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packages <- c("tidyverse", "fst", "modelsummary") # add any you need here

# Install packages if they aren't installed already
new_packages <- packages[!(packages %in% installed.packages()[,"Package"])]
if(length(new_packages)) install.packages(new_packages)

# Load the packages
lapply(packages, library, character.only = TRUE)
## ── Attaching core tidyverse packages ──────────────────────── tidyverse 2.0.0 ──
## ✔ dplyr     1.1.4     ✔ readr     2.1.5
## ✔ forcats   1.0.0     ✔ stringr   1.5.1
## ✔ ggplot2   3.4.4     ✔ tibble    3.2.1
## ✔ lubridate 1.9.3     ✔ tidyr     1.3.0
## ✔ purrr     1.0.2     
## ── Conflicts ────────────────────────────────────────── tidyverse_conflicts() ──
## ✖ dplyr::filter() masks stats::filter()
## ✖ dplyr::lag()    masks stats::lag()
## ℹ Use the conflicted package (<http://conflicted.r-lib.org/>) to force all conflicts to become errors
## [[1]]
##  [1] "lubridate" "forcats"   "stringr"   "dplyr"     "purrr"     "readr"    
##  [7] "tidyr"     "tibble"    "ggplot2"   "tidyverse" "stats"     "graphics" 
## [13] "grDevices" "utils"     "datasets"  "methods"   "base"     
## 
## [[2]]
##  [1] "fst"       "lubridate" "forcats"   "stringr"   "dplyr"     "purrr"    
##  [7] "readr"     "tidyr"     "tibble"    "ggplot2"   "tidyverse" "stats"    
## [13] "graphics"  "grDevices" "utils"     "datasets"  "methods"   "base"     
## 
## [[3]]
##  [1] "modelsummary" "fst"          "lubridate"    "forcats"      "stringr"     
##  [6] "dplyr"        "purrr"        "readr"        "tidyr"        "tibble"      
## [11] "ggplot2"      "tidyverse"    "stats"        "graphics"     "grDevices"   
## [16] "utils"        "datasets"     "methods"      "base"
belgium_data <- read.fst("belgium_data.fst")

estonia_data <- read.fst("estonia_data.fst")

france_data <- read.fst("france_data.fst")

norway_data <- read.fst("norway_data.fst")

ireland_data <- read.fst("ireland_data.fst")

portugal_data <- read.fst("portugal_data.fst")

serbia_data <- read.fst("serbia_data.fst")

italy_data <- read.fst("italy_data.fst")

Task 1 Provide code and answer.

Prompt and question: calculate the average for the variable ‘health’ for the countries of Serbia and Belgium. On average, based on the ESS data, who reports higher levels of subjective health: Serbia and Belgium?

belgium_health <- belgium_data %>% # note: if you work from belgium_data replace "ess" with belgium_data
  filter(cntry == "BE") %>% 
  select(health)

belgium_health$y <- belgium_health$health

table(belgium_health$y)
## 
##    1    2    3    4    5    7    8    9 
## 4033 9262 3338  674  137    2    3    2
mean_belgium <- mean(belgium_health$y, na.rm = TRUE)
cat("Mean of 'y' is:", mean_belgium, "\n")
## Mean of 'y' is: 2.063377
serbia_health <- serbia_data %>%
  select(health)

serbia_health$y <- serbia_health$health

table(serbia_health$y)
## 
##    1    2    3    4    5    8    9 
##  620 1051 1332  411  107    3   24
# Recode invalid values to NA
serbia_health$health[serbia_health$health %in% c(77, 88, 99)] <- NA

# Calculate mean
mean_serbia <- mean(serbia_health$health, na.rm = TRUE)
cat("Mean of 'y' is:", mean_serbia, "\n")
## Mean of 'y' is: 2.575254
mean_belgium
## [1] 2.063377
mean_serbia
## [1] 2.575254

Belgium = 2.06 and Serbia = 2.57 showing slightly higher signs of health

Task 2 Provide code and answer.

Prompt and question: what is the most common category selected, for Norwegian respondents, for how often they socially meet with friends or relatives. The variable of interest is: sclmeet.

Norway_meet <- norway_data %>%
  select(sclmeet)

norway_meet <- norway_data %>%
  filter(cntry == "NO") %>%
  select(sclmeet)
table(norway_meet$sclmeet)
## 
##    1    2    3    4    5    6    7   77   88 
##   44  356  746 2569 2119 5764 4458    2    7
norway_meet$sclmeet[norway_meet$sclmeet %in% 77:88] <- NA
table(norway_meet$sclmeet)
## 
##    1    2    3    4    5    6    7 
##   44  356  746 2569 2119 5764 4458
mode2 <- norway_meet%>%
  mutate(
    y_category = case_when(
      sclmeet == 1 ~ "Daily or almost daily",
      sclmeet == 2 ~ "Weekly",
      sclmeet == 3 ~ "Monthly",
      sclmeet == 4 ~ "Less than monthly",
      sclmeet == 5 ~ "Never",
      TRUE ~ NA_character_
    ),
    y_category = fct_relevel(factor(y_category),
                             "Daily or almost daily", 
                             "Weekly", 
                             "Monthly", 
                             "Less than monthly", 
                             "Never")
    
  
  )

table(mode2$y_category)
## 
## Daily or almost daily                Weekly               Monthly 
##                    44                   356                   746 
##     Less than monthly                 Never 
##                  2569                  2119
get_mode <- function(v) {
  tbl <- table(v)
  mode_vals <- as.character(names(tbl)[tbl == max(tbl)])
  return(mode_vals)
}

mode_norway_meet <- get_mode(norway_meet$sclmeet_category)
## Warning in max(tbl): no non-missing arguments to max; returning -Inf
mode_norway_meet
## character(0)

Task 3 Provide code and answer.

