require(ggplot2)
## Loading required package: ggplot2
## Warning: package 'ggplot2' was built under R version 4.3.3
require(plotly)
## Loading required package: plotly
## Warning: package 'plotly' was built under R version 4.3.3
##
## Attaching package: 'plotly'
## The following object is masked from 'package:ggplot2':
##
## last_plot
## The following object is masked from 'package:stats':
##
## filter
## The following object is masked from 'package:graphics':
##
## layout
require(datarium)
## Loading required package: datarium
## Warning: package 'datarium' was built under R version 4.3.3
data("marketing")
head(marketing)
## youtube facebook newspaper sales
## 1 276.12 45.36 83.04 26.52
## 2 53.40 47.16 54.12 12.48
## 3 20.64 55.08 83.16 11.16
## 4 181.80 49.56 70.20 22.20
## 5 216.96 12.96 70.08 15.48
## 6 10.44 58.68 90.00 8.64
promedio = mean(marketing$sales)
desviación = sd(marketing$sales)
data.frame(promedio, desviación)
## promedio desviación
## 1 16.827 6.260948
g1=ggplot(data = marketing,mapping = aes(x=sales))+geom_histogram(fill="blue")+theme_bw()
ggplotly(g1)
## `stat_bin()` using `bins = 30`. Pick better value with `binwidth`.
g2 = ggplot(data=marketing, mapping = aes(x=newspaper, y=sales))+geom_point()+theme_bw()+geom_smooth()
ggplotly(g2)
## `geom_smooth()` using method = 'loess' and formula = 'y ~ x'
cor(marketing$sales, marketing$newspaper)
## [1] 0.228299
g3 = ggplot(data=marketing, mapping = aes(x=facebook, y=sales))+geom_point()+theme_bw()+geom_smooth()
ggplotly(g3)
## `geom_smooth()` using method = 'loess' and formula = 'y ~ x'
cor(marketing$sales, marketing$facebook)
## [1] 0.5762226
g4 = ggplot(data=marketing, mapping = aes(x=youtube, y=sales))+geom_point()+theme_bw()+geom_smooth()
ggplotly(g4)
## `geom_smooth()` using method = 'loess' and formula = 'y ~ x'
cor(marketing$sales, marketing$youtube)
## [1] 0.7822244
mod_you=lm(sales~youtube, data = marketing)
mod_you
##
## Call:
## lm(formula = sales ~ youtube, data = marketing)
##
## Coefficients:
## (Intercept) youtube
## 8.43911 0.04754
summary(mod_you)
##
## Call:
## lm(formula = sales ~ youtube, data = marketing)
##
## Residuals:
## Min 1Q Median 3Q Max
## -10.0632 -2.3454 -0.2295 2.4805 8.6548
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 8.439112 0.549412 15.36 <2e-16 ***
## youtube 0.047537 0.002691 17.67 <2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 3.91 on 198 degrees of freedom
## Multiple R-squared: 0.6119, Adjusted R-squared: 0.6099
## F-statistic: 312.1 on 1 and 198 DF, p-value: < 2.2e-16
mod_face=lm(sales~facebook, data = marketing)
mod_face
##
## Call:
## lm(formula = sales ~ facebook, data = marketing)
##
## Coefficients:
## (Intercept) facebook
## 11.1740 0.2025
summary(mod_face)
##
## Call:
## lm(formula = sales ~ facebook, data = marketing)
##
## Residuals:
## Min 1Q Median 3Q Max
## -18.8766 -2.5589 0.9248 3.3330 9.8173
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 11.17397 0.67548 16.542 <2e-16 ***
## facebook 0.20250 0.02041 9.921 <2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 5.13 on 198 degrees of freedom
## Multiple R-squared: 0.332, Adjusted R-squared: 0.3287
## F-statistic: 98.42 on 1 and 198 DF, p-value: < 2.2e-16
mod_new=lm(sales~newspaper, data = marketing)
mod_new
##
## Call:
## lm(formula = sales ~ newspaper, data = marketing)
##
## Coefficients:
## (Intercept) newspaper
## 14.82169 0.05469
summary(mod_new)
##
## Call:
## lm(formula = sales ~ newspaper, data = marketing)
##
## Residuals:
## Min 1Q Median 3Q Max
## -13.473 -4.065 -1.007 4.207 15.330
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 14.82169 0.74570 19.88 < 2e-16 ***
## newspaper 0.05469 0.01658 3.30 0.00115 **
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 6.111 on 198 degrees of freedom
## Multiple R-squared: 0.05212, Adjusted R-squared: 0.04733
## F-statistic: 10.89 on 1 and 198 DF, p-value: 0.001148
predict(mod_you,list(youtube=65), interval = "confidence", level=0.95)
## fit lwr upr
## 1 11.52899 10.72462 12.33337
predict(mod_face,list(facebook=65), interval = "confidence", level=0.95)
## fit lwr upr
## 1 24.33619 22.68098 25.9914
##Paso 1 - Segmentar los Datos
id_modelar=sample(1:200,size = 160)
marketing_modelar=marketing[id_modelar,]
marketing_validar=marketing[-id_modelar,]
##Paso 2 - Estimar el Modelo Set de Modelar
mod_you_modelar=lm(sales~youtube,data=marketing_modelar)
##Paso 3 - Predeccir Set de Validación
sales_pred=predict(mod_you_modelar,list(youtube=marketing_validar$youtube))
##Paso 4 - Comparar Ventas del Modelo y Reales
sales_real=marketing_validar$sales
error=sales_real-sales_pred
res=data.frame(sales_real,sales_pred,error)
##Paso 5 - Calcular Indicador de Evaluación de la Predicción
MAE=mean(abs(error)) #Mean Absolut Error (Error Medio Absoluto)
MAE
## [1] 3.175392
##Paso 1 - Segmentar los Datos
id_modelar=sample(1:200,size = 160)
marketing_modelar=marketing[id_modelar,]
marketing_validar=marketing[-id_modelar,]
##Paso 2 - Estimar el Modelo Set de Modelar
mod_face_modelar=lm(sales~facebook,data=marketing_modelar)
##Paso 3 - Predeccir Set de Validación
sales_pred=predict(mod_face_modelar,list(facebook=marketing_validar$facebook))
##Paso 4 - Comparar Ventas del Modelo y Reales
sales_real=marketing_validar$sales
error=sales_real-sales_pred
res=data.frame(sales_real,sales_pred,error)
##Paso 5 - Calcular Indicador de Evaluación de la Predicción
MAE=mean(abs(error)) #Mean Absolut Error (Error Medio Absoluto)
MAE
## [1] 4.312971
Conclusión: El modelo de regresión lineal simple para predecir las ventas en base a la inversión en mercadeo vía YouTube más optima es: Ventas ~= 8.44 + 0.047 * Inversión en YouTube, lo cuál significa que la venta mínima, en caso que no haya inversión, será en promedio de 8.44 millones y que por cada millón extra invertido en publicidad en YouTube se observarán 0.047 millones más en ventas. Este modelo cuenta con un R^2 de 0.612, lo que significa que explica el comportamiento del 61% de las ventas, de lo cual se concluye que es un buen modelo.
De acuerdo a este modelo, las ventas predecidas para el próximo periodo en caso de que se inviertan 65 Millones de publicidad en YouTube serán de 11.53 Millones, en un rango de 10.72 - 12.33 millones, con un intérvalo de confianza del 95%.