广义最小二乘结果解释

时间:2018-01-27 10:57:30

标签: r linear-regression

我检查了我的线性回归模型(WMAN = Species,WDNE = sea surface temp)并找到了自相关,所以相反,我正在尝试使用以下脚本的广义最小二乘;

library(nlme)
modelwa <- gls(WMAN ~WDNE, data=dat, 
               correlation = corAR1(form=~MONTH),
               na.action=na.omit)
summary(modelwa) 

我比较了两种模型;

> library(MuMIn)
> model.sel(modelw,modelwa)
Model selection table 
        (Intrc)   WDNE class na.action  correlation df   logLik   AICc delta
modelwa   31.50 0.1874   gls   na.omit crAR1(MONTH)  4 -610.461 1229.2  0.00
modelw    11.31 0.7974    lm   na.excl               3 -658.741 1323.7 94.44
        weight
modelwa      1
modelw       0
Abbreviations:
na.action: na.excl = ‘na.exclude’
correlation: crAR1(MONTH) = ‘corAR1(~MONTH)’
Models ranked by AICc(x) 

我相信结果表明我应该使用gls,因为AIC较低。

我的问题是,我一直在报告F值/R²/ p值,但gls的输出没有这些?

如果有人能协助我解释这些结果,我将非常感激?

> summary(modelwa)
Generalized least squares fit by REML
  Model: WMAN ~ WDNE 
  Data: mp2017.dat 
       AIC      BIC    logLik
  1228.923 1240.661 -610.4614

Correlation Structure: ARMA(1,0)
 Formula: ~MONTH 
 Parameter estimate(s):
     Phi1 
0.4809973 

Coefficients:
                Value Std.Error  t-value p-value
(Intercept) 31.496911  8.052339 3.911524  0.0001
WDNE         0.187419  0.091495 2.048401  0.0424

 Correlation: 
     (Intr)
WDNE -0.339

Standardized residuals:
      Min        Q1       Med        Q3       Max 
-2.023362 -1.606329 -1.210127  1.427247  3.567186 

Residual standard error: 18.85341 
Degrees of freedom: 141 total; 139 residual
> 

1 个答案:

答案 0 :(得分:0)

我现在已经克服了自动关联的问题,所以我可以使用lm()

将残差的lag1作为X变量添加到原始模型中。这可以使用slide包中的DataCombine函数来完成。

library(DataCombine)
econ_data <- data.frame(economics, resid_mod1=lmMod$residuals)
econ_data_1 <- slide(econ_data, Var="resid_mod1", 
NewVar = "lag1", slideBy = -1)
econ_data_2 <- na.omit(econ_data_1)
lmMod2 <- lm(pce ~ pop + lag1, data=econ_data_2)

可以找到此脚本here