R
比較重複測量 FE 模型中的組，具有嵌套誤差分量，使用 plm 估計
我已經估計了一些重複測量的固定效應模型，具有嵌套的誤差分量，基於分組變量，即非嵌套模型，使用plm. 我現在有興趣
- 測試完整模型是否顯著不同，即在哪裡是和的完整
Females模型是和的完整Males模型- 隨後測試兩組之間選定的回歸係數，即在哪裡是女性在 的回歸係數
year1.5，並且是男性的回歸係數year1.5。我將使用以下工作示例來說明情況，
首先，需要一些包，
# install.packages(c("plm","texreg","tidyverse","lmtest"), dependencies = TRUE) library(plm); library(lmtest); require(tidyverse)二、一些資料準備，
data(egsingle, package = "mlmRev") dta <- egsingle %>% mutate(Female = recode(female,.default = 0L,`Female` = 1L))三、我為數據中的每個性別估計了一組模型
MoSpc <- as.formula(math ~ Female + size + year) dfMo = dta %>% group_by(female) %>% do(fitMo = plm(update(MoSpc, . ~ . -Female), data = ., index = c("childid", "year", "schoolid"), model="within") )第四，讓我們看一下兩個估計模型，
texreg::screenreg(dfMo[[2]], custom.model.names = paste0('FE: ', dfMo[[1]])) #> =================================== #> FE: Female FE: Male #> ----------------------------------- #> year-1.5 0.79 *** 0.88 *** #> (0.07) (0.10) #> year-0.5 1.80 *** 1.88 *** #> (0.07) (0.10) #> year0.5 2.51 *** 2.56 *** #> (0.08) (0.10) #> year1.5 3.04 *** 3.17 *** #> (0.08) (0.10) #> year2.5 3.84 *** 3.98 *** #> (0.08) (0.10) #> ----------------------------------- #> R^2 0.77 0.79 #> Adj. R^2 0.70 0.72 #> Num. obs. 3545 3685 #> =================================== #> *** p < 0.001, ** p < 0.01, * p < 0.05 #>現在，我想測試這兩個（線性 OLS）模型是否顯著不同，參見。上面的第 1 點。我環顧了 SO和互聯網，有些人建議我需要使用
plm::pFtest()，也建議在這裡，我已經嘗試過，但我不相信。我會想像一些非嵌套模型的測試，可能的 Cox 測試，lmtest::coxtest但我完全不確定。如果這裡有人可以幫助我。我試過了，
plm::pFtest(dfMo[[1,2]], dfMo[[2,2]]) # > # > F test for individual effects # > # >data: update(MoSpc, . ~ . - Female) # >F = -0.30494, df1 = 113, df2 = 2693, p-value = 1 # >alternative hypothesis: significant effects和，
lmtest::coxtest(dfMo[[1,2]], dfMo[[2,2]]) # > Cox test # > # > Model 1: math ~ size + year # > Model 2: math ~ size + year # > Estimate Std. Error z value Pr(>|z|) # > fitted(M1) ~ M2 0.32 1.66695 0.1898 0.8494 # > fitted(M2) ~ M1 -1222.87 0.13616 -8981.1963 <2e-16 *** # > --- # > Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 # > Warning messages: # > 1: In lmtest::coxtest(dfMo[[1, 2]], dfMo[[2, 2]]) : # > models fitted on different subsets # > 2: In lmtest::coxtest(dfMo[[1, 2]], dfMo[[2, 2]]) : # > different dependent variables specified其次，我有興趣比較兩組之間的回歸係數。比如說，
year1.53.04 的估計值是否與 3.17 顯著不同？參照。上述第 2 點。請詢問以上是否有任何不清楚的地方，我很樂意詳細說明。任何幫助將不勝感激！
我意識到這個問題有點像編程，但我最初是在 SO 中發布的。但是，DWin非常友好地指出該問題屬於 CrossValidated 並將其遷移到此處。
下面的代碼實現了在
Femaledummy 和 year 之間放置交互的做法。底部的 F 測試測試你的 null. 輸出中的 t 統計plm量測試您的空值. 特別是，對於year=1.5，p 值為 0.32。library(plm) # Use plm library(car) # Use F-test in command linearHypothesis library(tidyverse) data(egsingle, package = 'mlmRev') dta <- egsingle %>% mutate(Female = recode(female, .default = 0L, `Female` = 1L)) plm1 <- plm(math ~ Female * (year), data = dta, index = c('childid', 'year', 'schoolid'), model = 'within') # Output from `summary(plm1)` --- I deleted a few lines to save space. # Coefficients: # Estimate Std. Error t-value Pr(>|t|) # year-1.5 0.8842 0.1008 8.77 <2e-16 *** # year-0.5 1.8821 0.1007 18.70 <2e-16 *** # year0.5 2.5626 0.1011 25.36 <2e-16 *** # year1.5 3.1680 0.1016 31.18 <2e-16 *** # year2.5 3.9841 0.1022 38.98 <2e-16 *** # Female:year-1.5 -0.0918 0.1248 -0.74 0.46 # Female:year-0.5 -0.0773 0.1246 -0.62 0.53 # Female:year0.5 -0.0517 0.1255 -0.41 0.68 # Female:year1.5 -0.1265 0.1265 -1.00 0.32 # Female:year2.5 -0.1465 0.1275 -1.15 0.25 # --- xnames <- names(coef(plm1)) # a vector of all independent variables' names in 'plm1' # Use 'grepl' to construct a vector of logic value that is TRUE if the variable # name starts with 'Female:' at the beginning. This is generic, to pick up # every variable that starts with 'year' at the beginning, just write # 'grepl('^year+', xnames)'. picked <- grepl('^Female:+', xnames) linearHypothesis(plm1, xnames[picked]) # Hypothesis: # Female:year - 1.5 = 0 # Female:year - 0.5 = 0 # Female:year0.5 = 0 # Female:year1.5 = 0 # Female:year2.5 = 0 # # Model 1: restricted model # Model 2: math ~ Female * (year) # # Res.Df Df Chisq Pr(>Chisq) # 1 5504 # 2 5499 5 6.15 0.29