具有 R 的一般混合线性模型中截距的假设检验 [英] Hypothesis test for intercepts in general mixed linear models with R
问题描述
我有固定效应的数据:基因型 = C、E、K、M；年龄 = 30、45、60、75、90 天；随机效果:block = 1, 2, 3;和变量 = weight_DM.
I have data of fixed effects: genotypes = C, E, K, M; age = 30, 45, 60, 75, 90 days; random effects: block = 1, 2, 3; and variable = weight_DM.
文件位于:https://drive.google.com/open?id=1_H6YZbdesK7pk5H23mZtp5KhVRKz0Ozl
我有每个基因型的年龄的线性和二次方斜率，但我没有截距和标准误差.R 代码是:
I have the slopes linear and quadratic of ages for each genotype, but I do not have the intercepts and standard errors. The R code is:
library(nlme)
library(lme4)
library(car)
library(carData)
library(emmeans)
library(ggplot2)
library(Matrix)
library(multcompView)
datos_weight <- read.csv2("D:/investigacion/publicaciones/articulos-escribiendo/pennisetum/pennisetum-agronomicas/data_weight.csv",header=T, sep = ";", dec = ",")
parte_fija_3 <- formula(weight_DM
~ Genotypes
+ Age
+ I(Age^2)
+ Genotypes*Age
+ Genotypes*I(Age^2))
heterocedasticidad_5 <- varComb(varExp(form = ~fitted(.)))
correlacion_4 <- corCompSymm(form = ~ 1|Block/Genotypes)
modelo_43 <- gls(parte_fija_3,
weights = heterocedasticidad_5,
correlation = correlacion_4,
na.action = na.omit,
data = datos_weight)
anova(modelo_43)
#response
Denom. DF: 48
numDF F-value p-value
(Intercept) 1 597.3828 <.0001
Genotypes 3 2.9416 0.0424
Age 1 471.6933 <.0001
I(Age^2) 1 22.7748 <.0001
Genotypes:Age 3 5.9425 0.0016
Genotypes:I(Age^2) 3 0.7544 0.5253
#################################
#test whether the linear age slopes of each genotype is equal to zero
################################
(tendencias_em_lin <- emtrends(modelo_43,
"Genotypes",
var = "Age"))
#response
Genotypes Age.trend SE df lower.CL upper.CL
C 1.693331 0.2218320 48 1.247308 2.139354
E 1.459517 0.2135037 48 1.030239 1.888795
K 2.001097 0.2818587 48 1.434382 2.567811
M 1.050767 0.1301906 48 0.789001 1.312532
Confidence level used: 0.95
(tendencias_em_lin_prueba <- update(tendencias_em_lin,
infer = c(TRUE,TRUE),
null = 0))
#response
Genotypes Age.trend SE df lower.CL upper.CL t.ratio p.value
C 1.693331 0.2218320 48 1.247308 2.139354 7.633 <.0001
E 1.459517 0.2135037 48 1.030239 1.888795 6.836 <.0001
K 2.001097 0.2818587 48 1.434382 2.567811 7.100 <.0001
M 1.050767 0.1301906 48 0.789001 1.312532 8.071 <.0001
Confidence level used: 0.95
########################################
#test differences between slope of the age linear for each genotype
########################################
CLD(tendencias_em_lin,
adjust = "bonferroni",
alpha = 0.05)
#response
Genotypes Age.trend SE df lower.CL upper.CL .group
M 1.050767 0.1301906 48 0.7128801 1.388653 1
E 1.459517 0.2135037 48 0.9054057 2.013628 12
C 1.693331 0.2218320 48 1.1176055 2.269057 12
K 2.001097 0.2818587 48 1.2695822 2.732611 2
Confidence level used: 0.95
Conf-level adjustment: bonferroni method for 4 estimates
P value adjustment: bonferroni method for 6 tests
significance level used: alpha = 0.05
问题
- 如何检验每个基因型的年龄截距是否为零?
- 如何测试每个基因型的年龄截距之间的差异?
- 每种基因型截距的标准误差是多少?
感谢您的帮助.
推荐答案
您可以使用 emmeans() 以与使用 emtrends() 类似的方式回答这些问题代码>.
You can answer these questions by using emmeans() in similar ways to what you did with emtrends().
另请查看 summary.emmGrid 的文档，并注意您可以选择是进行 CI、测试还是两者兼而有之.例如，
Also look at the documentation for summary.emmGrid, and note that you can choose whether to do CIs, tests, or both. e.g.,
emm <- emmeans(...)
summary(emm, infer = c(TRUE,TRUE))
summary(emm, infer = c(TRUE,FALSE)) # or confint(emm)
summary(emm, infer = c(FALSE,TRUE)) # or test(emm)
真正的拦截
如果实际上你想要实际的 y 截距，你可以使用
True intercepts
If in fact you want the actual y intercepts, you can do that using
emm <- emmeans(..., at = list(age = 0))
预测是在 0 岁进行的，这是每组条件的回归方程中的截距.但是，我想劝阻你不要这样做，因为 (a) 这些预测是巨大的外推，因此它们的标准误差也很大；(b) 预测 0 岁时的反应没有实际意义.因此，我认为问题 #1 基本上毫无意义.
The, predictions are made at age 0, which are the intercepts in the regression equations for each set of conditions. However, I would like to try to dissuade you from doing that because this because (a) these predictions are huge extrapolations, hence their standard errors are huge as well; and (b) it makes no practical sense to predict responses at age 0. For that reason, I think question #1 is basically meaningless.
如果您不考虑 at 部分，则 emmeans() 会在数据集中的平均年龄进行预测.与截距相比，这些预测的标准误差要小得多.由于模型中存在涉及 age 的交互，因此每个年龄的预测比较不同.我建议把
If you leave that at part out, then emmeans() makes predictions at the mean age in the dataset. Those predictions will have a much smaller standard error than the intercepts do. Since you have interactions involving age in the model, the predictions compare differently at each age. I suggest it would be useful to put
emm <- emmeans(..., cov.reduce = FALSE, by = "age")
这相当于使用 at 指定数据集中出现的 age 值集，并对每个年龄值进行单独比较.
which is equivalent to using at to specify the set of age values that occur in the dataset, and doing separate comparisons at each of those age values.
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