Práctica 1
Franz Robles
2026-03-13
##problema 1
Mujer<-c(75,77,78,79,77,73,78,79,78,80)
Hombre<-c(74,72,77,76,76,73,75,73,74,75)
sexo<-c(rep("Mujer",10),rep("Hombre",10))
Temp<-c(Mujer,Hombre)
datos<-data.frame(sexo=sexo,T=Temp)
resultado<-t.test(Mujer,Hombre,alternative="two.sided",var.equal=FALSE)
print(resultado)##
## Welch Two Sample t-test
##
## data: Mujer and Hombre
## t = 3.5254, df = 16.851, p-value = 0.002626
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
## 1.163304 4.636696
## sample estimates:
## mean of x mean of y
## 77.4 74.5datos## sexo T
## 1 Mujer 75
## 2 Mujer 77
## 3 Mujer 78
## 4 Mujer 79
## 5 Mujer 77
## 6 Mujer 73
## 7 Mujer 78
## 8 Mujer 79
## 9 Mujer 78
## 10 Mujer 80
## 11 Hombre 74
## 12 Hombre 72
## 13 Hombre 77
## 14 Hombre 76
## 15 Hombre 76
## 16 Hombre 73
## 17 Hombre 75
## 18 Hombre 73
## 19 Hombre 74
## 20 Hombre 75datos$sexo<-factor(datos$sexo)
boxplot(T~sexo,data=datos,col=c("orange","green"))
library(car)## Cargando paquete requerido: carDataleveneTest(T~sexo,data=datos,center=mean)## Levene's Test for Homogeneity of Variance (center = mean)
## Df F value Pr(>F)
## group 1 0.2085 0.6534
## 18##problema 2
actual<-c(1.88, 1.84, 1.83, 1.90, 2.19, 1.89, 2.27, 2.03, 1.96,
1.98, 2.00, 1.92, 1.83, 1.94, 1.94, 1.95, 1.93, 2.01)
nuevo<-c(1.87, 1.90, 1.85, 1.88, 2.18, 1.87, 2.23, 1.97, 2.00,
1.98, 1.99, 1.89, 1.78, 1.92, 2.02, 2.00, 1.95, 2.05)
resultado<-t.test(actual,nuevo,alternative="two.sided",paired=TRUE,var.equal=FALSE)
print(resultado)##
## Paired t-test
##
## data: actual and nuevo
## t = -0.23874, df = 17, p-value = 0.8142
## alternative hypothesis: true mean difference is not equal to 0
## 95 percent confidence interval:
## -0.02186024 0.01741580
## sample estimates:
## mean difference
## -0.002222222desgaste<-c(264, 260, 258, 241, 262, 255,208, 220, 216, 200, 213, 206,220, 263, 219, 225, 230, 228,217, 226, 215, 227, 220, 222)
tipocuero<-c(rep("A",6),rep("B",6),rep("C",6),rep("D",6))
datos<-data.frame(tipocuero=tipocuero,desgaste=desgaste)datos## tipocuero desgaste
## 1 A 264
## 2 A 260
## 3 A 258
## 4 A 241
## 5 A 262
## 6 A 255
## 7 B 208
## 8 B 220
## 9 B 216
## 10 B 200
## 11 B 213
## 12 B 206
## 13 C 220
## 14 C 263
## 15 C 219
## 16 C 225
## 17 C 230
## 18 C 228
## 19 D 217
## 20 D 226
## 21 D 215
## 22 D 227
## 23 D 220
## 24 D 222datos$tipocuero<-factor(datos$tipocuero)
modelo<-aov(desgaste~tipocuero,data=datos)
summary(modelo)## Df Sum Sq Mean Sq F value Pr(>F)
## tipocuero 3 7019 2339.8 22.75 1.18e-06 ***
## Residuals 20 2056 102.8
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1boxplot(desgaste~tipocuero,data=datos)
plot(modelo)
shapiro.test(modelo$residuals)##
## Shapiro-Wilk normality test
##
## data: modelo$residuals
## W = 0.88326, p-value = 0.00967hist(modelo$residuals,breaks=5)
TukeyHSD(modelo)## Tukey multiple comparisons of means
## 95% family-wise confidence level
##
## Fit: aov(formula = desgaste ~ tipocuero, data = datos)
##
## $tipocuero
## diff lwr upr p adj
## B-A -46.166667 -62.552998 -29.780336 0.0000008
## C-A -25.833333 -42.219664 -9.447002 0.0014117
## D-A -35.500000 -51.886331 -19.113669 0.0000349
## C-B 20.333333 3.947002 36.719664 0.0118160
## D-B 10.666667 -5.719664 27.052998 0.2926431
## D-C -9.666667 -26.052998 6.719664 0.3742863##Regresión Lineal
x1<-seq(0,10,length=10)
y<-0.8+2.3*x1
y<-y+rnorm(10,0,1.5)
plot(x1,y)
modelo<-lm(y~x1)
summary(modelo)##
## Call:
## lm(formula = y ~ x1)
##
## Residuals:
## Min 1Q Median 3Q Max
## -2.662 -1.091 -0.567 1.613 2.698
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 0.1585 1.1110 0.143 0.89
## x1 2.5349 0.1873 13.534 8.53e-07 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 1.89 on 8 degrees of freedom
## Multiple R-squared: 0.9582, Adjusted R-squared: 0.9529
## F-statistic: 183.2 on 1 and 8 DF, p-value: 8.528e-07yajustado<-modelo$fitted.values
plot(x1,y)
lines(x1,yajustado,col="red",lwd=2)