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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4.18~4.19 R语言解题。主要涉及方差分析，拉丁方。

batch <- c(rep("batch1",5), rep("batch2",5), rep("batch3",5), rep("batch4",5), rep("batch5",5))

  oper <- rep(c("oper1", "oper2", "oper3", "oper4", "oper5"),5)

  seed <- c("A", "B", "D", "C", "E", "C", "E", "A", "D", "B", "B", "A", "C", "E", "D", "D", "C",

             "E", "B", "A", "E", "D", "B", "A", "C")

  y1 <- c(8,7,1,7,3)

  y2 <- c(11,2,7,3,8)

  y3 <- c(4,9,10,1,5)

  y4 <- c(6,8,6,6,10)

  y5 <- c(4,2,3,8,8)

  y <- c(y1,y2,y3,y4,y5)

  rocket.data <- data.frame(batch, oper, seed, y)

  matrix(rocket.data$seed, nrow=5, byrow=T)

  rocket.data

  fit <- lm(y~seed+batch+oper, data=rocket.data)

  anova(fit)

> anova(fit)Analysis of Variance Table

Response: y

          Df Sum Sq Mean Sq F value    Pr(>F)   

seed       4 141.44  35.360 11.3092 0.0004877 ***

batch      4  15.44   3.860  1.2345 0.3476182   

oper       4  12.24   3.060  0.9787 0.4550143   

Residuals 12  37.52   3.127                     

---

Signif. codes: 

0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

batch <- c(rep("batch1",4), rep("batch2",4), rep("batch3",4), rep("batch4",4))

  oper <- rep(c("oper1", "oper2", "oper3", "oper4"),4)

  seed <- c("C", "D", "A", "B", "B", "C", "D", "A", "A", "B", "C", "D", "D", "A", "B", "C")

  y1 <- c(10,14,7,8)

  y2 <- c(7,18,11,8)

  y3 <- c(5,10,11,9)

  y4 <- c(10,10,12,14)

  y <- c(y1,y2,y3,y4)

  rocket.data <- data.frame(batch, oper, seed, y)

  matrix(rocket.data$seed, nrow=4, byrow=T)

  rocket.data

  fit <- lm(y~seed+batch+oper, data=rocket.data)

  anova(fit)

> anova(fit)Analysis of Variance Table

Response: y

          Df Sum Sq Mean Sq F value   Pr(>F)  

seed       3   72.5 24.1667 13.8095 0.004213 **

batch      3   18.5  6.1667  3.5238 0.088519 .

oper       3   51.5 17.1667  9.8095 0.009926 **

Residuals  6   10.5  1.7500                   

---

Signif. codes: 

0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1