Prompt and question: when you use the summary() function for the variable wrclmch (i.e., how worried about climate change) for both the countries of Estonia and Portugal – what do you notice? What stands out as different when you compare the two countries (note: look up the variable information on the ESS website to help with interpretation)? Explain while referring to the output generated.

portugal_wrclmch <- portugal_data %>%
  filter(cntry == "PT") %>%
  select(wrclmch)


estonia_wrclmch <- estonia_data %>%
  filter(cntry == "ES") %>%
  select(wrclmch)
summary(portugal_wrclmch)
##     wrclmch     
##  Min.   :1.000  
##  1st Qu.:3.000  
##  Median :4.000  
##  Mean   :3.583  
##  3rd Qu.:4.000  
##  Max.   :8.000  
##  NA's   :14773
summary(estonia_wrclmch)
##     wrclmch   
##  Min.   : NA  
##  1st Qu.: NA  
##  Median : NA  
##  Mean   :NaN  
##  3rd Qu.: NA  
##  Max.   : NA
table(portugal_wrclmch$wrclmch)
## 
##    1    2    3    4    5    6    7    8 
##   55  251 1171 1203  380   13    4   31
table(estonia_wrclmch$wrclmch)
## < table of extent 0 >
ggplot(estonia_wrclmch, aes(x = wrclmch)) +
  geom_histogram(bins = 10, fill = "blue", color = "black") +
  theme_minimal() +
  labs(title = "Distribution for estonia",
       x = "Scale",
       y = "Count")

ggplot(portugal_wrclmch, aes(x = wrclmch)) +
  geom_histogram(bins = 10, fill = "blue", color = "black") +
  theme_minimal() +
  labs(title = "Distribution for Portugal",
       x = "Scale",
       y = "Count")
## Warning: Removed 14773 rows containing non-finite values (`stat_bin()`).

Task 4 Provide code and answer.

Prompt and question: using the variables stfhlth and gndr, answer the following: on average, who is more sastified with health services in Italy, men or women? Explain while referring to the output generated.

table(italy_data$gndr)
## 
##    1    2    9 
## 4836 5329   13
table(italy_data$stfhlth)
## 
##    0    1    2    3    4    5    6    7    8    9   10   77   88   99 
##  455  279  553  778  958 1470 1718 1869 1363  411  192   24  105    3
italy_data$gndr[italy_data$gndr %in% 9] <- NA
table(italy_data$stfdem)
## 
##    0    1    2    3    4    5    6    7    8    9   10   77   88   99 
##  712  365  733  962 1221 1785 1724 1300  733  138  102   57  344    2
italy_data$stfdem[italy_data$stfdem %in% 77:99] <- NA
italy_data <- italy_data %>%
  mutate(
    gndr = case_when(
      gndr == 1 ~ "Male",
      gndr == 2 ~ "Female",
      TRUE ~ as.character(gndr)
    )
  )
italy_data <- italy_data %>% filter(!is.na(gndr))
italy_data <- italy_data %>% filter(!is.na(stfdem))
table(italy_data$gndr)
## 
## Female   Male 
##   5084   4679
table(italy_data$stfdem)
## 
##    0    1    2    3    4    5    6    7    8    9   10 
##  710  365  731  958 1220 1785 1722 1299  733  138  102
italy_data %>%
  group_by(gndr) %>%
  summarize(meanstfdem = mean(stfdem, na.rm = TRUE))
## # A tibble: 2 × 2
##   gndr   meanstfdem
##   <chr>       <dbl>
## 1 Female       4.66
## 2 Male         4.78

Task 5 Create boxplots, in the same graph, for stfdem and stfgov for the country of France. Provide an appropriate title and labels. Further, tell us: according to ESS data, would we say that the median French person is more satisfied with the democracy or the national government? Explain.

france_data %>%
  # Setting values to NA
  mutate(stfedu = ifelse(stfdem %in% c(77, 88, 99), NA, stfdem),
         stfhlth = ifelse(stfgov %in% c(77, 88, 99), NA, stfgov)) %>%
  # Reshaping the data
  select(stfdem, stfgov) %>%
  gather(variable, value, c(stfdem, stfgov)) %>%
  # Creating the boxplot
  ggplot(aes(x = variable, y = value)) +
  geom_boxplot() +
  labs(y = "Y-axis", x = "X-axis", title = "Boxplot of stfdem vs. stfgov") +
  theme_minimal()

median(france_data$stfdem)
## [1] 5
median(france_data$stfgov)
## [1] 4