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线性混合模型中系数线性组合的p值

[英]p-values of linear combination of coefficients in linear mixed model

I fitted a linear mixed model using lme function of nlme package in R. My goal is to find a p-value of a linear combination of coefficients of fixed effects part. 我在R中使用nlme包的lme函数拟合了线性混合模型。我的目标是找到固定效果部分系数的线性组合的p值。

                 Value   Std.Error    DF   t-value p-value
(Intercept) -1.5897634 0.019578359 41547 -81.20004  0.0000
group_k      0.0866493 0.027662455   594   3.13238  0.0018
x1           0.1621744 0.006963385 41547  23.28960  0.0000

For example, p-value of the hypothesis $ \\beta_1 = 0 $ is provided in this result (0.0018). 例如,在此结果中提供了假设$ \\ beta_1 = 0 $的p值(0.0018)。 But if I want to find a p-value of a hypothesis such as $ \\beta_1 + 3 * \\beta_2 = $, what should I do? 但是,如果我想找到一个假设的p值,例如$ \\ beta_1 + 3 * \\ beta_2 = $,我该怎么办? Is there any ready-made command to do this calculation? 是否有现成的命令可以进行此计算?

It would be nice to have a reproducible example. 有一个可重现的例子会很好。 It's not very hard to set up your own test for this: 为此,设置您自己的测试不是很困难:

ct <- c(0,1,3)        ## set up contrast (linear comb. of coefficients)
m <- fixef(m) %*% ct  ## mean of contrast
v <- c(t(ct) %*% vcov(m) %*% ct)   ## variance of contrast
stder <- sqrt(v)      ## standard error
tstat <- m/stder      ## t statistic
2*pt(abs(tstat),df=594,lower.tail=FALSE)

I'm not sure about the degrees of freedom calculation, but all of the df are so large that it won't make any practical difference; 我不确定自由度的计算,但是所有df太大,以至于不会产生任何实际的变化。 you can use pnorm() instead of pt . 您可以使用pnorm()代替pt

If you want the theory, take a look at eg the section on Wald tests of contrasts in Dobson and Barnett Generalized Linear Models . 如果需要理论,请查看例如Dobson和Barnett 广义线性模型中的对比Wald检验一节。

You might also take a look at the effects , lsmeans , and/or multcomp packages. 您还可以查看effectslsmeans和/或multcomp软件包。

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