# List of packages
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     2.1.4
## ✔ 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"
ess <- read_fst("All-ESS-Data.fst")
unique(ess$cntry)
##  [1] "AT" "BE" "CH" "CZ" "DE" "DK" "ES" "FI" "FR" "GB" "GR" "HU" "IE" "IL" "IT"
## [16] "LU" "NL" "NO" "PL" "PT" "SE" "SI" "EE" "IS" "SK" "TR" "UA" "BG" "CY" "RU"
## [31] "HR" "LV" "RO" "LT" "AL" "XK" "ME" "RS" "MK"

Task 1

Provide code and answer.

Prompt and question: calculate the average for the variable ‘happy’ for the country of Norway. On average, based on the ESS data, who reports higher levels of happiness: Norway or Belgium?

Note: we already did it for Belgium. You just need to compare to Norway’s average, making sure to provide the code for both.

belgium_happy <- ess %>% 
  filter(cntry == "BE") %>% 
  select(happy)
belgium_happy$y <- belgium_happy$happy

table(belgium_happy$y)
## 
##    0    1    2    3    4    5    6    7    8    9   10   77   88   99 
##   50   27  104  194  234  830  999 3503 6521 3402 1565    3   16    3
# Recode values 77 through 99 to NA
belgium_happy$y[belgium_happy$y %in% 77:99] <- NA

# checking again
table(belgium_happy$y)
## 
##    0    1    2    3    4    5    6    7    8    9   10 
##   50   27  104  194  234  830  999 3503 6521 3402 1565
norway_happy <- ess %>% 
  filter(cntry == "NO") %>% 
  select(happy)
norway_happy$y <- norway_happy$happy

table(norway_happy$y)
## 
##    0    1    2    3    4    5    6    7    8    9   10   77   88 
##   15   29   59  163  238  730  817 2617 5235 3796 2344   12   10
norway_happy$y[norway_happy$y %in% 77:99] <- NA

table(norway_happy$y)
## 
##    0    1    2    3    4    5    6    7    8    9   10 
##   15   29   59  163  238  730  817 2617 5235 3796 2344
mean_y <- mean(belgium_happy$y, na.rm = TRUE)
cat("Mean of 'y' is:", mean_y, "\n")
## Mean of 'y' is: 7.737334
mean_y <- mean(norway_happy$y, na.rm = TRUE)
cat("Mean of 'y' is:", mean_y, "\n")
## Mean of 'y' is: 7.975005

On average, based on the ESS data, Norway reports higher levels of happiness.

Task 2

Provide code and answer.

Prompt and question: what is the most common category selected, for Irish respondents, for frequency of binge drinking? The variable of interest is: alcbnge.

More info here: https://ess-search.nsd.no/en/variable/0c65116e-7481-4ca6-b1d9-f237db99a694.

Hint: need to convert numeric value entries to categories as specified in the variable information link. We did similar steps for Estonia and the climate change attitude variable.

ireland_ccbinge <- ess %>%
  filter(cntry == "IE") %>%
  select(alcbnge)

ireland_ccbinge$y <- ireland_ccbinge$alcbnge

table(ireland_ccbinge$y)
## 
##   1   2   3   4   5   6   7   8 
##  65 650 346 417 239 641  26   6
# Recode values 6 through 8 to NA
ireland_ccbinge$y[ireland_ccbinge$y %in% 6:8] <- NA
# Converting to categories to get mode as a category instead of a number
df <- ireland_ccbinge %>%
  mutate(
    y_category = case_when(
      y == 1 ~ "Daily",
      y == 2 ~ "Weekly",
      y == 3 ~ "Monthly",
      y == 4 ~ "Less Than Monthly",
      y == 5 ~ "Never",
      TRUE ~ NA_character_
    ),
    y_category = fct_relevel(factor(y_category),  ### here you would put the categories in order you want them to appear or else it will appear alphabetically
                             "Daily", 
                             "Weekly", 
                             "Monthly", 
                             "Less Than Monthly", 
                             "Never")
  )

# To confirm the conversion:
table(df$y_category)
## 
##             Daily            Weekly           Monthly Less Than Monthly 
##                65               650               346               417 
##             Never 
##               239
# Let's determine the mode of our newly created category:

get_mode <- function(v) {
  tbl <- table(v)
  mode_vals <- as.character(names(tbl)[tbl == max(tbl)])
  return(mode_vals)
}

mode_values <- get_mode(df$y_category)
cat("Mode of y category:", paste(mode_values, collapse = ", "), "\n")
## Mode of y category: Weekly

The most common category selected for Irish respondents, for frequency of binge drinking is “Weekly”.

