Childhood

rm(list = ls())
date()

## [1] "Sun Dec  1 17:50:18 2024"

sessionInfo()

## R version 4.4.1 (2024-06-14)
## Platform: x86_64-apple-darwin20
## Running under: macOS Ventura 13.7.1
## 
## Matrix products: default
## BLAS:   /Library/Frameworks/R.framework/Versions/4.4-x86_64/Resources/lib/libRblas.0.dylib 
## LAPACK: /Library/Frameworks/R.framework/Versions/4.4-x86_64/Resources/lib/libRlapack.dylib;  LAPACK version 3.12.0
## 
## locale:
## [1] en_US.UTF-8/en_US.UTF-8/en_US.UTF-8/C/en_US.UTF-8/en_US.UTF-8
## 
## time zone: Europe/Moscow
## tzcode source: internal
## 
## attached base packages:
## [1] stats     graphics  grDevices utils     datasets  methods   base     
## 
## loaded via a namespace (and not attached):
##  [1] digest_0.6.36     R6_2.5.1          fastmap_1.2.0     xfun_0.46        
##  [5] cachem_1.1.0      knitr_1.48        htmltools_0.5.8.1 rmarkdown_2.27   
##  [9] lifecycle_1.0.4   cli_3.6.3         sass_0.4.9        jquerylib_0.1.4  
## [13] compiler_4.4.1    rstudioapi_0.16.0 tools_4.4.1       evaluate_0.24.0  
## [17] bslib_0.7.0       yaml_2.3.9        rlang_1.1.4       jsonlite_1.8.8

options(scipen = 999) # Убирает научную запись чисел

#Library

library(ggplot2)
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

Загрузка таблицы

load(file = "Data.RData")

Описательная статистика

summary(Data$WellBeing)

##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
##   258.0   303.0   326.0   328.7   343.0   447.0

summary(Data$EnvIdentity)

##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
##   26.00   49.00   61.00   59.85   70.00   85.00

table(Data$Nationality, Data$ResidenceChild)

##         
##          BigCity SmallTown Village SmallVillage
##   France      22        16      14           10
##   Russia      27        10       6            4

summary(Data$Age)

##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
##   19.00   22.00   24.00   24.81   27.00   34.00

Data |> 
  group_by(Nationality) |> 
  summarise(
    Mean_Age = mean(Age, na.rm = TRUE),
    SD_Age = sd(Age, na.rm = TRUE),
  )

## # A tibble: 2 × 3
##   Nationality Mean_Age SD_Age
##   <fct>          <dbl>  <dbl>
## 1 France          24.8   4.54
## 2 Russia          24.8   3.99

Создаю переменную с сокращением числа групп в “место жительства в детстве”

library(dplyr)

Data$ResidenceChild_3 <- recode(Data$ResidenceChild,
                              "Village" = "Small",
                              "SmallVillage" = "Small")

Data <- Data %>%
  select(1:4, ResidenceChild_3, 5:(ncol(Data) - 1))  # Перемещаем 'ResidenceChild_3' между 4 и 5 столбцом


table(Data$Nationality, Data$ResidenceChild_3)

##         
##          BigCity SmallTown Small
##   France      22        16    24
##   Russia      27        10    10

Plots

boxplot(WellBeing ~ ResidenceChild
        , data = Data, main = "Well-Being по месту жительства в детстве")

boxplot(EnvIdentity ~ ResidenceChild
        , data = Data, main = "Идентичность с природой по месту жительства в детстве")

boxplot(WellBeing ~ ResidenceChild_3
        , data = Data, main = "Well-Being по месту жительства в детстве")

boxplot(EnvIdentity ~ ResidenceChild_3
        , data = Data, main = "Идентичность с природой по месту жительства в детстве")

ggplot(Data, aes(x = ResidenceChild, y = WellBeing, fill = Nationality)) +
    geom_boxplot(outlier.color = "red", outlier.shape = 1) +
    labs(title = "WellBeing по месту жительства и национальности",
         x = "Место жительства в детстве",
         y = "WellBeing") +
    scale_fill_manual(values = c("skyblue", "lightgreen")) +
    theme_minimal()

