#"Maricely Zelada Alvarado", "Rodrigo Mendivil Rodriguez", "Liz Hinostroza Vidal", "Piero Morales Romero"
data2 <- data.frame(ÁrbolC =arbol, Red_N = redn, Regresion = regresion, SActual = actual)
data2
## ÁrbolC Red_N Regresion SActual
## 1 23.81 23.24 16.13 17.09
## 2 22.13 20.08 17.84 15.77
## 3 22.64 18.01 18.28 18.45
## 4 21.69 23.28 15.61 16.55
## 5 23.58 19.23 17.62 22.23
## 6 22.14 21.22 16.12 22.11
## 7 18.73 21.47 17.29 18.26
## 8 21.59 20.60 16.13 18.04
## 9 20.36 21.11 16.64 19.66
## 10 20.53 21.27 15.03 19.76
## 11 20.11 21.03 18.16 18.74
## 12 20.34 17.34 16.82 19.02
## 13 19.19 22.80 17.44 18.54
## 14 22.92 21.85 16.76 16.70
## 15 18.65 17.85 17.26 17.57
## 16 20.60 23.15 15.55 19.89
## 17 19.83 19.57 17.49 19.06
## 18 20.09 19.56 18.42 18.70
## 19 19.43 20.79 17.54 19.39
## 20 22.06 18.04 17.13 19.68
## 21 21.15 20.95 15.50 19.20
## 22 19.26 21.83 16.80 16.85
## 23 18.08 18.17 18.47 19.91
## 24 20.24 22.66 18.42 19.82
## 25 18.75 18.29 18.43 18.08
## 26 20.69 18.89 15.56 19.38
## 27 21.62 19.49 16.03 20.30
## 28 23.69 19.19 15.39 21.60
## 29 23.93 26.47 15.12 23.39
## 30 23.19 25.25 17.77 19.33
summary(data2$SActual)
## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 15.77 18.12 19.13 19.10 19.80 23.39
hist(data2$SActual)
cv_SActual <- sd(data2$SActual)/mean(data2$SActual)
cv_SActual
## [1] 0.0905464
summary(data2$ÁrbolC)
## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 18.08 19.89 20.64 21.03 22.14 23.93
hist(data2$ÁrbolC)
cv_ÁrbolC <- sd(data2$ÁrbolC)/mean(data2$ÁrbolC)
cv_ÁrbolC
## [1] 0.08108817
Prop1 <- t.test(data2$SActual, data2$ÁrbolC, conf.level = 0.95)
Prop1
##
## Welch Two Sample t-test
##
## data: data2$SActual and data2$ÁrbolC
## t = -4.3555, df = 57.989, p-value = 5.479e-05
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
## -2.819433 -1.043900
## sample estimates:
## mean of x mean of y
## 19.10233 21.03400
summary(data2$Red_N)
## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 17.34 19.20 20.87 20.76 21.84 26.47
hist(data2$Red_N)
cv_Red_N <- sd(data2$Red_N)/mean(data2$Red_N)
cv_Red_N
## [1] 0.1071663
Prop2 <- t.test(data2$SActual, data2$Red_N, conf.level = 0.95)
Prop2
##
## Welch Two Sample t-test
##
## data: data2$SActual and data2$Red_N
## t = -3.2145, df = 54.681, p-value = 0.002194
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
## -2.684758 -0.622575
## sample estimates:
## mean of x mean of y
## 19.10233 20.75600
summary(data2$Regresion)
## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 15.03 16.05 16.98 16.89 17.73 18.47
hist(data2$Regresion)
cv_Regresion <- sd(data2$Regresion)/mean(data2$Regresion)
cv_Regresion
## [1] 0.06531693
Prop3 <- t.test(data2$SActual, data2$Regresion, conf.level = 0.95)
Prop3
##
## Welch Two Sample t-test
##
## data: data2$SActual and data2$Regresion
## t = 5.902, df = 49.248, p-value = 3.26e-07
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
## 1.458045 2.963288
## sample estimates:
## mean of x mean of y
## 19.10233 16.89167
Se implementará el Metodo de regresión por lo siguiente:
p-value < 0.05
IC 95% = (1.46 , 2.96)
Media del Método Actual = 19.10
Media del Método Regresión = 16.89
El método de regresión reduce el tiempo promedio de fallas en
aproximadamente 2.21 minutos, respecto al sistema actual, lo que
representa una mejora cercana al 11.6%.
Nota: Las alternativas de Árbol de clasificación y Redes neuronales
presentan tiempos de fallas significativamente mayores que el sistema
actual, por lo que no se recomienda su implementación.
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