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웹에서 하는 R 통계

회귀모형 자동해석

2020.03.08 22:25

cardiomoon 조회 수:35

안녕하세요? 문건웅입니다. 이번에 웹R프로그램을 대대적으로 다시 짜고 있습니다. 프로그램을 손보는 김에 우선 회귀모형에 대한 설명을 해주는 함수를 만들어 보았습니다. 제가 영어도 짧고 통계가 전공이 아니어서 오류가 있을까 걱정됩니다. 한번 보시고 평가해주시면 고맙겠습니다. 

Loading the packages

library(webrSub)
library(predict3d)

Simple linear regression

fit=lm(mpg~wt,data=mtcars)
ggPredict(fit)

cat(result=explain(fit))
The predictor variable 'wt' accounts for 75.28 % of the total variation of your dependent variable 'mpg'. Your R^2 is quite good. The coefficient for 'wt' is smaller than zero. This means that the value of your dependent variable 'mpg' decreases by 5.34 for any increase of 1 in your independent variable 'wt'. The coefficient is  significantly different from zero(p< 0.001). The predicted value of dependent variable 'mpg' is calculated by the equation: '37.29 - 5.34*wt'.

Polynomial regression

fit=lm(mpg~wt+I(wt^2),data=mtcars)
ggPredict(fit)

cat(explain(fit))
The predictor variables 'wt','I(wt^2)' account for 81.91 % of the total variation of your dependent variable 'mpg'. Your R^2 is quite good. The coefficient for 'wt' is  significantly different from zero(p< 0.001) The coefficient for 'I(wt^2)' is  significantly different from zero(p =0.003) The predicted value of dependent variable 'mpg' is calculated by the equation: '49.93 - 13.38*wt + 1.17*I(wt^2)'.

Multiple linear regression

fit=lm(mpg~hp+wt+am,data=mtcars)
ggPredict(fit)

cat(explain(fit))
The predictor variables 'hp','wt','am' account for 83.99 % of the total variation of your dependent variable 'mpg'. Your R^2 is quite good. The coefficient for 'hp' is smaller than zero. This means that the value of your dependent variable 'mpg' decreases by 0.037 for any increase of 1 in your independent variable 'hp', if 'wt','am' remain(s) constant. The coefficient is  significantly different from zero(p< 0.001), controlling for the other variable(s). The coefficient for 'wt' is smaller than zero. This means that the value of your dependent variable 'mpg' decreases by 2.88 for any increase of 1 in your independent variable 'wt', if 'hp','am' remain(s) constant. The coefficient is  significantly different from zero(p =0.004), controlling for the other variable(s). The coefficient for 'am' is larger than zero. This means that the value of your dependent variable 'mpg' increases by 2.08 for any increase of 1 in your independent variable 'am', if 'hp','wt' remain(s) constant. The coefficient is  not significantly different from zero(p =0.141) suggesting that 'am' and 'mpg' are not linearly related when controlling for other predictor variable(s). The predicted value of dependent variable 'mpg' is calculated by the equation: '34.00 - 0.04*hp - 2.88*wt + 2.08*am'.

Multiple linear regression with interaction

fit=lm(mpg~hp*wt,data=mtcars)
ggPredict(fit)

cat(explain(fit))
The predictor variables 'hp','wt' and the interaction 'hp:wt' account for 88.48 % of the total variation of your dependent variable 'mpg'. Your R^2 is quite good. The interaction between 'hp' and 'wt' is significant. This means that the relationship between one predictor and the response variable depends on the level of the other variable(s). Here it means that the relationship between 'mpg' and 'hp' varies by 'wt'. The predicted value of dependent variable 'mpg' is calculated by the equation: '49.81 - 0.12*hp - 8.22*wt + 0.03*hp*wt' which simplifies to '49.81 - 8.22*wt + (-0.12 + 0.03*wt)*hp'. If 'wt' is 2.24, the value of your dependent variable 'mpg' decreases by 0.058 for any increase of 1 in your independent variable 'hp' If 'wt' is 3.22, the value of your dependent variable 'mpg' decreases by 0.031 for any increase of 1 in your independent variable 'hp' If 'wt' is 4.2, the value of your dependent variable 'mpg' decreases by 0.0033 for any increase of 1 in your independent variable 'hp'