解释R中的Probit模型估计

Question

这是我的数据框（请复制并粘贴以复制）：

Control <- replicate(2, c("112", "113", "116", "118", "127", "131", "134", "135", "136", "138", "143", "148", "149", "152", "153", "155", "162", "163"))
EPD <- replicate(2, c("101", "102", "103", "104", "105", "106", "107", "108", "109", "110", "114", "115", "117", "119", "120", "122", "124", "125", "126", "128", "130", "133", "137", "139", "140", "141", "142", "144", "145", "147"))
Subject <- c(Control, EPD)
Control_FA_L <- c(0.43, 0.39, 0.38, 0.58, 0.37, 0.5, 0.35, 0.36, 0.72, 0.38, 0.45, 0.30, 0.47, 0.30, 0.67, 0.34, 0.42, 0.29)
Control_FA_R <- c(0.36, 0.49, 0.55, 0.59, 0.33, 0.41, 0.32, 0.50, 0.59, 0.52, 0.32, 0.40, 0.49, 0.33, 0.46, 0.39, 0.37, 0.33)
EPD_FA_L <- c(0.25, 0.39, 0.36, 0.42, 0.21, 0.40, 0.43, 0.16, 0.31, 0.41, 0.39, 0.40, 0.35, 0.29, 0.31, 0.24, 0.39, 0.36, 0.54, 0.38, 0.34, 0.28, 0.42, 0.33, 0.40, 0.36, 0.42, 0.28, 0.40, 0.41)
EPD_FA_R <- c(0.26, 0.36, 0.36, 0.61, 0.22, 0.33, 0.36, 0.34, 0.35, 0.37, 0.39, 0.45, 0.30, 0.31, 0.50, 0.31, 0.29, 0.43, 0.41, 0.21, 0.38, 0.28, 0.66, 0.33, 0.50, 0.27, 0.46, 0.37, 0.26, 0.39)
FA <- c(Control_FA_L, Control_FA_R, EPD_FA_L, EPD_FA_R)
Control_Volume_L <- c(99, 119, 119, 146, 127, 96, 100, 132, 103, 103, 107, 142, 140, 134, 117, 117, 133, 143)
Control_Volume_R <- c(93, 123, 114, 152, 122, 105, 98, 138, 111, 110, 115, 137, 142, 140, 124, 102, 153, 143)
EPD_Volume_L <- c(132, 115, 140, 102, 130, 131, 110, 124, 102, 111, 93, 92, 94, 104, 92, 115, 144, 118, 104, 132, 90, 102, 94, 112, 106, 105, 79, 114, 104, 108)
EPD_Volume_R <- c(136, 116, 143, 105, 136, 137, 103, 121, 105, 115, 97, 97, 93, 108, 91, 117, 147, 111, 97, 129, 85, 107, 91, 116, 113, 101, 75, 108, 95, 98)
Volume <- c(Control_Volume_L, Control_Volume_R, EPD_Volume_L, EPD_Volume_R)
Group <- c(replicate(36, "Control"), replicate(60, "Patient"))

data <- data.frame(Subject, FA, Volume, Group)

我正在对变量FA建模，该变量被约束为[0,1]，但不代表比例，并且是非二进制的。我测试了几个链接，发现AIC的概率最低。因此，我选择使用glmmTMB软件包使用概率链接对模型进行拟合：

library(glmmTMB)
lmm <- glmmTMB(FA ~ Volume + Group + (1 | Subject), data = data, family = beta_family(link = "probit"))
summary(lmm)

据我了解，概率模型中的回归系数代表z得分的差异。从我们的概率模型的summary（）输出中，我们看到GroupPatient系数估计为-0.228。因此，患者组的FA的z评分比对照组的FA的z评分低0.228。我们还看到，根据我们的概率模型，体积系数为-0.00398。因此，每增加1单位体积，FA的z得分就降低-0.00398单位。

我主要担心的是，我想在更直观的背景下报告这些结果。我想知道是否有任何方法可以对这些估计值进行逆变换，以使它们与响应变量处于相同的规模。换句话说，我想报告模型估计的FA值，而不是z得分的差异。任何帮助将不胜感激！