Homework 1

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.

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.

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.

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

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).

Hint: copy the boxplot code above and just replace or cut what is asked.

First things first: setting up your environment

In the list following c(, you would be any package you need in between quotation marks.

# 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"

rm(list=ls()); gc()

##           used (Mb) gc trigger  (Mb) limit (Mb) max used (Mb)
## Ncells 1079655 57.7    2196882 117.4         NA  1462128 78.1
## Vcells 1860170 14.2    8388608  64.0      16384  2380965 18.2

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

belgium_happy$y[belgium_happy$y %in% 77:99] <- NA

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

belgium_happy %>%
  summarize(
    mean_y = mean(y, na.rm = TRUE),
    median_y = median(y, na.rm = TRUE)
  ) %>%
  print()

##     mean_y median_y
## 1 7.737334        8

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

norway_happy %>%
  summarize(
    mean_y = mean(y, na.rm = TRUE),
    median_y = median(y, na.rm = TRUE)
  ) %>%
  print()

##     mean_y median_y
## 1 7.975005        8

##Answer: Norway reports their happiness as “7.97”, “0.24” highier than Belgiums “7.73”. Therefore, Norway experiences higher levels of happiness than Belgium.

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.

ireland_ccdrinking <- ess %>%
  filter(cntry == "IL") %>%
  select(alcbnge)

ireland_ccdrinking$y <- ireland_ccdrinking$alcbnge

table(ireland_ccdrinking$y)

## 
##    1    2    3    4    5    7    8    9 
##   15  153  149  264  362    5   69 1545

ireland_ccdrinking$y[ireland_ccdrinking$y %in% 6:8] <- NA

table(ireland_ccdrinking$y)

## 
##    1    2    3    4    5    9 
##   15  153  149  264  362 1545

mean_y <- mean(ireland_ccdrinking$y, na.rm = TRUE)
median_y <- median(ireland_ccdrinking$y, na.rm = TRUE)

cat("Mean of 'y':", mean_y, "\n")

## Mean of 'y': 7.049437

cat("Median of 'y':", median_y, "\n")

## Median of 'y': 9

df <- ireland_ccdrinking %>%
  mutate(
    y_category = case_when(
      y == 1 ~ "Don't Binge Drink",
      y == 2 ~ "Rarely Binge Drink",
      y == 3 ~ "Somewhat Binge Drink",
      y == 4 ~ "Frequently Binge Drink",
      y == 5 ~ "Extreme Binge Drink ",
      TRUE ~ NA_character_
    ),
    y_category = fct_relevel(factor(y_category),
                             
                             "Don't Binge Drink", 
                             "Rarely Binge Drink", 
                             "Somewhat Binge Drink", 
                             "Frequently Binge Drink", 
                             "Extreme Binge Drink")
  )

## Warning: There was 1 warning in `mutate()`.
## ℹ In argument: `y_category = fct_relevel(...)`.
## Caused by warning:
## ! 1 unknown level in `f`: Extreme Binge Drink

table(df$y_category)

## 
##      Don't Binge Drink     Rarely Binge Drink   Somewhat Binge Drink 
##                     15                    153                    149 
## Frequently Binge Drink   Extreme Binge Drink  
##                    264                    362

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: Extreme Binge Drink

##Answer: Irish respondents most frequently choose “Extreme Binge Drinking” as the predominant category for their binge drinking behaviour.

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.

result <- ess %>%
  
  filter(cntry %in% c("PT", "SE")) %>%
  

  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 SE           5.02

##Answer: A distinction arises as Portugal leans slightly more towards the “taking every day as it comes” perspective compared to Serbia, with respective averages of 5.41 and 5.01.

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.

italy_data <- ess %>% 
  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) 
  )

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 9        3.25
## 2 Female   4.69
## 3 Male     4.78

##Answer: Upon examining the data, it becomes apparent that in Italy, men exhibit higher satisfaction with democracy than women, evident in their greater average score of 4.78.

Task 5

Provide code and answer.

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

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()`).

Homework 1

Luna Marin

2024-01-22

Task 1

Task 2

Task 3

Task 4

Task 5

First things first: setting up your environment

Task 1

Task 2

Task 3

Prompt and question:

Task 4

Task 5