问题描述
我正在尝试用R解决练习，我在《 农业和自然科学实验设计》 一书中找到了这一点（第164页）
我已将表格传递给R软件
library(tibble)
tb = tibble("replica" = factor(x = rep(1:4,12),labels = c("I","II","III","IV")),"row" = factor(x = rep(1:4,c(12,12,12)),labels = c("12P","25P","12B","25B")),"hybrid" = factor(x = rep(1:2,c(24,24)),labels = c("P3730","B70XH55")),"density" = factor(x = c(rep(1:3,c(4,4,4)),rep(1:3,4)) ),labels = c("12000","16000","20000")),"valor" = c(140,138,130,142,145,146,150,147,149,136,132,134,140,128,141,148,136))
但是我很难创建aov模型
model = aov(valor ~ replica + hybrid + replica/hybrid + row + replica/row + density + replica/density + row:density,data = tb)
应用anova(model)时，它与下表的结果不一致。
如果还有其他方法可以应用该模型，请尽我所能。我之所以申请anova()是因为我可以提取msres anova(model)['Residuals','Mean Sq']，并且可以帮助我计算模型的可靠性（cv） >
解决方法
我认为输出是使用agricolae库中的ssp.plot获得的，但是首先您需要将行间距转换为连续变量：
tb$row = as.numeric(gsub("[A-Z]*","",as.character(tb$row)))
library(agricolae)
with(tb,ssp.plot(replica,hybrid,row,density,Y=valor))
ANALYSIS SPLIT-SPLIT PLOT: valor
Class level information
hybrid : P3730 B70XH55
row : 12 25
density : 12000 16000 20000
replica : I II III IV
Number of observations: 48
Analysis of Variance Table
Response: valor
Df Sum Sq Mean Sq F value Pr(>F)
replica 3 237.73 79.24 7.0135 0.0719692 .
hybrid 1 238.52 238.52 21.1106 0.0193734 *
Ea 3 33.90 11.30
row 1 475.02 475.02 71.6262 0.0001487 ***
hybrid:row 1 1.69 1.69 0.2545 0.6319437
Eb 6 39.79 6.63
density 2 350.04 175.02 19.9233 7.959e-06 ***
density:hybrid 2 37.04 18.52 2.1083 0.1433767
density:row 2 87.79 43.90 4.9968 0.0153376 *
density:hybrid:row 2 1.63 0.81 0.0925 0.9119816
Ec 24 210.83 8.78
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
cv(a) = 2.4 %,cv(b) = 1.9 %,cv(c) = 2.1 %,Mean = 138.8542