ggplot(Data, aes(x = ResidenceChild, y = EnvIdentity, fill = Nationality)) +
    geom_boxplot(outlier.color = "red", outlier.shape = 1) +
    labs(title = "EnvIdentity по месту жительства и национальности",
         x = "Место жительства в детстве",
         y = "EnvIdentity") +
    scale_fill_manual(values = c("pink", "lightblue")) +
    theme_minimal()

ggplot(Data, aes(x = ResidenceChild_3, y = WellBeing, fill = Nationality)) +
    geom_boxplot(outlier.color = "red", outlier.shape = 1) +
    labs(title = "WellBeing по месту жительства и национальности",
         x = "Место жительства в детстве",
         y = "WellBeing") +
    scale_fill_manual(values = c("skyblue", "lightgreen")) +
    theme_minimal()

ggplot(Data, aes(x = ResidenceChild_3, y = EnvIdentity, fill = Nationality)) +
    geom_boxplot(outlier.color = "red", outlier.shape = 1) +
    labs(title = "EnvIdentity по месту жительства и национальности",
         x = "Место жительства в детстве",
         y = "EnvIdentity") +
    scale_fill_manual(values = c("pink", "lightblue")) +
    theme_minimal()

Статистика 4

Data %>%
  group_by(Nationality, ResidenceChild) %>%
  summarise(
    Mean_WellBeing = mean(WellBeing, na.rm = TRUE),
    SD_WellBeing = sd(WellBeing, na.rm = TRUE),
    Mean_EnvIdentity = mean(EnvIdentity, na.rm = TRUE),
    SD_EnvIdentity = sd(EnvIdentity, na.rm = TRUE)
  )

## `summarise()` has grouped output by 'Nationality'. You can override using the
## `.groups` argument.

## # A tibble: 8 × 6
## # Groups:   Nationality [2]
##   Nationality ResidenceChild Mean_WellBeing SD_WellBeing Mean_EnvIdentity
##   <fct>       <ord>                   <dbl>        <dbl>            <dbl>
## 1 France      BigCity                  315.         25.5             57.2
## 2 France      SmallTown                301.         26.2             54.9
## 3 France      Village                  313.         13.4             64.7
## 4 France      SmallVillage             310.         22.2             66.6
## 5 Russia      BigCity                  354.         42.0             58.5
## 6 Russia      SmallTown                366.         36.8             60.9
## 7 Russia      Village                  345.         39.5             57.7
## 8 Russia      SmallVillage             322          23.6             69.8
## # ℹ 1 more variable: SD_EnvIdentity <dbl>

Статистика 3

Data %>%
  group_by(Nationality, ResidenceChild_3) %>%
  summarise(
    Mean_WellBeing = mean(WellBeing, na.rm = TRUE),
    SD_WellBeing = sd(WellBeing, na.rm = TRUE),
    Mean_EnvIdentity = mean(EnvIdentity, na.rm = TRUE),
    SD_EnvIdentity = sd(EnvIdentity, na.rm = TRUE)
  )

## `summarise()` has grouped output by 'Nationality'. You can override using the
## `.groups` argument.

## # A tibble: 6 × 6
## # Groups:   Nationality [2]
##   Nationality ResidenceChild_3 Mean_WellBeing SD_WellBeing Mean_EnvIdentity
##   <fct>       <ord>                     <dbl>        <dbl>            <dbl>
## 1 France      BigCity                    315.         25.5             57.2
## 2 France      SmallTown                  301.         26.2             54.9
## 3 France      Small                      312.         17.3             65.5
## 4 Russia      BigCity                    354.         42.0             58.5
## 5 Russia      SmallTown                  366.         36.8             60.9
## 6 Russia      Small                      336          34.6             62.5
## # ℹ 1 more variable: SD_EnvIdentity <dbl>

Проверка на нормальность

shapiro.test(Data$WellBeing[Data$Nationality == "France"])

