实验设计与分析(第6版,Montgomery)第5章析因设计引导5.7节思考题5.6 R语言解题

发布于:2025-05-30 ⋅ 阅读:(18) ⋅ 点赞:(0)

本文是实验设计与分析(第6版,Montgomery著,傅珏生译) 第5章析因设计引导5.7节思考题5.6 R语言解题。主要涉及方差分析,正态假设检验,残差分析,交互作用图,等值线图。

dataframe <-data.frame(

strength=c(109,110,110,112,116,114,110,115,110,111,112,115,108,109,111,109,114,119,110,108,114,112,120,117),

machine=gl(4,6,24),

operator=gl(3,2,24))

summary (dataframe)

dataframe.aov2 <- aov(strength~operator*machine,data=dataframe)

summary (dataframe.aov2)

> summary (dataframe.aov2)

                 Df Sum Sq Mean Sq F value   Pr(>F)   

operator          2 160.33   80.17  21.143 0.000117 ***

machine           3  12.46    4.15   1.095 0.388753   

operator:machine  6  44.67    7.44   1.963 0.150681   

Residuals        12  45.50    3.79                    

---

Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

with(dataframe,interaction.plot(machine,operator,strength,type="b",pch=19,fixed=T,xlab="operator",ylab="strength"))

plot.design(strength~machine*operator,data=dataframe)

fit <-lm(strength~operator*machine,data=dataframe)

anova(fit)

> anova(fit)

Analysis of Variance Table

Response: strength

                 Df  Sum Sq Mean Sq F value    Pr(>F)   

operator          2 160.333  80.167 21.1429 0.0001167 ***

machine           3  12.458   4.153  1.0952 0.3887526   

operator:machine  6  44.667   7.444  1.9634 0.1506807   

Residuals        12  45.500   3.792                     

---

Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

summary(fit)

> summary(fit)

Call:

lm(formula = strength ~ operator * machine, data = dataframe)

Residuals:

   Min     1Q Median     3Q    Max

  -2.5   -1.0    0.0    1.0    2.5

Coefficients:

                     Estimate Std. Error t value Pr(>|t|)   

(Intercept)         1.095e+02  1.377e+00  79.527   <2e-16 ***

operator2           1.500e+00  1.947e+00   0.770   0.4560   

operator3           5.500e+00  1.947e+00   2.825   0.0153 * 

machine2            3.000e+00  1.947e+00   1.541   0.1493   

machine3           -1.000e+00  1.947e+00  -0.514   0.6169   

machine4           -5.000e-01  1.947e+00  -0.257   0.8017   

operator2:machine2 -3.500e+00  2.754e+00  -1.271   0.2278   

operator3:machine2 -4.500e+00  2.754e+00  -1.634   0.1282   

operator2:machine3  8.808e-14  2.754e+00   0.000   1.0000   

operator3:machine3  2.500e+00  2.754e+00   0.908   0.3818   

operator2:machine4  2.500e+00  2.754e+00   0.908   0.3818   

operator3:machine4  4.000e+00  2.754e+00   1.453   0.1720   

---

Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 1.947 on 12 degrees of freedom

Multiple R-squared:  0.827,     Adjusted R-squared:  0.6684

F-statistic: 5.214 on 11 and 12 DF,  p-value: 0.004136

par(mfrow=c(2,2))

plot(fit)

par(mfrow=c(2,2))

plot(as.numeric(dataframe$machine), fit$residuals, xlab="machine", ylab="Residuals", type="p", pch=16)

plot(as.numeric(dataframe$operator), fit$residuals, xlab="operator", ylab="Residuals", pch=16)


网站公告

今日签到

点亮在社区的每一天
去签到