实验设计与分析（第6版，Montgomery)第4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

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                    

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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                    

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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.