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