Kim_Eric_Homework_1
Seokang Eric Kim
2024-01-17
Before I start
# 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"
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## [1] "fst" "lubridate" "forcats" "stringr" "dplyr" "purrr"
## [7] "readr" "tidyr" "tibble" "ggplot2" "tidyverse" "stats"
## [13] "graphics" "grDevices" "utils" "datasets" "methods" "base"
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## [[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
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?
Levels of happiness for Belgium
belgium_happy <- ess %>%
filter(cntry == "BE") %>%
select(happy)belgium_happy$y <- belgium_happy$happy
table(belgium_happy$y)##
## 0 1 2 3 4 5 6 7 8 9 10 77 88 99
## 50 27 104 194 234 830 999 3503 6521 3402 1565 3 16 3belgium_happy$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 1565mean_y <- mean(belgium_happy$y, na.rm = TRUE)
cat("Mean of 'y' is:", mean_y, "\n")## Mean of 'y' is: 7.737334median_y <- median(belgium_happy$y, na.rm = TRUE)
cat("Median of 'y' is:", median_y, "\n")## Median of 'y' is: 8mode_y <- belgium_happy %>%
filter(!is.na(y)) %>%
count(y) %>%
arrange(desc(n)) %>%
slice(1) %>%
pull(y)
cat("\nMode of Y:", mode_y, "\n")##
## Mode of Y: 8sd_y <- sd(belgium_happy$y, na.rm = TRUE)
cat("Standard Deviation of 'y':", sd_y, "\n")## Standard Deviation of 'y': 1.52045summary(belgium_happy)## happy y
## Min. : 0.000 Min. : 0.000
## 1st Qu.: 7.000 1st Qu.: 7.000
## Median : 8.000 Median : 8.000
## Mean : 7.839 Mean : 7.737
## 3rd Qu.: 9.000 3rd Qu.: 9.000
## Max. :99.000 Max. :10.000
## NA's :22Levels of happiness for Norway
norway_happy <- ess %>%
filter(cntry == "NO") %>%
select(happy)norway_happy$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 10norway_happy$y[norway_happy$y %in% 77:99] <- NA
table(norway_happy$y)##
## 0 1 2 3 4 5 6 7 8 9 10
## 15 29 59 163 238 730 817 2617 5235 3796 2344mean_y <- mean(norway_happy$y, na.rm = TRUE)
cat("Mean of 'y' is:", mean_y, "\n")## Mean of 'y' is: 7.975005median_y <- median(norway_happy$y, na.rm = TRUE)
cat("Median of 'y' is:", median_y, "\n")## Median of 'y' is: 8mode_y <- norway_happy %>%
filter(!is.na(y)) %>%
count(y) %>%
arrange(desc(n)) %>%
slice(1) %>%
pull(y)
cat("\nMode of Y:", mode_y, "\n")##
## Mode of Y: 8sd_y <- sd(norway_happy$y, na.rm = TRUE)
cat("Standard Deviation of 'y':", sd_y, "\n")## Standard Deviation of 'y': 1.539186summary(norway_happy)## happy y
## Min. : 0.000 Min. : 0.000
## 1st Qu.: 7.000 1st Qu.: 7.000
## Median : 8.000 Median : 8.000
## Mean : 8.076 Mean : 7.975
## 3rd Qu.: 9.000 3rd Qu.: 9.000
## Max. :88.000 Max. :10.000
## NA's :22Answer: Norway (mean 7.975 ± SD 1.539) reports higher levels of happiness than Belgium (mean 7.737 ± SD 1.520).
Task 2
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_alcbnge <- ess %>%
filter(cntry == "IE") %>%
select(alcbnge)
ireland_alcbnge$y <- ireland_alcbnge$alcbnge
table(ireland_alcbnge$y)##
## 1 2 3 4 5 6 7 8
## 65 650 346 417 239 641 26 6ireland_alcbnge$y[ireland_alcbnge$y %in% 6:8] <- NAmean_y <- mean(ireland_alcbnge$y, na.rm = TRUE)
median_y <- median(ireland_alcbnge$y, na.rm = TRUE)
cat("Mean of 'y':", mean_y, "\n")## Mean of 'y': 3.066977cat("Median of 'y':", median_y, "\n")## Median of 'y': 3df <- ireland_alcbnge %>%
mutate(
y_category = case_when(
y == 1 ~ "Daily or almost daily",
y == 2 ~ "Weekly",
y == 3 ~ "Monthly",
y == 4 ~ "Less than monthly",
y == 5 ~ "Never",
TRUE ~ NA_character_
),
y_category = fct_relevel(factor(y_category),
"Daily or almost daily",
"Weekly",
"Monthly",
"Less than monthly",
"Never")
)
table(df$y_category)##
## Daily or almost daily Weekly Monthly
## 65 650 346
## Less than monthly Never
## 417 239get_mode <- function(v) {
tbl <- table(v)
mode_vals <- as.character(names(tbl)[tbl == max(tbl)])
return(mode_vals)
}
mode_values <- get_mode(df$y_category)
cat("Mode of y category:", paste(mode_values, collapse = ", "), "\n")## Mode of y category: WeeklyAnswer: The most common category selected, for Irish respondents, for frequency of binge drinking is 2. “Weekly” (650 responds).
