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GG AMB GGXAMB ATF6.M
1 COBB CONFORTO CC 1.7391386
2 COBB CONFORTO CC 0.8269537
3 COBB CONFORTO CC 0.3464495
4 COBB CONFORTO CC 1.3126458
5 COBB CONFORTO CC 1.3938351
6 COBB CONFORTO CC 1.0969472
7 COBB STRESS CS 3.1431619
8 COBB STRESS CS 0.9023480
9 COBB STRESS CS 2.5106332
10 COBB STRESS CS 1.2833235
11 COBB STRESS CS 0.4485298
12 COBB STRESS CS 0.3553028
13 PELOCO CONFORTO PC 0.3481456
14 PELOCO CONFORTO PC 2.5095779
15 PELOCO CONFORTO PC 0.8871572
16 PELOCO CONFORTO PC 2.3148108
17 PELOCO CONFORTO PC 73.2463832
18 PELOCO CONFORTO PC 16.0056771
19 PELOCO STRESS PS 15.4836898
20 PELOCO STRESS PS 1.2041695
21 PELOCO STRESS PS 1.8424005
22 PELOCO STRESS PS 0.9193776
23 PELOCO STRESS PS 0.9451780
24 PELOCO STRESS PS 0.9715508
最佳答案
使用您的数据:
data = structure(list(GG = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L
), .Label = c("COBB", "PELOCO"), class = "factor"), AMB = structure(c(1L,
1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 1L, 1L, 1L, 1L, 1L,
1L, 2L, 2L, 2L, 2L, 2L, 2L), .Label = c("CONFORTO", "STRESS"), class = "factor"),
GGXAMB = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L,
2L, 2L, 2L, 3L, 3L, 3L, 3L, 3L, 3L, 4L, 4L, 4L, 4L, 4L, 4L
), .Label = c("CC", "CS", "PC", "PS"), class = "factor"),
ATF6.M = c(1.7391386, 0.8269537, 0.3464495, 1.3126458, 1.3938351,
1.0969472, 3.1431619, 0.902348, 2.5106332, 1.2833235, 0.4485298,
0.3553028, 0.3481456, 2.5095779, 0.8871572, 2.3148108, 73.2463832,
16.0056771, 15.4836898, 1.2041695, 1.8424005, 0.9193776,
0.945178, 0.9715508)), class = "data.frame", row.names = c("1",
"2", "3", "4", "5", "6", "7", "8", "9", "10", "11", "12", "13",
"14", "15", "16", "17", "18", "19", "20", "21", "22", "23", "24"
))
f1 = aov(ATF6.M ~ GG + AMB + GGXAMB, data=data)
f2 = aov(ATF6.M ~ GG * AMB, data=data)
summary(f1)
Df Sum Sq Mean Sq F value Pr(>F)
GG 1 428 427.7 1.990 0.174
AMB 1 216 216.1 1.005 0.328
GGXAMB 1 240 239.9 1.116 0.303
Residuals 20 4299 214.9
summary(f2)
Df Sum Sq Mean Sq F value Pr(>F)
GG 1 428 427.7 1.990 0.174
AMB 1 216 216.1 1.005 0.328
GG:AMB 1 240 239.9 1.116 0.303
Residuals 20 4299 214.9
f1$coefficients
(Intercept) GGPELOCO AMBSTRESS GGXAMBCS GGXAMBPC GGXAMBPS
1.119328 14.765964 -12.324231 12.645452 NA NA
f2$coefficients
(Intercept) GGPELOCO AMBSTRESS GGPELOCO:AMBSTRESS
1.1193283 14.7659637 0.3212216 -12.6454525
GGXAMB
的组合可以返回
GG
的系数,例如 CC + CS 在
COBB
中为您提供
GG
,使您的 3 个系数变得多余。这将导致估计系数时出现问题。在这种情况下,效果是
AMBSTRESS
获得一个小值,其余为 NA。
aov(ATF6.M ~ GG*AMB, data)
或
aov(ATF6.M ~ GG+AMB+GG:AMB, data)
,它来自在完整排名矩阵上拟合线性模型,并且所有系数都是可估计的(如您从上面看到的)。
关于r - 方差分析模型中的 "+"与 "*"有什么区别?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/60789698/
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