Homework 1
2024-01-17
First, set the environment.
# 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"
##
## [[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"Then, load the data.
ess <- read_fst("All-ESS-Data.fst")Subsetting
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.
##Creating a table of all unique entries.
The countries in the ESS dataset are stored in the “cntry” column (for country) under their two letter country codes, so the code should be:
unique(ess$cntry)## [1] "AT" "BE" "CH" "CZ" "DE" "DK" "ES" "FI" "FR" "GB" "GR" "HU" "IE" "IL" "IT"
## [16] "LU" "NL" "NO" "PL" "PT" "SE" "SI" "EE" "IS" "SK" "TR" "UA" "BG" "CY" "RU"
## [31] "HR" "LV" "RO" "LT" "AL" "XK" "ME" "RS" "MK"BE stands for Belgium. And NO stands for Norway.
Filter to a specific value.
Then,we need filter to a specific value (Belgium) within the cntry column and “select” our variable of interest –> happy.
belgium_happy <- ess %>%
filter(cntry == "BE") %>%
select(happy)belgium_happy$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$y)##
## 0 1 2 3 4 5 6 7 8 9 10
## 50 27 104 194 234 830 999 3503 6521 3402 1565Cauculating Mean (Average) for the variable ‘happy’ for the country of Belgium
mean_y <- mean(belgium_happy$y, na.rm = TRUE)
cat("Mean of 'y' is:", mean_y, "\n")## Mean of 'y' is: 7.737334Filter to a specific value for Norway.
norway_happy <- ess %>%
filter(cntry == "NO") %>%
select(happy)norway_happy$z <- norway_happy$happy
table(norway_happy$z)##
## 0 1 2 3 4 5 6 7 8 9 10 77 88
## 15 29 59 163 238 730 817 2617 5235 3796 2344 12 10norway_happy$z[norway_happy$z %in% 77:99] <- NA
table(norway_happy$z)##
## 0 1 2 3 4 5 6 7 8 9 10
## 15 29 59 163 238 730 817 2617 5235 3796 2344Cauculating Mean (Average) for the variable ‘happy’ for the country of Norway
mean_z <- mean(norway_happy$z, na.rm = TRUE)
cat("Mean of 'z' is:", mean_y, "\n")## Mean of 'z' is: 7.737334Answer:
On average, based on the ESS data, Norway reports higher levels of happiness than Belgium.Because mena of ‘z’(Norway) is higher than mean of “y”(Belgium).
Task 2
Provide code and answer.
Prompt and question: what is the most common category selected, for Irish respondents, for frequency of binge drinking? The variable of interest is: alcbnge.
More info here: https://ess-search.nsd.no/en/variable/0c65116e-7481-4ca6-b1d9-f237db99a694.
Hint: need to convert numeric value entries to categories as specified in the variable information link. We did similar steps for Estonia and the climate change attitude variable.
Filter to a specific value.
irland_alcbnge <- ess %>%
filter(cntry == "IL") %>%
select(alcbnge)
irland_alcbnge$y <- irland_alcbnge$alcbnge
irland_alcbnge$y[irland_alcbnge$y %in% 6:9] <- NA
table(irland_alcbnge$y)##
## 1 2 3 4 5
## 15 153 149 264 362!! I don’t know why it is different from the website… Really confusing and I think on the website the 2 should be the biggest number…
df <- irland_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$y_category)##
## Daily or almost daily Weekly Monthly
## 15 153 149
## Less than monthly Never
## 264 362Determine the mode of our created category:
get_mode <- function(v) {
tbl <- table(v)
mode_vals <- as.character(names(tbl)[tbl == max(tbl)])
return(mode_vals)
}
mode_values <- get_mode(df$y_category)
cat("Mode of y category:", paste(mode_values, collapse = ", "), "\n")## Mode of y category: Never!!I think the mode should be WEEKLY due to the website… But due to the code…
Answer
The most common category selected, for frequency of binge drinking is 5 (Never)
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 5.02Answer
I notice that Portugal is on the top of the list and Serbia is on the bottom of the list (ordered by mean). But their mean are close to each other.
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")
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 )
)Compute mean for male
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 4.782646means_by_gender <- italy_data %>%
group_by(gndr) %>%
summarize(stfdem = mean(stfdem, na.rm = TRUE)) italy_data$gndr[italy_data$gndr %in% 9] <- NA
italy_data <- italy_data %>% filter(!is.na(gndr))print(means_by_gender)## # A tibble: 3 × 2
## gndr stfdem
## <chr> <dbl>
## 1 9 3.25
## 2 Female 4.66
## 3 Male 4.78Answer
On average men is more dissastified with democracy in Italy
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.
#france_data <- read.fst("france_data.fst")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(x = variable, y = value)) +
geom_boxplot() +
labs(y = "Satisfaction (0-10)", x = " ", title = "oxplot of satisfaction with the state of education vs. health services") +
theme_minimal()## Warning: Removed 364 rows containing non-finite values (`stat_boxplot()`).
Answer
According to ESS data, the median French person is more satisfied with the health services. Because the median of stfhlth is higher than stfedu. And the boxplot graph has been changed.