🎯 Objetivo

Aplicar y comprender distintas pruebas de hipótesis (para una y dos muestras), utilizando variables cuantitativas y cualitativas del conjunto Credit Score Classification Dataset.csv.

1. 📊 Introducción a vectores en R

# Vectores numéricos
variable <- c(12, 0.23, 4/5)
variable
## [1] 12.00  0.23  0.80
# Vectores de caracteres (no factores aún)
factor <- c("Manuel", "2", "red", "este espacio es para aprender Rstudio")
factor
## [1] "Manuel"                               
## [2] "2"                                    
## [3] "red"                                  
## [4] "este espacio es para aprender Rstudio"
# Operaciones básicas
Var1 <- c(20,25,25,12,28,26,27,29,29,29,30,40)

length(Var1)
## [1] 12
sum(Var1)
## [1] 320
prod(Var1)
## [1] 8.629023e+16
max(Var1)
## [1] 40
min(Var1)
## [1] 12
diff(range(Var1))  # rango
## [1] 28
mean(Var1)
## [1] 26.66667
median(Var1)
## [1] 27.5
var(Var1)
## [1] 42.9697
sd(Var1)
## [1] 6.555128

2. 📥 Carga de datos y muestra aleatoria

library(readr)
## Warning: package 'readr' was built under R version 4.5.1
library(BSDA)
## Warning: package 'BSDA' was built under R version 4.5.1
## Cargando paquete requerido: lattice
## 
## Adjuntando el paquete: 'BSDA'
## The following object is masked from 'package:datasets':
## 
##     Orange
Credit <- read_csv("Credit Score Classification Dataset.csv")
## Rows: 164 Columns: 8
## ── Column specification ────────────────────────────────────────────────────────
## Delimiter: ","
## chr (5): Gender, Education, Marital Status, Home Ownership, Credit Score
## dbl (3): Age, Income, Number of Children
## 
## ℹ Use `spec()` to retrieve the full column specification for this data.
## ℹ Specify the column types or set `show_col_types = FALSE` to quiet this message.
Credit <- as.data.frame(unclass(Credit), stringsAsFactors=TRUE)

str(Credit)
## 'data.frame':    164 obs. of  8 variables:
##  $ Age               : num  25 30 35 40 45 50 26 31 36 41 ...
##  $ Gender            : Factor w/ 2 levels "Female","Male": 1 2 1 2 1 2 1 2 1 2 ...
##  $ Income            : num  50000 100000 75000 125000 100000 150000 40000 60000 80000 105000 ...
##  $ Education         : Factor w/ 5 levels "Associate's Degree",..: 2 5 3 4 2 5 1 2 5 3 ...
##  $ Marital.Status    : Factor w/ 2 levels "Married","Single": 2 1 1 2 1 1 2 2 1 2 ...
##  $ Number.of.Children: num  0 2 1 0 3 0 0 0 2 0 ...
##  $ Home.Ownership    : Factor w/ 2 levels "Owned","Rented": 2 1 1 1 1 1 2 2 1 1 ...
##  $ Credit.Score      : Factor w/ 3 levels "Average","High",..: 2 2 2 2 2 2 1 1 2 2 ...
# Selección de muestra aleatoria
set.seed(1020)
n <- 80
N <- nrow(Credit)
Muestra.x1 <- sample(1:N, n, replace = FALSE)
data.m <- Credit[Muestra.x1, ]
head(data.m, 10)
##     Age Gender Income           Education Marital.Status Number.of.Children
## 80   28 Female  32500  Associate's Degree         Single                  0
## 114  38 Female  67500   Bachelor's Degree        Married                  2
## 112  28 Female  32500  Associate's Degree         Single                  0
## 47   52   Male 130000 High School Diploma        Married                  0
## 33   40   Male 130000 High School Diploma         Single                  0
## 127  30   Male 117500     Master's Degree        Married                  2
## 81   33   Male  52500 High School Diploma         Single                  0
## 35   50   Male 155000     Master's Degree        Married                  0
## 19   28 Female  30000  Associate's Degree         Single                  0
## 70   36 Female  90000     Master's Degree        Married                  2
##     Home.Ownership Credit.Score
## 80          Rented          Low
## 114          Owned         High
## 112         Rented          Low
## 47           Owned         High
## 33           Owned         High
## 127          Owned         High
## 81          Rented      Average
## 35           Owned         High
## 19          Rented          Low
## 70           Owned         High

