
2.某型号玻璃纸的横向延伸率要求不低于65%，且其服从正态分布，现对一批该批号的玻璃纸测得100个数据如下：
上机实验步骤：
(1)设置假设：H0：u=65, H1：u<65.
(2)确定自由度为n=100-1=99；显著性水平a=0.05
(3) 输入数据x<-
c(35.5,35.5,35.5,35.5,35.5,35.5,35.5,37.5,37.5,37.5,37.5,37.5,37.5,37.5,37.5,39.5,39.5,
3
39.5,39.5,39.5,39.5,39.5,39.5,39.5,39.5,39.5,41.5,41.5,41.5,41.5,41.5,41.5,41.5,41.5,4 1.5,43.5,43.5,43.5,43.5,43.5,43.5,43.5,43.5,43.5,45.5,45.5,45.5,45.5,45.5,45.5,45.5,45. 5,45.5,45.5,45.5,45.5,47.5,47.5,47.5,47.5,47.5,47.5,47.5,47.5,47.5,47.5,47.5,47.5,47.5, 47.5,47.5,47.5,47.5,49.5,49.5,49.5,49.5,49.5,49.5,49.5,49.5,49.5,49.5,49.5,49.5,49.5,4 9.5,51.5,51.5,51.5,51.5,51.5,53.5,53.5,53.5,55.5,55.5,59.5,59.5,63.5)
（4）用R软件计算临界值
（5）比较临界值和统计量的观测值，并作出推断
实例计算结果及分析：
计算过程如下：
alpha<-0.05;
n<-100;
x<-
c(35.5,35.5,35.5,35.5,35.5,35.5,35.5,37.5,37.5,37.5,37.5,37.5,37.5,37.5,37.5,39.5,39.5, 39.5,39.5,39.5,39.5,39.5,39.5,39.5,39.5,39.5,41.5,41.5,41.5,41.5,41.5,41.5,41.5,41.5,4 1.5,43.5,43.5,43.5,43.5,43.5,43.5,43.5,43.5,43.5,45.5,45.5,45.5,45.5,45.5,45.5,45.5,45. 5,45.5,45.5,45.5,45.5,47.5,47.5,47.5,47.5,47.5,47.5,47.5,47.5,47.5,47.5,47.5,47.5,47.5, 47.5,47.5,47.5,47.5,49.5,49.5,49.5,49.5,49.5,49.5,49.5,49.5,49.5,49.5,49.5,49.5,49.5,4 9.5,51.5,51.5,51.5,51.5,51.5,53.5,53.5,53.5,55.5,55.5,59.5,59.5,63.5)
sd1<-sd(x);
xbar<-mean(x);
t<-(xbar-65.0)/(sd1/sqrt(n));
tvalue<-qt(alpha,n-1);
sd1
[1] 5.815896
xbar
[1] 45.06
t
[1] -34.28534
tvalue
[1] -1.660391
分析推断：因为t=-34.28534<tvalue=-1.660391所以拒绝原假设。

即该批玻璃纸的横向延伸率不符合要求
（三）
实验实例和数据资料：
4
为了检验一种杂交作物的两种新处理方案，在同一地区随机选择16块地段在各实验地段，按两种方案处理作物，这8块地段的单位面积产量（单位：公斤）是：
一号方案产量：86 87 56 93 84 93 75 79 81 78 79 90 68 65 87 90；
二号方案产量：80 79 58 91 77 82 74 66 58 59 64 78 76 80 82 55；
假设两种方案的产量都服从正态分布，分别为N（u1，a^2）,N(u2,a^2),a^2未知，求均值差u1-u2的置信区间；
实例计算结果及分析：
利用R软件求解过程如下：
>alpha<-0.05;
> x<-c(86,87,56,93,84,93,75,79,81,78,79,90,68,65,87,90);
> y<-c(80,79,58,91,77,82,74,66,58,59,64,78,76,80,82,55);
> n1<-length(x);
> n2<-length(y);
> xbar=mean(x);
> ybar=mean(y);
> sw<-sqrt((n1-1)*sqrt(var(x))+(n2-1)*sqrt(var(y))) /(n1+n2-2);
> q<-qt(1-alpha/2,(n1+n2-2));
> left<-xbar-ybar-q*sw*sqrt(1/n1+1/n2);
> right<-xbar-ybar+q*sw*sqrt(1/n1+1/n2);
> n1
[1] 16
> n2
[1] 16
> left
[1] 7.819162
> right
[1] 8.680838
所以置信区间【7.819162，,8.680838】
5
6。