library(openxlsx)
d5.7 = read.xlsx('C:/Users/12290/Desktop/多元统计表格/mvexer5.xlsx',sheet = 'E5.7')
d5.7
##   实验号 充磁量 定位角度 线圈匝数 输出力矩 是否启动
## 1      1    900       10       70      160        0
## 2      2    900       11       80      215        1
## 3      3    900       12       90      180        0
## 4      4   1100       10       70      168        0
## 5      5   1100       11       80      236        1
## 6      6   1100       12       90      190        0
## 7      7   1300       10       70      157        1
## 8      8   1300       11       80      205        1
## 9      9   1300       12       90      140        0
attach(d5.7)
x1=d5.7$充磁量
x2=d5.7$定位角度
x3=d5.7$线圈匝数
y=d5.7$输出力矩
Y=d5.7$是否启动

#相关系数的计算与假设检验

cor(x1,y)
## [1] -0.2473469
cor.test(x1,y)
## 
##  Pearson's product-moment correlation
## 
## data:  x1 and y
## t = -0.67541, df = 7, p-value = 0.5211
## alternative hypothesis: true correlation is not equal to 0
## 95 percent confidence interval:
##  -0.7828681  0.4986946
## sample estimates:
##        cor 
## -0.2473469
cor(x2,y)
## [1] 0.1166731
cor.test(x2,y)
## 
##  Pearson's product-moment correlation
## 
## data:  x2 and y
## t = 0.31081, df = 7, p-value = 0.765
## alternative hypothesis: true correlation is not equal to 0
## 95 percent confidence interval:
##  -0.5934307  0.7246455
## sample estimates:
##       cor 
## 0.1166731
cor(x3,y)
## [1] 0.1166731
cor.test(x3,y)
## 
##  Pearson's product-moment correlation
## 
## data:  x3 and y
## t = 0.31081, df = 7, p-value = 0.765
## alternative hypothesis: true correlation is not equal to 0
## 95 percent confidence interval:
##  -0.5934307  0.7246455
## sample estimates:
##       cor 
## 0.1166731
plot(d5.7[-1],gap=0)

根据P值和散点图来看，每一个单独变量与输出力矩之间都没有呈现线性关系

fm=lm(y~x1+x2+x3+x1*x2+x1*x3+x2*x3+x1*x2*x3)
#整体方差分析
anova(fm)
## Analysis of Variance Table
## 
## Response: y
##           Df Sum Sq Mean Sq F value  Pr(>F)  
## x1         1  468.2   468.2  1.2644 0.34268  
## x2         1  104.2   104.2  0.2813 0.63259  
## x1:x2      1  342.2   342.2  0.9243 0.40727  
## x2:x3      1 5582.7  5582.7 15.0771 0.03026 *
## x1:x2:x3   1   44.1    44.1  0.1191 0.75284  
## Residuals  3 1110.8   370.3                  
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

P（x2：x3）<0.05，说明定位角度和线圈匝数对输出力矩有显著影响

#将输出力矩与是否启动成功建立一个logistic回归模型

logit.glm<-glm(Y~y,family = binomial)
pre1<-predict(logit.glm,data.frame(y=291))
p1<-exp(pre1)/(1+exp(pre1))
p1
##         1 
## 0.9972859
library(MASS)
ld=lda(Y~y) #线性判别分析函数
ld
## Call:
## lda(Y ~ y)
## 
## Prior probabilities of groups:
##         0         1 
## 0.5555556 0.4444444 
## 
## Group means:
##        y
## 0 167.60
## 1 203.25
## 
## Coefficients of linear discriminants:
##          LD1
## y 0.03807743
lp=predict(ld) #根据线性判别模型预测所属类别
Y1=lp$class  #预测的所属类别结果
data.frame(Y,Y1)
##   Y Y1
## 1 0  0
## 2 1  1
## 3 0  0
## 4 0  0
## 5 1  1
## 6 0  1
## 7 1  0
## 8 1  1
## 9 0  0

九个预测值中只有一个出错，因此该判别函数可靠度较高