[英]One p-value for glm model
我正在尋找一種獲得一個p值的方法,該值描述了glm模型的擬合優度。 以下是來自lm
聯機幫助頁的略微修改的示例:
ctl <- c(4.17,5.58,5.18,6.11,4.50,4.61,5.17,4.53,5.33,5.14)
trt <- c(4.81,4.17,4.41,3.59,5.87,3.83,6.03,4.89,4.32,4.69)
conf<- c(rnorm(mean=-1, sd=1, n=10), rnorm(mean=1, sd=1, n=10))
group <- gl(2,10,20, labels=c("Ctl","Trt"))
weight <- c(ctl, trt)
lm.D9 <- lm(weight ~ group + conf)
隨着summary(lm.D9)
得到
Call:
lm(formula = weight ~ group + conf)
Residuals:
Min 1Q Median 3Q Max
-1.17619 -0.40373 -0.05262 0.24987 1.40777
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 4.97416 0.25153 19.775 3.6e-13 ***
groupTrt -0.23724 0.41117 -0.577 0.572
conf -0.07044 0.13725 -0.513 0.614
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 0.7111 on 17 degrees of freedom
Multiple R-squared: 0.08722, Adjusted R-squared: -0.02017
F-statistic: 0.8122 on 2 and 17 DF, p-value: 0.4604
如果id對glm做同樣的事情
glm.D9 <- glm(weight ~ group + conf)
summary(glm.D9)
我明白了
Call:
glm(formula = weight ~ group + conf)
Deviance Residuals:
Min 1Q Median 3Q Max
-1.17619 -0.40373 -0.05262 0.24987 1.40777
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 4.97416 0.25153 19.775 3.6e-13 ***
groupTrt -0.23724 0.41117 -0.577 0.572
conf -0.07044 0.13725 -0.513 0.614
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
(Dispersion parameter for gaussian family taken to be 0.5056514)
Null deviance: 9.4175 on 19 degrees of freedom
Residual deviance: 8.5961 on 17 degrees of freedom
AIC: 47.869
Number of Fisher Scoring iterations: 2
lm
有F統計量作為整個模型的總結, glm
沒有。 再問一個問題:如何從描述擬合的glm模型中得到一個p值?
謝謝
您可以像這樣計算F統計數據:
glm.D9 <- glm(weight ~ group + conf)
glm.0 <- glm(weight ~ 1)
anova(glm.D9, glm.0, test="F")
# Analysis of Deviance Table
#
# Model 1: weight ~ group + conf
# Model 2: weight ~ 1
# Resid. Df Resid. Dev Df Deviance F Pr(>F)
# 1 17 8.5868
# 2 19 9.4175 -2 -0.8307 0.8223 0.4562
有關詳細信息和其他可用測試,請參閱?anova.glm
。
聲明:本站的技術帖子網頁,遵循CC BY-SA 4.0協議,如果您需要轉載,請注明本站網址或者原文地址。任何問題請咨詢:yoyou2525@163.com.