basepl<- read.csv("https://docs.google.com/spreadsheets/d/e/2PACX-1vR1rmwvUzt8Udd9cpwqLwjcf72FJZ4NCEwb9hLE6VRxcncx2kU9ALtc24VcxlGh4mIvHXtpsWmyI7hb/pub?output=csv",T, sep = ",")
attach(basepl)
problin<- lm(ins~retire+age+hstatusg+hhincome+educyear+married+hisp)
summary(problin$fitted.values)
## Min. 1st Qu. Median Mean 3rd Qu. Max.
## -0.1557 0.3055 0.4074 0.3871 0.4736 1.1972
summary(problin)
##
## Call:
## lm(formula = ins ~ retire + age + hstatusg + hhincome + educyear +
## married + hisp)
##
## Residuals:
## Min 1Q Median 3Q Max
## -1.1233 -0.4065 -0.2291 0.5298 1.0341
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 0.1270857 0.1605628 0.792 0.42871
## retire 0.0408508 0.0182197 2.242 0.02502 *
## age -0.0028955 0.0024189 -1.197 0.23138
## hstatusg 0.0655583 0.0194531 3.370 0.00076 ***
## hhincome 0.0004921 0.0001375 3.579 0.00035 ***
## educyear 0.0233686 0.0028672 8.150 5.15e-16 ***
## married 0.1234699 0.0193618 6.377 2.07e-10 ***
## hisp -0.1210059 0.0336660 -3.594 0.00033 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 0.4671 on 3198 degrees of freedom
## Multiple R-squared: 0.08262, Adjusted R-squared: 0.08061
## F-statistic: 41.14 on 7 and 3198 DF, p-value: < 2.2e-16
#####
logit<- glm(ins~ retire+age+hstatusg+hhincome+educyear+married+hisp, family=(binomial(link=logit)))
logit
##
## Call: glm(formula = ins ~ retire + age + hstatusg + hhincome + educyear +
## married + hisp, family = (binomial(link = logit)))
##
## Coefficients:
## (Intercept) retire age hstatusg hhincome educyear
## -1.715578 0.196930 -0.014596 0.312265 0.002304 0.114263
## married hisp
## 0.578636 -0.810306
##
## Degrees of Freedom: 3205 Total (i.e. Null); 3198 Residual
## Null Deviance: 4280
## Residual Deviance: 3990 AIC: 4006
exp(logit$coefficients)
## (Intercept) retire age hstatusg hhincome educyear
## 0.1798597 1.2176584 0.9855105 1.3665173 1.0023063 1.1210464
## married hisp
## 1.7836040 0.4447220
##efectos marginales
logitvec <- mean(dlogis(predict(logit, type = "link")))
logitvec * coef(logit)
## (Intercept) retire age hstatusg hhincome
## -0.3725232720 0.0427616020 -0.0031692953 0.0678057706 0.0005002078
## educyear married hisp
## 0.0248111432 0.1256458981 -0.1759510453
eta=(-1.715578+ ( 0.196930 *0) +(62*-0.014596) +(0*0.312265) +(0.002304*0)+(0.114263*12)+(0.578636 *0)+(-0.810306* 0) )
PX=(exp(eta)/(1+exp(eta)))
###Si está retirado la probabilidad de tener un seguro médico aumenta en 4.2%
logitvec * coef(logit)
## (Intercept) retire age hstatusg hhincome
## -0.3725232720 0.0427616020 -0.0031692953 0.0678057706 0.0005002078
## educyear married hisp
## 0.0248111432 0.1256458981 -0.1759510453
####Probit
probit<- glm(ins~ retire+age+hstatusg+hhincome+educyear+married+hisp, family=(binomial(link=probit)))
(probit$coefficients)*1.6
## (Intercept) retire age hstatusg hhincome educyear
## -1.710947627 0.189364238 -0.014191073 0.316385757 0.001972254 0.113198684
## married hisp
## 0.579740240 -0.756982952
(problin$coefficients)*2.5
## (Intercept) retire age hstatusg hhincome educyear
## 0.317714239 0.102127043 -0.007238866 0.163895854 0.001230219 0.058421574
## married hisp
## 0.308674700 -0.302514837
probit$coefficients
## (Intercept) retire age hstatusg hhincome educyear
## -1.069342267 0.118352649 -0.008869421 0.197741098 0.001232659 0.070749177
## married hisp
## 0.362337650 -0.473114345
head(predict(probit, type = "response"))
## 1 2 3 4 5 6
## 0.2205740 0.2285447 0.3573040 0.1809490 0.1810033 0.4634186
###INDIVIDUO1
###probabilidad de los individuos
z1=( -1.069342 -(0.008869 *62)+(0.070749*12))
dnorm(z1)
## [1] 0.2965418
###
dnorm(0.008869421)# interpretación de la maginitud del beta,
## [1] 0.3989266
###INDIVIDUO2
z2= (-1.069342267 -0.008869421*59 +0.070749177*12)
dnorm(z2)
## [1] 0.3025694
## INDIVIDUO3
z3=(-1.069342267 +0.118352649 -0.008869421*60 +0.197741098 + 0.070749177 *13 )
dnorm(z3)
## [1] 0.3731415
## INDIVIDUO4
z4= (-1.069342267- 0.008869421*62 +0.070749177*10)
dnorm(z4)
## [1] 0.2632669
###Interpretacion resultados
#Para poder hayar la probabilidad de exito, necesitamos la informacion de las x de ese individuo, solo debemos reemplazar donde existe un valor de x, esto quiere decir donde sea diferente a 0. Para el individuo 1, 2 y 4 tomamos el intercepto, mas el coeficiente de la edad, por la edad del individuo, mas el coeficeinte de la educacion por la educacion. Para el individuo #3 es difernete, ya que el si esta retirado, tambien tiene hstatusg y educacion, así que se le incluyen estas x.
#Individuo1: 0.2965418 Para el individuo #1 la probabilidad de tener un seguro medico es de 0,29
#Individuo2: 0.3025694 Para el individuo #2 la probabilidad de tener un seguro medico es de 0,30
#Individuo3: 0.3731415 Para el individuo #3 la probabilidad de tener un seguro medico es de 0,37
#Individuo4: 0.2632669 Para el individuo #4 la probabilidad de tener un seguro medico es de 0,26
##Conclusion: Podemos observar que el individuo 3 es el que tiene mayor probabilidad de tener un seguro medico, esto se debe a que se tomaron mas variables.
summary(cars)
## speed dist
## Min. : 4.0 Min. : 2.00
## 1st Qu.:12.0 1st Qu.: 26.00
## Median :15.0 Median : 36.00
## Mean :15.4 Mean : 42.98
## 3rd Qu.:19.0 3rd Qu.: 56.00
## Max. :25.0 Max. :120.00