本文是实验设计与分析（第6版，Montgomery著，傅珏生译) 第章随机化区组，拉丁方， 及有关设计4.5节思考题4.1~4.4 R语言解题。主要涉及方差分析，随机化区组。
chemical<-data.frame(
X=c(73,68,74,71,67,73,67,75,72,70,75,68,78,73,68,73,71,75,75,69),
A = gl(4,5,20), #注释1 主因子chemical types
B = gl(5,1,20)) #区组
chemical.aov<-aov(X~A+B, data= chemical)
summary(chemical.aov)
> summary(chemical.aov)
Df Sum Sq Mean Sq F value Pr(>F)
A 3 12.95 4.32 2.376 0.121
B 4 157.00 39.25 21.606 2.06e-05 ***
Residuals 12 21.80 1.82
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
There is no difference among the chemical types at α = 0.05 level.
baterial<-data.frame(X=c(13,22,18,39,16,24,17,44,5,4,1,22),
A = gl(3,4,12), #注释1 主因子chemical types
B = gl(4,1,12)) #区组
baterial.aov<-aov(X~A+B, data= baterial)
summary(baterial.aov)
> summary(baterial.aov)
Df Sum Sq Mean Sq F value Pr(>F)
A 2 703.5 351.8 40.72 0.000323 ***
B 3 1106.9 369.0 42.71 0.000192 ***
Residuals 6 51.8 8.6
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
There is a difference between the means of the three solutions. The Fisher LSD procedure indicates that solution 3 is significantly different than the other two.