## 
##  Shapiro-Wilk normality test
## 
## data:  Data$WellBeing[Data$Nationality == "France"]
## W = 0.98671, p-value = 0.7408

shapiro.test(Data$WellBeing[Data$Nationality == "Russia"])

## 
##  Shapiro-Wilk normality test
## 
## data:  Data$WellBeing[Data$Nationality == "Russia"]
## W = 0.9697, p-value = 0.258

shapiro.test(Data$EnvIdentity[Data$Nationality == "France"])

## 
##  Shapiro-Wilk normality test
## 
## data:  Data$EnvIdentity[Data$Nationality == "France"]
## W = 0.9758, p-value = 0.2582

shapiro.test(Data$EnvIdentity[Data$Nationality == "Russia"])

## 
##  Shapiro-Wilk normality test
## 
## data:  Data$EnvIdentity[Data$Nationality == "Russia"]
## W = 0.98051, p-value = 0.6136

Распределение не отличается от нормального

ggplot(Data, aes(x = WellBeing)) + geom_histogram(binwidth = 10) + facet_wrap(~Nationality)

ggplot(Data, aes(x = EnvIdentity)) + geom_histogram(binwidth = 10) + facet_wrap(~Nationality)

ggplot(Data, aes(x = EnvIdentity)) + geom_histogram(binwidth = 10) + facet_wrap(~Nationality)

# Проверка на равенство дисперсий

bartlett.test(WellBeing ~ Nationality, data = Data)

## 
##  Bartlett test of homogeneity of variances
## 
## data:  WellBeing by Nationality
## Bartlett's K-squared = 15.588, df = 1, p-value = 0.00007876

bartlett.test(EnvIdentity ~ Nationality, data = Data)

## 
##  Bartlett test of homogeneity of variances
## 
## data:  EnvIdentity by Nationality
## Bartlett's K-squared = 0.027032, df = 1, p-value = 0.8694

t.test

Благополучие - равенство дисперсий не соблюдается. Флаг - для неравных дисперсий Идентичность - равенство дисперсий соблюдается. Использую тест для равных дисперсий.

t.test(WellBeing ~ Nationality, data = Data, var.equal = FALSE)

## 
##  Welch Two Sample t-test
## 
## data:  WellBeing by Nationality
## t = -6.5518, df = 68.958, p-value = 0.000000008566
## alternative hypothesis: true difference in means between group France and group Russia is not equal to 0
## 95 percent confidence interval:
##  -55.81866 -29.76061
## sample estimates:
## mean in group France mean in group Russia 
##             310.2742             353.0638

t.test(EnvIdentity ~ Nationality, data = Data, var.equal = TRUE)

## 
##  Two Sample t-test
## 
## data:  EnvIdentity by Nationality
## t = -0.013705, df = 107, p-value = 0.9891
## alternative hypothesis: true difference in means between group France and group Russia is not equal to 0
## 95 percent confidence interval:
##  -4.898282  4.831020
## sample estimates:
## mean in group France mean in group Russia 
##             59.83871             59.87234

Вилкоксон 4

Сравнение по отдельным группам при четырех местах жительства

Благополучие 4

unique(Data$ResidenceChild) |>
  lapply(function(res) {
    subset_data <- Data[Data$ResidenceChild == res, ]
    test <- wilcox.test(WellBeing ~ Nationality, data = subset_data
                        #, exact = FALSE
                        )
    data.frame(
      ResidenceChild = res,
      p_value = test$p.value,
      statistic = as.numeric(test$statistic) # Преобразуем для совместимости
    )
  }) |>
  (\(results) do.call(rbind, results))()

## Warning in wilcox.test.default(x = DATA[[1L]], y = DATA[[2L]], ...): cannot
## compute exact p-value with ties
## Warning in wilcox.test.default(x = DATA[[1L]], y = DATA[[2L]], ...): cannot
## compute exact p-value with ties
## Warning in wilcox.test.default(x = DATA[[1L]], y = DATA[[2L]], ...): cannot
## compute exact p-value with ties