3. 🧪 Prueba de Hipótesis para una muestra

3.1 Para la media de Income

media.income <- mean(data.m$Income)
sd.income <- sd(data.m$Income)

# Ho: media = 75000  vs Ha: media ≠ 75000
z.test(x = data.m$Income,
       sigma.x = sd.income,
       mu = 75000,
       alternative = "two.sided",
       conf.level = 0.95)
## 
##  One-sample z-Test
## 
## data:  data.m$Income
## z = 2.3358, p-value = 0.0195
## alternative hypothesis: true mean is not equal to 75000
## 95 percent confidence interval:
##  76308.21 89954.29
## sample estimates:
## mean of x 
##  83131.25

3.2 Para la proporción de Gender == Female

table(data.m$Gender)
## 
## Female   Male 
##     44     36
x_female <- sum(data.m$Gender == "Female")

# Ho: p = 0.5  vs Ha: p ≠ 0.5
prop.test(x = x_female,
          n = 80,
          p = 0.5,
          alternative = "two.sided",
          conf.level = 0.95,
          correct = TRUE)
## 
##  1-sample proportions test with continuity correction
## 
## data:  x_female out of 80, null probability 0.5
## X-squared = 0.6125, df = 1, p-value = 0.4338
## alternative hypothesis: true p is not equal to 0.5
## 95 percent confidence interval:
##  0.4351113 0.6600504
## sample estimates:
##    p 
## 0.55

4. 🧪 Pruebas para dos muestras

4.1 Diferencia de medias: Income por Gender

set.seed(1234)
data.m1 <- Credit[sample(1:N, 110), ]
data.Gender <- split(data.m1, data.m1$Gender)

z.test(x = data.Gender$Female$Income,
       y = data.Gender$Male$Income,
       sigma.x = sd(data.Gender$Female$Income),
       sigma.y = sd(data.Gender$Male$Income),
       mu = 0,
       alternative = "two.sided",
       conf.level = 0.95)
## 
##  Two-sample z-Test
## 
## data:  data.Gender$Female$Income and data.Gender$Male$Income
## z = -5.3184, p-value = 1.047e-07
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
##  -39288.09 -18128.57
## sample estimates:
## mean of x mean of y 
##  69591.67  98300.00

4.2 Diferencia de proporciones: Marital.Status == Married por Gender

table(data.Gender$Female$Marital.Status)
## 
## Married  Single 
##      43      17
table(data.Gender$Male$Marital.Status)
## 
## Married  Single 
##      18      32
# Ejemplo con datos observados: 43 mujeres casadas de 60, 18 hombres casados de 50
prop.test(x = c(43, 18),
          n = c(60, 50),
          alternative = "two.sided",
          conf.level = 0.95,
          correct = TRUE)
## 
##  2-sample test for equality of proportions with continuity correction
## 
## data:  c(43, 18) out of c(60, 50)
## X-squared = 12.638, df = 1, p-value = 0.000378
## alternative hypothesis: two.sided
## 95 percent confidence interval:
##  0.1631137 0.5502197
## sample estimates:
##    prop 1    prop 2 
## 0.7166667 0.3600000

5. 📝 Actividad final: Taller de Aplicación

Objetivo: Aplicar lo aprendido para formular y probar hipótesis con otras variables del mismo conjunto de datos.

Instrucciones:

  1. Seleccione una variable cuantitativa distinta a Income (por ejemplo: Age o Loan.Amount).
  2. Plantee y realice una prueba de hipótesis para una muestra (media).
  3. Seleccione una variable cualitativa (por ejemplo: Occupation o Credit.Score) y realice una prueba para proporción.
  4. Luego, realice pruebas de diferencia de medias y proporciones entre grupos definidos por alguna variable categórica (ej: Occupation, Gender, etc.).
  5. Interprete los resultados (valor p, intervalos, decisiones sobre H0).