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5.14 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。
dataframe<-data.frame(
strength=c(9.60,9.69,8.43,9.98,11.28,10.10,11.01,10.44,9.00,9.57,9.03,9.80),
Temperature=gl(3,4,12),
pressure=gl(4,1,12))
summary (dataframe)
dataframe.aov2 <- aov(strength~pressure+Temperature,data=dataframe)
summary (dataframe.aov2)
> summary (dataframe.aov2)
Df Sum Sq Mean Sq F value Pr(>F)
pressure 3 0.581 0.1936 0.539 0.6727
Temperature 2 4.658 2.3288 6.487 0.0316 *
Residuals 6 2.154 0.3590
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
with(dataframe,interaction.plot(Temperature,pressure,strength,type="b",pch=19,fixed=T,xlab="Temperature (°F)",ylab="strength"))
plot.design(strength~pressure+Temperature,data=dataframe)
fit <-lm(strength~pressure+Temperature,data=dataframe)
anova(fit)
> anova(fit)
Analysis of Variance Table
Response: strength
Df Sum Sq Mean Sq F value Pr(>F)
pressure 3 0.5807 0.19356 0.5392 0.67270
Temperature 2 4.6576 2.32882 6.4873 0.03162 *
Residuals 6 2.1539 0.35898
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
summary(fit)
> summary(fit)
Call:
lm(formula = strength ~ pressure + Temperature, data = dataframe)
Residuals:
Min 1Q Median 3Q Max
-0.6575 -0.4902 0.1233 0.3067 0.6400
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 9.5575 0.4237 22.559 4.97e-07 ***
pressure2 -0.1733 0.4892 -0.354 0.7352
pressure3 -0.4700 0.4892 -0.961 0.3738
pressure4 0.1133 0.4892 0.232 0.8245
Temperature2 1.2825 0.4237 3.027 0.0232 *
Temperature3 -0.0750 0.4237 -0.177 0.8653
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 0.5991 on 6 degrees of freedom
Multiple R-squared: 0.7086, Adjusted R-squared: 0.4658
F-statistic: 2.918 on 5 and 6 DF, p-value: 0.1124
par(mfrow=c(2,2))
plot(fit)
par(mfrow=c(2,2))
plot(as.numeric(dataframe$pressure), fit$residuals, xlab="pressure", ylab="Residuals", type="p", pch=16)
plot(as.numeric(dataframe$Temperature), fit$residuals, xlab="Temperature", ylab="Residuals", pch=16)
