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4.26~4.27 R语言解题。主要涉及方差分析，正交拉丁方。
batch <- c(rep("batch1",5), rep("batch2",5), rep("batch3",5), rep("batch4",5), rep("batch5",5))
acid <- rep(c("oper1", "oper2", "oper3", "oper4", "oper5"),5)
time <- c("A", "B", "C", "D", "E", "B", "C", "D", "E", "A", "C", "D", "E", "A", "B", "D", "E",
"A", "B", "C", "E", "A", "B", "C", "D")
catalyst <- c("a", "b", "y", "d", "e", "y", "d", "e", "a", "b", "e", "a", "b", "y", "d", "b", "y",
"d", "e", "a", "d", "e", "a", "b", "y")
y1 <- c(26,16,19,16,13)
y2 <- c(18,21,18,11,21)
y3 <- c(20,12,16,25,13)
y4 <- c(15,15,22,14,17)
y5 <- c(10,24,17,17,14)
y <- c(y1,y2,y3,y4,y5)
rocket.data <- data.frame(batch, acid,time,catalyst, y)
fit <- lm(y~ acid +batch+ time + catalyst, data=rocket.data)
anova(fit)
> anova(fit)
Analysis of Variance Table
Response: y
Df Sum Sq Mean Sq F value Pr(>F)
acid 4 24.4 6.10 1.0427 0.442543
batch 4 10.0 2.50 0.4274 0.785447
time 4 342.8 85.70 14.6496 0.000941 ***
catalyst 4 12.0 3.00 0.5128 0.728900
Residuals 8 46.8 5.85
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
order <- c(rep("order 1",4), rep("order 2",4), rep("order 3",4), rep("order 4",4))
oper <- rep(c("oper1", "oper2", "oper3", "oper4"),4)
method <- c("C", "B", "D", "A", "B", "C", "A", "D", "A", "D", "B", "C", "D", "A", "C", "B")
place <- c("b", "y", "d", "a", "a", "d", "y", "b", "d", "a", "b", "y", "y", "b", "a", "d")
y1 <- c(11,10,14,8)
y2 <- c(8,12,10,12)
y3 <- c(9,11,7,15)
y4 <- c(9,8,18,6)
y <- c(y1,y2,y3,y4)
rocket.data <- data.frame(order, oper, method, place, y)
fit <- lm(y~ method + order +oper+ place, data=rocket.data)
anova(fit)
> anova(fit)
Analysis of Variance Table
Response: y
Df Sum Sq Mean Sq F value Pr(>F)
method 3 95.5 31.833 3.4727 0.1669
order 3 0.5 0.167 0.0182 0.9960
oper 3 19.0 6.333 0.6909 0.6157
place 3 7.5 2.500 0.2727 0.8429
Residuals 3 27.5 9.167