Task 3

Provide code and answer.

Prompt and question: when you use the summary() function for the variable plnftr (about planning for future or taking every each day as it comes from 0-10) for both the countries of Portugal and Serbia, 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_future <- ess %>% 
  filter(cntry == "PT") %>% 
  select(plnftr)
summary(portugal_future)
##      plnftr      
##  Min.   : 0.000  
##  1st Qu.: 3.000  
##  Median : 5.000  
##  Mean   : 6.426  
##  3rd Qu.: 8.000  
##  Max.   :88.000  
##  NA's   :14604
serbia_future <- ess %>% 
  filter(cntry == "RS") %>% 
  select(plnftr)
summary(serbia_future)
##      plnftr      
##  Min.   : 0.000  
##  1st Qu.: 0.000  
##  Median : 4.000  
##  Mean   : 4.983  
##  3rd Qu.: 8.000  
##  Max.   :88.000  
##  NA's   :1505

When comparing Portugal and Serbia, the output generated for Portugal’s mean is 6.426 and the output generated for Serbia’s mean is 4.983 meaning that on average, when it comes to planning for future or taking every each day as it comes from 0-10 (0 being planning for the future as much as possible and 10 being taking each day as it comes), people in Portugal generally take each day as it comes while people from Serbia plan for the future more. ## Task 4

Provide code and answer.

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

Info on variable here: https://ess.sikt.no/en/variable/query/stfdem/page/1

italy_data <- ess %>% 
  filter(cntry == "IT")

# Convert gender and stfdem 
italy_data <- italy_data %>%
  mutate(
    gndr = case_when(
      gndr == 1 ~ "Male",
      gndr == 2 ~ "Female",
      TRUE ~ as.character(gndr)
    ),
    gndr = ifelse(gndr %in% c(9),NA, gndr)  
  )
# Compute average of stfdem by gender
means_by_gender <- italy_data %>%
  group_by(gndr) %>% 
  summarize(stfdem = mean(stfdem, na.rm = TRUE)) 

print(means_by_gender)
## # A tibble: 3 × 2
##   gndr   stfdem
##   <chr>   <dbl>
## 1 Female   8.43
## 2 Male     7.43
## 3 <NA>     8.92

Since 0 represents extremely dissatisfied and 10 represents extremely satisfied, on average Italian men are more dissatified with democracy in Italy. The output generated for male satisfaction is lower as the average is 7.425145 while the women’s is 8.923077 meaning they are more satisfied.

Task 5

Provide code and answer.

Prompt: Interpret the boxplot graph of stfedu and stfhlth that we generated already: according to ESS data, would we say that the median French person is more satisfied with the education system or health services? Explain.

According to the ess data, the average French person is more satisfied with the health system compared to education services because the median for satisfaction with the health system is around 7 while the median for satisfaction of education systems is 5.

Change the boxplot graph: provide the code to change some of the key labels: (1) Change the title to: Boxplot of satisfaction with the state of education vs. health services; (2) Remove the x-axis label; (3) Change the y-axis label to: Satisfaction (0-10).

france_data <- ess %>% 
  filter(cntry == "FR")
france_data %>%
  # Setting values to NA
  mutate(stfedu = ifelse(stfedu %in% c(77, 88, 99), NA, stfedu),
         stfhlth = ifelse(stfhlth %in% c(77, 88, 99), NA, stfhlth)) %>%
  # Reshaping the data
  select(stfedu, stfhlth) %>%
  gather(variable, value, c(stfedu, stfhlth)) %>%
  # Creating the boxplot
  ggplot(aes(x = variable, y = value)) +
  geom_boxplot() +
  labs(y = "Satisfaction (0-10)", x = "", title = "Boxplot of satisfaction with the state of education vs. health services") +
  theme_minimal()
## Warning: Removed 364 rows containing non-finite values (`stat_boxplot()`).