Sohn_Jio_Homework1
2024-01-22
# 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 2.1.5
## ✔ 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"
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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")belgium_data <- read.fst("belgium_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")france_data <- read.fst("france_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?
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 <- belgium_data %>%
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_belgium <- mean(belgium_happy$y, na.rm = TRUE)
cat("average for belgium is:", mean_belgium, "\n")## average for belgium is: 7.737334norway_happy <- norway_data %>%
filter(cntry == "NO") %>%
select(happy)norway_happy$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% 77:88] <- NA
table(norway_happy$y)##
## 0 1 2 3 4 5 6 7 8 9 10
## 15 29 59 163 238 730 817 2617 5235 3796 2344mean_norway <- mean(norway_happy$y, na.rm = TRUE)
cat("average for norway is:", mean_norway, "\n")## average for norway is: 7.975005A: 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_drinking <- ireland_data %>%
filter(cntry == "IE") %>%
select(alcbnge)
ireland_drinking$y <- ireland_drinking$alcbnge
table(ireland_drinking$y)##
## 1 2 3 4 5 6 7 8
## 65 650 346 417 239 641 26 6ireland_drinking$y[ireland_drinking$y %in% 6:8] <- NA
table(ireland_drinking$y)##
## 1 2 3 4 5
## 65 650 346 417 239df <- ireland_drinking %>%
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("Most common category:", paste(mode_values, collapse = ", "), "\n")## Most common category: WeeklyTask 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_planning <- portugal_data %>%
filter(cntry == "PT") %>%
select(plnftr)
portugal_planning$y <- portugal_planning$plnftr
table(portugal_planning$y)##
## 0 1 2 3 4 5 6 7 8 9 10 88
## 114 184 313 356 264 481 262 382 345 166 370 40portugal_planning$y[portugal_planning$y %in% 88] <- NA
table(portugal_planning$y)##
## 0 1 2 3 4 5 6 7 8 9 10
## 114 184 313 356 264 481 262 382 345 166 370summary(portugal_planning)## 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 :14644serbia_planning <- serbia_data %>%
filter(cntry == "RS") %>%
select(plnftr)
serbia_planning$y <- serbia_planning$plnftr
table(serbia_planning$y)##
## 0 1 2 3 4 5 6 7 8 9 10 77 88
## 587 133 152 138 95 246 70 87 103 47 364 4 17serbia_planning$y[serbia_planning$y %in% 77:88] <- NA
table(serbia_planning$y)##
## 0 1 2 3 4 5 6 7 8 9 10
## 587 133 152 138 95 246 70 87 103 47 364summary(serbia_planning)## plnftr y
## Min. : 0.000 Min. : 0.000
## 1st Qu.: 0.000 1st Qu.: 0.000
## Median : 4.000 Median : 4.000
## Mean : 4.983 Mean : 4.143
## 3rd Qu.: 8.000 3rd Qu.: 8.000
## Max. :88.000 Max. :10.000
## NA's :1505 NA's :1526A: By looking at the summary, we can see that the median and mean value is higher is Portugal than in Serbia. This indicates that more people in Serbia lean towards planning for the future than people in Portugal. Additionally, no Portuguese participant refused to answer.
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
trial_gndr <- italy_data %>%
filter(cntry == "IT") %>%
select(gndr)
trial_gndr$y <- trial_gndr$gndr
table(trial_gndr$y)##
## 1 2 9
## 4836 5329 13italy_data <- italy_data %>%
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(77,88,99), NA, stfdem),
gndr = ifelse(gndr %in% c(9), NA, gndr)
)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 4.66
## 2 Male 4.78
## 3 <NA> 3.25A: According to the output, females are more dissatisfied. A lower number indicates greater dissatisfaction and the mean stfdem value is lower in females than in males.
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.
A: The median French person is more satisfied with the health system than the education system. The box plot shows that the median line is lower for the education system than the median line for the health system. Since a score of 10 represents extreme satisfaction, a higher median indicates greater satisfaction.
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()`).