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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5.11 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(
density=c(570,565,583,528,547,521,1063,1080,1043,988,1026,1004,565,510,590,526,538,532),
Temperature=gl(3,6,18),
position=gl(2,3,18))
summary (dataframe)
dataframe.aov2 <- aov(density~position+Temperature,data=dataframe)
summary (dataframe.aov2)
> summary (dataframe.aov2)
Df Sum Sq Mean Sq F value Pr(>F)
position 1 7160 7160 16.2 0.00125 **
Temperature 2 945342 472671 1069.3 4.92e-16 ***
Residuals 14 6189 442
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
with(dataframe,interaction.plot(Temperature,position,density,type="b",pch=19,fixed=T,xlab="Temperature (°F)",ylab="density"))
plot.design(density~position*Temperature,data=dataframe)
fit <-lm(density~position+Temperature,data=dataframe)
anova(fit)
> anova(fit)
Analysis of Variance Table
Response: density
Df Sum Sq Mean Sq F value Pr(>F)
position 1 7160 7160 16.197 0.001254 **
Temperature 2 945342 472671 1069.257 4.924e-16 ***
Residuals 14 6189 442
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
summary(fit)
> summary(fit)
Call:
lm(formula = density ~ position + Temperature, data = dataframe)
Residuals:
Min 1Q Median 3Q Max
-53.444 -9.361 2.000 11.639 26.556
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 572.278 9.911 57.740 < 2e-16 ***
position2 -39.889 9.911 -4.025 0.00125 **
Temperature2 481.667 12.139 39.680 8.69e-16 ***
Temperature3 -8.833 12.139 -0.728 0.47880
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 21.03 on 14 degrees of freedom
Multiple R-squared: 0.9935, Adjusted R-squared: 0.9922
F-statistic: 718.2 on 3 and 14 DF, p-value: 1.464e-15
par(mfrow=c(2,2))
plot(fit)
par(mfrow=c(2,2))
plot(as.numeric(dataframe$position), fit$residuals, xlab="position", ylab="Residuals", type="p", pch=16)
plot(as.numeric(dataframe$Temperature), fit$residuals, xlab="Temperature", ylab="Residuals", pch=16)