Task 3
Prompt and question: when you use the summary() function for the variable plnftr (about planning for future or taking every each day as it comes from 0-10) for both the countries of Portugal and Serbia, what do you notice? What stands out as different when you compare the two countries (note: look up the variable information on the ESS website to help with interpretation)? Explain while referring to the output generated.
Obtaining data for the variable plnftr for Portugal
portugal_plnftr <- ess %>%
filter(cntry == "PT") %>%
select(plnftr)portugal_plnftr$y <- portugal_plnftr$plnftr
table(portugal_plnftr$y)##
## 0 1 2 3 4 5 6 7 8 9 10 88
## 114 184 313 356 264 481 262 382 345 166 370 40portugal_plnftr$y[portugal_plnftr$y %in% 88] <- NA
table(portugal_plnftr$y)##
## 0 1 2 3 4 5 6 7 8 9 10
## 114 184 313 356 264 481 262 382 345 166 370mean_y <- mean(portugal_plnftr$y, na.rm = TRUE)
cat("Mean of 'y' is:", mean_y, "\n")## Mean of 'y' is: 5.418289median_y <- median(portugal_plnftr$y, na.rm = TRUE)
cat("Median of 'y' is:", median_y, "\n")## Median of 'y' is: 5Obtaining data for the variable plnftr for Serbia
serbia_plnftr <- ess %>%
filter(cntry == "SE") %>%
select(plnftr)serbia_plnftr$y <- serbia_plnftr$plnftr
table(serbia_plnftr$y)##
## 0 1 2 3 4 5 6 7 8 9 10 77 88 99
## 151 155 398 517 301 503 270 405 374 145 237 2 7 1serbia_plnftr$y[serbia_plnftr$y %in% 88] <- NA
table(serbia_plnftr$y)##
## 0 1 2 3 4 5 6 7 8 9 10 77 99
## 151 155 398 517 301 503 270 405 374 145 237 2 1mean_y <- mean(serbia_plnftr$y, na.rm = TRUE)
cat("Mean of 'y' is:", mean_y, "\n")## Mean of 'y' is: 5.087019median_y <- median(serbia_plnftr$y, na.rm = TRUE)
cat("Median of 'y' is:", median_y, "\n")## Median of 'y' is: 5Comparing the summaries for two countries
summary(portugal_plnftr)## plnftr y
## Min. : 0.000 Min. : 0.000
## 1st Qu.: 3.000 1st Qu.: 3.000
## Median : 5.000 Median : 5.000
## Mean : 6.426 Mean : 5.418
## 3rd Qu.: 8.000 3rd Qu.: 8.000
## Max. :88.000 Max. :10.000
## NA's :14604 NA's :14644summary(serbia_plnftr)## plnftr y
## Min. : 0.000 Min. : 0.000
## 1st Qu.: 3.000 1st Qu.: 3.000
## Median : 5.000 Median : 5.000
## Mean : 5.254 Mean : 5.087
## 3rd Qu.: 7.000 3rd Qu.: 7.000
## Max. :99.000 Max. :99.000
## NA's :14750 NA's :14757Answer: The degree to which people tend to either plan for their future or take each day as it comes is very similar between those two countries and it stands out. The Portuguese population (mean 5.418), however, tends to plan for their future slightly more than the Serbian population (mean 5.018).
Task 4
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.
Loading datasets for the variables stfdem and gndr
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)
)mean_male_stfdem <- italy_data %>%
filter(gndr == "Male") %>%
summarize(mean_stfdem_men = mean(stfdem, na.rm = TRUE))
print(mean_male_stfdem)## mean_stfdem_men
## 1 4.782646mean_female_stfdem <- italy_data %>%
filter(gndr == "Female") %>%
summarize(mean_stfdem_women = mean(stfdem, na.rm = TRUE))
print(mean_female_stfdem)## mean_stfdem_women
## 1 4.694652Answer: On average, female population (mean_female_stfdem = 4.694) is slightly more dissatisfied with democracy in Italy than male population (mean_female_stfdem = 4.782).
Task 5
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.
Answer: As the provided boxplot graph shows, the median of the variable stfhlth is apparently higher than that of the variable stfedu. Thus, we could say that the median French person is more satisfied with the health services than with the education system.
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 %>%
mutate(stfedu = ifelse(stfedu %in% c(77, 88, 99), NA, stfedu),
stfhlth = ifelse(stfhlth %in% c(77, 88, 99), NA, stfhlth)) %>%
select(stfedu, stfhlth) %>%
gather(variable, value, c(stfedu, stfhlth)) %>%
ggplot(aes(y = value)) +
geom_boxplot() +
labs(y = "Satisfaction (0-10)", title = "Boxplot of satisfaction with stfedu vs. stfhlth") +
theme_minimal()## Warning: Removed 364 rows containing non-finite values (`stat_boxplot()`).
This homework is done by Seokang “Eric” Kim at 9:11 PM, January 16 2024.