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library(dplyr)

## 
## Attaching package: 'dplyr'

## The following objects are masked from 'package:stats':
## 
##     filter, lag

## The following objects are masked from 'package:base':
## 
##     intersect, setdiff, setequal, union

library(readxl)
library(coin)

## Loading required package: survival

library(DT)
data <- read_excel("/Users/pavelponomarev/Downloads/datadownload.xlsx")
data1 = data.frame(wwage = data$`Weekly male`, gender = "Male")
data2 = data.frame(wwage = data$`Weekly female`, gender = "Female")
data_test = rbind(data1,data2)
data_test$wwage =data_test$wwage %>% as.numeric()

## Warning in data_test$wwage %>% as.numeric(): в результате преобразования созданы
## NA

data_test$gender =data_test$gender %>% as.factor()

По итогу вот такая таблица и будем смотреть зависимость распределения wage от переменной gender

datatable(data_test)

chisq.test(data_test$wwage,data_test$gender)

## Warning in chisq.test(data_test$wwage, data_test$gender): Chi-squared
## approximation may be incorrect

## 
##  Pearson's Chi-squared test
## 
## data:  data_test$wwage and data_test$gender
## X-squared = 699.05, df = 666, p-value = 0.1816

С хи-квадрат какое-то огромное p-value(((((

independence_test(wwage ~ gender, data = data_test)

## 
##  Asymptotic General Independence Test
## 
## data:  wwage by gender (Female, Male)
## Z = -6.1371, p-value = 8.403e-10
## alternative hypothesis: two.sided

t.test(wwage ~ gender, data = data_test)

## 
##  Welch Two Sample t-test
## 
## data:  wwage by gender
## t = -6.3441, df = 751.36, p-value = 3.864e-10
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
##  -133.42858  -70.36598
## sample estimates:
## mean in group Female   mean in group Male 
##             487.6926             589.5899