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5.7 R语言解题

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5.7 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(

force=c(2.70,2.78,2.83,2.86,2.45,2.49,2.85,2.80,2.60,2.72,2.86,2.87,2.75,2.86,2.94,2.88),

feed=gl(4,4,16),

speed=gl(2,2,16))

summary (dataframe)

dataframe.aov2 <- aov(force~feed*speed,data=dataframe)

summary (dataframe.aov2)

> summary (dataframe.aov2)

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

feed 3 0.09250 0.03083 11.859 0.00258 **

speed 1 0.14822 0.14822 57.010 6.61e-05 ***

feed:speed 3 0.04187 0.01396 5.369 0.02557 *

Residuals 8 0.02080 0.00260


Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

with(dataframe,interaction.plot(feed,speed,force,type="b",pch=19,fixed=T,xlab="feed",ylab="force"))

plot.design(force~feed*speed,data=dataframe)

fit <-lm(force~feed*speed,data=dataframe)

anova(fit)

> anova(fit)

Analysis of Variance Table

Response: force

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

feed 3 0.092500 0.030833 11.8590 0.002582 **

speed 1 0.148225 0.148225 57.0096 6.605e-05 ***

feed:speed 3 0.041875 0.013958 5.3686 0.025567 *

Residuals 8 0.020800 0.002600


Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

summary(fit)

> summary(fit)

Call:

lm(formula = force ~ feed * speed, data = dataframe)

Residuals:

Min 1Q Median 3Q Max

-0.06000 -0.02625 0.00000 0.02625 0.06000

Coefficients:

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

(Intercept) 2.740e+00 3.606e-02 75.994 1e-12 ***

feed2 -2.700e-01 5.099e-02 -5.295 0.000733 ***

feed3 -8.000e-02 5.099e-02 -1.569 0.155303

feed4 6.500e-02 5.099e-02 1.275 0.238172

speed2 1.050e-01 5.099e-02 2.059 0.073449 .

feed2:speed2 2.500e-01 7.211e-02 3.467 0.008482 **

feed3:speed2 1.000e-01 7.211e-02 1.387 0.202934

feed4:speed2 -5.912e-16 7.211e-02 0.000 1.000000


Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual standard error: 0.05099 on 8 degrees of freedom

Multiple R-squared: 0.9314, Adjusted R-squared: 0.8715

F-statistic: 15.53 on 7 and 8 DF, p-value: 0.0004502

par(mfrow=c(2,2))

plot(fit)

par(mfrow=c(2,2))

plot(as.numeric(dataframefeed), fitresiduals, xlab="feed", ylab="Residuals", type="p", pch=16)

plot(as.numeric(dataframespeed), fitresiduals, xlab="speed", ylab="Residuals", pch=16)

dataframe<-data.frame(

force=c(2.70,2.78,2.83,2.86,2.45,2.49,2.85,2.80,2.60,2.72,2.86,2.87,2.75,2.86,2.94,2.88),

feed=c(0.015,0.015,0.015,0.015,0.030,0.030,0.030,0.030,0.045,0.045,0.045,0.045,0.060,0.060,0.060,0.060),

speed=c(125,125,200,200,125,125,200,200,125,125,200,200,125,125,200,200))

fit <-lm(force~feed*speed+feed*I(speed^2)+I(feed^2)*speed+I(feed^2)+I(speed^2),data=dataframe)

tmp.speed <- seq(125,200,by=.5)

tmp.feed <- seq(0.015,0.060,by=.005)

tmp <- list(feed=tmp.feed,speed=tmp.speed)

new <- expand.grid(tmp)

new$fit <- c(predict(fit,new))

require(lattice)

contourplot (fit~feed*speed ,data=new, cuts=8,region=T,col.regions=gray(7:16/16))

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