我正在执行一次ANCOVA，以测试每种性别（分类变量，logLCC）的体型（协变量，logLP）和不同的头部测量值（响应变量，sexo）之间的关系是什么。
我在lm中获得了每个性别的斜率，我想将其与1进行比较。更具体地说，我想知道斜率是显着高于还是小于1，或者等于1，例如这在它们的异形关系中将具有不同的生物学意义。
这是我的代码：
#Modelling my lm#
> lm.logLP.sexo.adu<-lm(logLP~logLCC*sexo, data=ADU)
> anova(lm.logLP.sexo.adu)
Analysis of Variance Table
Response: logLP
Df Sum Sq Mean Sq F value Pr(>F)
logLCC 1 3.8727 3.8727 3407.208 < 2.2e-16 ***
sexo 1 0.6926 0.6926 609.386 < 2.2e-16 ***
logLCC:sexo 1 0.0396 0.0396 34.829 7.563e-09 ***
Residuals 409 0.4649 0.0011
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
#Obtaining slopes#
> lm.logLP.sexo.adu$coefficients
(Intercept) logLCC sexoM logLCC:sexoM
-0.1008891 0.6725818 -1.0058962 0.2633595
> lm.logLP.sexo.adu1<-lstrends(lm.logLP.sexo.adu,"sexo",var="logLCC")
> lm.logLP.sexo.adu1
sexo logLCC.trend SE df lower.CL upper.CL
H 0.6725818 0.03020017 409 0.6132149 0.7319487
M 0.9359413 0.03285353 409 0.8713585 1.0005241
Confidence level used: 0.95
#Comparing slopes#
> pairs(lm.logLP.sexo.adu1)
contrast estimate SE df t.ratio p.value
H - M -0.2633595 0.04462515 409 -5.902 <.0001
#Checking whether the slopes are different than 1#
#Computes Summary with statistics
> s1<-summary(lm.logLP.sexo.adu)
> s1
Call:
lm(formula = logLP ~ logLCC * sexo, data = ADU)
Residuals:
Min 1Q Median 3Q Max
-0.13728 -0.02202 -0.00109 0.01880 0.12468
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) -0.10089 0.12497 -0.807 0.42
logLCC 0.67258 0.03020 22.271 < 2e-16 ***
sexoM -1.00590 0.18700 -5.379 1.26e-07 ***
logLCC:sexoM 0.26336 0.04463 5.902 7.56e-09 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 0.03371 on 409 degrees of freedom
Multiple R-squared: 0.9083, Adjusted R-squared: 0.9076
F-statistic: 1350 on 3 and 409 DF, p-value: < 2.2e-16
#Computes t-student H0: intercept=1. The estimation of coefficients and their s.d. are in s1$coefficients
> t1<-(1-s1$coefficients[2,1])/s1$coefficients[2,2]
#Calculates two tailed probability
> pval<- 2 * pt(abs(t1), df = df.residual(lm.logLP.sexo.adu), lower.tail = FALSE)
> print(pval)
[1] 3.037231e-24
我在这里的几个线程中看到了整个过程。但是我所能理解的是我的斜率与1。
如何检查它们是否大于或小于1？
已编辑
解决了！
#performs one-side test H0=slope bigger than 1
pval<-pt(t1, df = df.residual(lm.logLP.sexo.adu), lower.tail = FALSE)
#performs one-side test H0=slope smaller than 1
pval<-pt(t1, df = df.residual(lm.logLP.sexo.adu), lower.tail = TRUE)
另外，应在单性别模型中进行测试。
最佳答案
如何检查它们是否大于或小于1？
像在this post，this post中一样，并且有疑问，您可以进行Wald测试，通过
t1<-(1-s1$coefficients[2,1])/s1$coefficients[2,2]
或者，使用
vcov和coef函数使代码更具可读性fit <- lm.logLP.sexo.adu
t1<-(1-coef(fit)[1])/vcov(fit)[1, 1]
Wald检验为您提供t统计量，该统计量可用于制作双面test或双面。因此，您可以放下
abs并根据要测试的尾巴设置lower.tail参数。关于r - 如何比较R中的斜率，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/47591911/