R Notebook de Semana04: Modelo Lineal, por Cancho-Rodríguez, Ernesto

1. Consulta de datasets

#ver librerias
library("datasets")
#    modelo lineal
dataframe1 = Orange
dataframe1

library(readr)
write_csv( dataframe1 , "./salidas/df_Orange.csv")

attach(Orange)

The following objects are masked from Orange (pos = 3):

    age, circumference, Tree

The following objects are masked from Orange (pos = 4):

    age, circumference, Tree

#el primer paso para verificar si se debe implementar
#modelo lineal es mediante el grafico
plot(Orange)

#existe una relacion lineal entre
#age y circumference
plot(age,circumference)

#siguiente para validar si implemento o no el modelo
# es evaluar el cor solo se acepta de 1-0.6
cor.test(circumference,age)

    Pearson's product-moment correlation

data:  circumference and age
t = 12.9, df = 33, p-value = 1.931e-14
alternative hypothesis: true correlation is not equal to 0
95 percent confidence interval:
 0.8342364 0.9557955
sample estimates:
      cor 
0.9135189 

#creo modelo lineal
modelo1=lm(circumference ~ age)
modelo1

Call:
lm(formula = circumference ~ age)

Coefficients:
(Intercept)          age  
    17.3997       0.1068  

#ver sus coeficiente   circumference=age*a+b
modelo1

Call:
lm(formula = circumference ~ age)

Coefficients:
(Intercept)          age  
    17.3997       0.1068  

#analizar modelo p-value<0.05    y R2>0.75
summary(modelo1)

Call:
lm(formula = circumference ~ age)

Residuals:
    Min      1Q  Median      3Q     Max 
-46.310 -14.946  -0.076  19.697  45.111 

Coefficients:
             Estimate Std. Error t value Pr(>|t|)    
(Intercept) 17.399650   8.622660   2.018   0.0518 .  
age          0.106770   0.008277  12.900 1.93e-14 ***
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 23.74 on 33 degrees of freedom
Multiple R-squared:  0.8345,    Adjusted R-squared:  0.8295 
F-statistic: 166.4 on 1 and 33 DF,  p-value: 1.931e-14

circumference=agea+b circumference = age 0.106770 + 17.399650

#  circumference=age*0.1068+17.3997
#arbol de  700 años
circumference_2 = 700*0.1068+17.3997
circumference_2

[1] 92.1597

#arbol de  800 años
800*0.1068+17.3997

[1] 102.8397

#grafico
plot(age,circumference,pch=19)
abline(modelo1,col="red")

#predecir para arboldes de 950 a 1000
predecir=predict(modelo1,list(age=c(950:1000)))
predecir

       1        2        3        4        5        6        7        8        9 
118.8315 118.9382 119.0450 119.1518 119.2585 119.3653 119.4721 119.5789 119.6856 
      10       11       12       13       14       15       16       17       18 
119.7924 119.8992 120.0059 120.1127 120.2195 120.3262 120.4330 120.5398 120.6466 
      19       20       21       22       23       24       25       26       27 
120.7533 120.8601 120.9669 121.0736 121.1804 121.2872 121.3939 121.5007 121.6075 
      28       29       30       31       32       33       34       35       36 
121.7143 121.8210 121.9278 122.0346 122.1413 122.2481 122.3549 122.4617 122.5684 
      37       38       39       40       41       42       43       44       45 
122.6752 122.7820 122.8887 122.9955 123.1023 123.2090 123.3158 123.4226 123.5294 
      46       47       48       49       50       51 
123.6361 123.7429 123.8497 123.9564 124.0632 124.1700 

#ver por años los resultados
(matriz=data.frame(c(950:1000),predecir))