R Markdown

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 <- rep(c("Mujer","Hombre"), each = 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      74.5

#ANNOVA

modelo <- aov(Temp~ sexo, data = datos)
summary(modelo)
##             Df Sum Sq Mean Sq F value  Pr(>F)   
## sexo         1  42.05   42.05   12.43 0.00242 **
## Residuals   18  60.90    3.38                   
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
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("cadetblue"))

library("car")
## Loading required package: 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
hist(datos$T,
     main   = "Distribución de temperatura",
     xlab   = "Temperatura (°F)",
     ylab   = "Frecuencia",
     col    = "cadetblue",
     border = "white",
     breaks = 10)

##PROBLEMA2

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, paired = TRUE)
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
diferencias <- actual - nuevo
diferencias
##  [1]  0.01 -0.06 -0.02  0.02  0.01  0.02  0.04  0.06 -0.04  0.00  0.01  0.03
## [13]  0.05  0.02 -0.08 -0.05 -0.02 -0.04
boxplot(diferencias,
        col  = "cadetblue",
        main = "Diferencias entre método actual y nuevo",
        ylab = "Diferencia de densidad",
        horizontal = TRUE)
abline(v = 0, col = "red", lwd = 2, lty = 2)

#PROBLEMA3

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("cadetblue","darkseagreen","lightblue3","lightpink3"),
        main = "Desgaste por tipo de cuero",
        ylab = "Desgaste",
        xlab = "Tipo de cuero")

#pruebadeTukey
tukey <- TukeyHSD(modelo)
tukey   
##   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
plot(modelo)

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

##Tregresion 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 
## -1.3335 -0.9854 -0.2061  0.6402  1.9678 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept)   0.3319     0.7102   0.467    0.653    
## x1            2.3032     0.1197  19.236 5.53e-08 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 1.208 on 8 degrees of freedom
## Multiple R-squared:  0.9788, Adjusted R-squared:  0.9762 
## F-statistic:   370 on 1 and 8 DF,  p-value: 5.531e-08
yajustado<-modelo$fitted.values
plot(x1,y)
lines(x1,yajustado,col="cadetblue",lwd=2)