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# 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
## `modelsummary` has built-in support to draw text-only (markdown) tables.
##   To generate tables in other formats, you must install one or more of
##   these libraries:
##   
## install.packages(c(
##     "kableExtra",
##     "gt",
##     "flextable",
##    
##   "huxtable",
##     "DT"
## ))
## 
##   Alternatively, you can set markdown as the default table format to
##   silence this alert:
##   
## config_modelsummary(factory_default = "markdown")
## [[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")

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?

norway_happy <- ess %>% # asssigned the 'happy' variable to the country Norway 
  filter(cntry == "NO") %>% 
  select(happy)
norway_happy$y <- norway_happy$happy # filtered to Norway with 'happy' as the variable of interest 

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 # removed 77 and 88 

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

The average for the variable ‘happy’ in Norway is 7.975005

belgium_happy <- ess %>% # assigned the 'happy' variable to the country of Belgium 
  filter(cntry == "BE") %>% 
  select(happy)
belgium_happy$y <- belgium_happy$happy # filitered to Belgium with 'happy' has the variable of interest 

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

table(belgium_happy$y)
## 
##    0    1    2    3    4    5    6    7    8    9   10 
##   50   27  104  194  234  830  999 3503 6521 3402 1565
mean_y <- mean(belgium_happy$y, na.rm = TRUE) 
cat("Mean of 'y' is:", mean_y, "\n")
## Mean of 'y' is: 7.737334

The average for the variable ‘happy’ in Belgium is 7.737334

Therefore, 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.

#  # This chunk filters out the country Ireland through its binge drinking levels. THe Ireland database must be selected and select the alcbnge varibale.  '
ireland_alcbnge <- ess %>%  
  filter(cntry == "IE") %>%           
  select(alcbnge)

ireland_alcbnge$y <- ireland_alcbnge$alcbnge

table(ireland_alcbnge$y)
## 
##   1   2   3   4   5   6   7   8 
##  65 650 346 417 239 641  26   6
ireland_alcbnge$y[ireland_alcbnge$y %in% 6:8] <- NA
# Following the code above, this chunk categorizes the names taken from the ESS website. 

 ireland_alcbnge <- ireland_alcbnge %>%
  mutate(
   alcbnge_category = case_when(
      alcbnge == 1 ~ "Daily or almost daily", 
      alcbnge == 2 ~ "Weekly",
      alcbnge == 3 ~ "Monthly",
      alcbnge == 4 ~ "Less than monthly",
      alcbnge == 5 ~ "Never",
      TRUE ~ NA_character_
    ),
 alcbnge_category = fct_relevel(factor(alcbnge_category),  
                             "Daily or almost daily", 
                             "Weekly", 
                             "Monthly", 
                             "Less than monthly", 
                             "Never")
)
  table(ireland_alcbnge$alcbnge_category)
## 
## Daily or almost daily                Weekly               Monthly 
##                    65                   650                   346 
##     Less than monthly                 Never 
##                   417                   239

The data shows that weekly is the most prominent frequency of binge drinking.

task 3

result <- ess %>% # Here, we are gathering the mean for the countries of Portugal and Serbia for the variable plnftr 
  filter(cntry %in% c("PT", "RS")) %>%
  
  mutate(plnftr = recode(plnftr, `77` = NA_real_, `88` = NA_real_, `99` = NA_real_)) %>%
  
  group_by(cntry) %>%
  summarize(mean_plnftr = mean(plnftr, na.rm = TRUE))

print(result)
## # A tibble: 2 × 2
##   cntry mean_plnftr
##   <chr>       <dbl>
## 1 PT           5.42
## 2 RS           4.14

We can see that Portugal has a higher rate of planning for the future compared to Serbia.

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 %>%  # Here, we are filtering Italy through the ESS database. 
  filter(cntry == "IT")

italy_data <- italy_data %>%
  mutate(
    gndr = case_when(
      gndr == 1 ~ "Male",
      gndr == 2 ~ "Female",
      TRUE ~ as.character(gndr)
    ),
  stfdem = ifelse(stfdem %in% c(77, 88), NA, stfdem)  
  )
mean_male_stfdem <- italy_data %>% # After filering, we found the mean for the variable stfdem for men. 
  filter(gndr == "Male") %>%
  summarize(mean_stfdem_men = mean(stfdem, na.rm = TRUE))

print(mean_male_stfdem)
##   mean_stfdem_men
## 1        4.782646
means_by_gender <- italy_data %>% # To compare, this line of code groups the average for the disatisfaction of democracy.
  group_by(gndr) %>%
  summarize(stfdem = mean(stfdem, na.rm = TRUE)) 

print(means_by_gender) 
## # A tibble: 3 × 2
##   gndr   stfdem
##   <chr>   <dbl>
## 1 9        3.25
## 2 Female   4.69
## 3 Male     4.78

We can see that females in Italy are more dissatisfied with their democracy.

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.

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).

# The following chunk of code produces the boxplot according to the satisfaction with the state of satisfaction with the state of education vs. health services in France
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)", title =  "Boxplot of satisfaction with the state of education vs. health services") +
  theme_minimal()
## Warning: Removed 364 rows containing non-finite values (`stat_boxplot()`).

We can see that the median French person is more satisfied with their health care because the variable ‘stfhlth’ indicates a higher median suggesting that based on the spread of data, the median French person is more satisfied with their healthcare over their education.