#PRACTICA 1

Mujeres = c(75,77,78,79,77,73,78,79,78,80)
Hombres= c(74,72,77,76,76,73,75,73,74,75)
datos<-data.frame(Mujeres=Mujeres, Hombres=Hombres)
resultado <-t.test(Mujeres,Hombres,
                   alternative="two.sided",
                   var.equal=FALSE)
resultado
## 
##  Welch Two Sample t-test
## 
## data:  Mujeres and Hombres
## 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      74.5
Mujeres = c(75,77,78,79,77,73,78,79,78,80)
Hombres= c(74,72,77,76,76,73,75,73,74,75)
Sexo<-c(rep("Mujer",10), rep("Hombre",10))
Temp<-c(Mujeres,Hombres)
datos<-data.frame(Sexo=Sexo,T=Temp)
resultado<-t.test(Mujeres,Hombres,
                   alternative="two.sided",
                   var.equal=FALSE)
print(resultado)
## 
##  Welch Two Sample t-test
## 
## data:  Mujeres and Hombres
## 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      74.5
datos
##      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 75
datos$Sexo<-factor(datos$Sexo)
boxplot(T~Sexo,data=datos,col=c("orange","purple"))

library(car)
## Cargando paquete requerido: carData
leveneTest(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(1.87,1.9,1.85,1.88,2.18,1.87,2.23,1.97,2.00,1.98,1.99,1.89,1.78,1.92,2.02,2.00,1.95,2.05)

resultado<-t.test(actual,nuevo,
                  alternative="two.sided", paired=T,
                  var.equal=F)
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 
##    -0.002222222

PROBLEMA 3

desgaste<-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      222
datos$tipoCuero<-factor(datos$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 ' ' 1
boxplot(desgaste~tipoCuero, data=datos, col=c("yellow","black", "green","red"))

plot(modelo)

shapiro.test(modelo$residuals)
## 
##  Shapiro-Wilk normality test
## 
## data:  modelo$residuals
## W = 0.88326, p-value = 0.00967
hist(modelo$residuals,breaks = 8)

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

esta prueba solo se utiliza cuando hay diferencias significativas debe ser igual con el boxplot

#RegresionLineal

x1<-seq(0,10,length=10)
y<-0.8+2.3*x1
y<-y+rnorm(1,0,0.8)
plot(x1,y)

modelo<-lm(y~x1)
summary(modelo)
## Warning in summary.lm(modelo): essentially perfect fit: summary may be
## unreliable
## 
## Call:
## lm(formula = y ~ x1)
## 
## Residuals:
##        Min         1Q     Median         3Q        Max 
## -1.833e-15 -4.576e-16 -1.742e-16  7.768e-16  1.449e-15 
## 
## Coefficients:
##              Estimate Std. Error   t value Pr(>|t|)    
## (Intercept) 1.705e+00  6.493e-16 2.625e+15   <2e-16 ***
## x1          2.300e+00  1.095e-16 2.101e+16   <2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 1.105e-15 on 8 degrees of freedom
## Multiple R-squared:      1,  Adjusted R-squared:      1 
## F-statistic: 4.415e+32 on 1 and 8 DF,  p-value: < 2.2e-16