##   ResidenceChild      p_value statistic
## 1        BigCity 0.0008143039       130
## 2      SmallTown 0.0002016445         9
## 3        Village 0.0186078164        13
## 4   SmallVillage 0.3736263736        13

exact = FALSE - предотвращает появление предупреждения. Но в целом на результатах не сказывается. Просто предупреждает, что не может показать точную статистику

Идентичность 4

unique(Data$ResidenceChild) |>
  lapply(function(res) {
    subset_data <- Data[Data$ResidenceChild == res, ]
    test <- wilcox.test(EnvIdentity ~ Nationality, data = subset_data
                        #, exact = FALSE
                        )
    data.frame(
      ResidenceChild = res,
      p_value = test$p.value,
      statistic = as.numeric(test$statistic)
    )
  }) |>
  (\(results) do.call(rbind, results))()

## Warning in wilcox.test.default(x = DATA[[1L]], y = DATA[[2L]], ...): cannot
## compute exact p-value with ties
## Warning in wilcox.test.default(x = DATA[[1L]], y = DATA[[2L]], ...): cannot
## compute exact p-value with ties
## Warning in wilcox.test.default(x = DATA[[1L]], y = DATA[[2L]], ...): cannot
## compute exact p-value with ties
## Warning in wilcox.test.default(x = DATA[[1L]], y = DATA[[2L]], ...): cannot
## compute exact p-value with ties

##   ResidenceChild   p_value statistic
## 1        BigCity 0.7022136     277.5
## 2      SmallTown 0.3836088      63.0
## 3        Village 0.2644301      56.0
## 4   SmallVillage 1.0000000      20.0

Вилкоксон 3

Сравнение по отдельным группам при четырех местах жительства

Благополучие

unique(Data$ResidenceChild_3) |>
  lapply(function(res) {
    subset_data <- Data[Data$ResidenceChild_3 == res, ]
    test <- wilcox.test(WellBeing ~ Nationality, data = subset_data
                        #, exact = FALSE
                        )
    data.frame(
      ResidenceChild_3 = res,
      p_value = test$p.value,
      statistic = as.numeric(test$statistic)
    )
  }) |>
  (\(results) do.call(rbind, results))()

## Warning in wilcox.test.default(x = DATA[[1L]], y = DATA[[2L]], ...): cannot
## compute exact p-value with ties
## Warning in wilcox.test.default(x = DATA[[1L]], y = DATA[[2L]], ...): cannot
## compute exact p-value with ties
## Warning in wilcox.test.default(x = DATA[[1L]], y = DATA[[2L]], ...): cannot
## compute exact p-value with ties

##   ResidenceChild_3      p_value statistic
## 1          BigCity 0.0008143039       130
## 2        SmallTown 0.0002016445         9
## 3            Small 0.0163343073        56

Идентичность 3

unique(Data$ResidenceChild_3) |>
  lapply(function(res) {
    subset_data <- Data[Data$ResidenceChild_3 == res, ]
    test <- wilcox.test(EnvIdentity ~ Nationality, data = subset_data#, exact = FALSE
                        )
    data.frame(
      ResidenceChild_3 = res,
      p_value = test$p.value,
      statistic = as.numeric(test$statistic)
    )
  }) |>
  (\(results) do.call(rbind, results))()

## Warning in wilcox.test.default(x = DATA[[1L]], y = DATA[[2L]], ...): cannot
## compute exact p-value with ties
## Warning in wilcox.test.default(x = DATA[[1L]], y = DATA[[2L]], ...): cannot
## compute exact p-value with ties
## Warning in wilcox.test.default(x = DATA[[1L]], y = DATA[[2L]], ...): cannot
## compute exact p-value with ties

##   ResidenceChild_3   p_value statistic
## 1          BigCity 0.7022136     277.5
## 2        SmallTown 0.3836088      63.0
## 3            Small 0.4156133     142.0

Круксал уолис

# Франция
Data %>%
  subset(Nationality == "France") %>%
  kruskal.test(WellBeing ~ ResidenceChild, data = .)

## 
##  Kruskal-Wallis rank sum test
## 
## data:  WellBeing by ResidenceChild
## Kruskal-Wallis chi-squared = 2.6401, df = 3, p-value = 0.4505

Data %>%
  subset(Nationality == "France") %>%
  kruskal.test(EnvIdentity ~ ResidenceChild, data = .)

## 
##  Kruskal-Wallis rank sum test
## 
## data:  EnvIdentity by ResidenceChild
## Kruskal-Wallis chi-squared = 8.8257, df = 3, p-value = 0.0317

# Россия
Data %>%
  subset(Nationality == "Russia") %>%
  kruskal.test(WellBeing ~ ResidenceChild, data = .)

## 
##  Kruskal-Wallis rank sum test
## 
## data:  WellBeing by ResidenceChild
## Kruskal-Wallis chi-squared = 4.5871, df = 3, p-value = 0.2047

Data %>%
  subset(Nationality == "Russia") %>%
  kruskal.test(EnvIdentity ~ ResidenceChild, data = .)

## 
##  Kruskal-Wallis rank sum test
## 
## data:  EnvIdentity by ResidenceChild
## Kruskal-Wallis chi-squared = 4.0903, df = 3, p-value = 0.2519

# Общий анализ (без разделения по странам)
Data %>%
  kruskal.test(WellBeing ~ ResidenceChild, data = .)

## 
##  Kruskal-Wallis rank sum test
## 
## data:  WellBeing by ResidenceChild
## Kruskal-Wallis chi-squared = 5.0176, df = 3, p-value = 0.1705

Data %>%
  kruskal.test(EnvIdentity ~ ResidenceChild, data = .)

## 
##  Kruskal-Wallis rank sum test
## 
## data:  EnvIdentity by ResidenceChild
## Kruskal-Wallis chi-squared = 8.7986, df = 3, p-value = 0.03209

Круксал уолис_3

# Франция
Data %>%
  subset(Nationality == "France") %>%
  kruskal.test(WellBeing ~ ResidenceChild_3, data = .)

## 
##  Kruskal-Wallis rank sum test
## 
## data:  WellBeing by ResidenceChild_3
## Kruskal-Wallis chi-squared = 2.2096, df = 2, p-value = 0.3313

Data %>%
  subset(Nationality == "France") %>%
  kruskal.test(EnvIdentity ~ ResidenceChild_3, data = .)

## 
##  Kruskal-Wallis rank sum test
## 
## data:  EnvIdentity by ResidenceChild_3
## Kruskal-Wallis chi-squared = 8.6499, df = 2, p-value = 0.01323

# Россия
Data %>%
  subset(Nationality == "Russia") %>%
  kruskal.test(WellBeing ~ ResidenceChild_3, data = .)

## 
##  Kruskal-Wallis rank sum test
## 
## data:  WellBeing by ResidenceChild_3
## Kruskal-Wallis chi-squared = 3.7001, df = 2, p-value = 0.1572

Data %>%
  subset(Nationality == "Russia") %>%
  kruskal.test(EnvIdentity ~ ResidenceChild_3, data = .)

## 
##  Kruskal-Wallis rank sum test
## 
## data:  EnvIdentity by ResidenceChild_3
## Kruskal-Wallis chi-squared = 0.76418, df = 2, p-value = 0.6824

# Общий анализ (без разделения по странам)
Data %>%
  kruskal.test(WellBeing ~ ResidenceChild_3, data = .)

## 
##  Kruskal-Wallis rank sum test
## 
## data:  WellBeing by ResidenceChild_3
## Kruskal-Wallis chi-squared = 4.2913, df = 2, p-value = 0.117

Data %>%
  kruskal.test(EnvIdentity ~ ResidenceChild_3, data = .)

## 
##  Kruskal-Wallis rank sum test
## 
## data:  EnvIdentity by ResidenceChild_3
## Kruskal-Wallis chi-squared = 7.2376, df = 2, p-value = 0.02681