Task1 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.
# 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"
ess <- read_fst("All-ESS-Data .fst")
unique(ess$cntry)
## [1] "AT" "BE" "CH" "CZ" "DE" "DK" "ES" "FI" "FR" "GB" "GR" "HU" "IE" "IL" "IT"
## [16] "LU" "NL" "NO" "PL" "PT" "SE" "SI" "EE" "IS" "SK" "TR" "UA" "BG" "CY" "RU"
## [31] "HR" "LV" "RO" "LT" "AL" "XK" "ME" "RS" "MK"
Task1 Provide code and answer.
Prompt and question: calculate the average for the variable ‘happy’ for the country of Norway. On average, based on the ESS data, who reports higher levels of happiness: Norway or Belgium?
Note: we already did it for Belgium. You just need to compare to Norway’s average, making sure to provide the code for both.
belgium_happy <- ess %>%
filter(cntry == "BE") %>%
select(happy)
belgium_happy$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$y)
##
## 0 1 2 3 4 5 6 7 8 9 10
## 15 29 59 163 238 730 817 2617 5235 3796 2344
norway_happy %>%
summarize(
mean_y = mean(y, na.rm = TRUE),
median_y = median(y, na.rm = TRUE)
) %>%
print()
## mean_y median_y
## 1 7.975005 8
##Answer: Norway displays thier happiness as“7.97”, “0.24” highier than Belgiums “7.73”. further showcasing that, Norway experiences higher levels of happiness than Belgium.
Task 2 Provide code and answer.
Prompt and question: what is the most common category selected, for Irish respondents, for frequency of binge drinking? The variable of interest is: alcbnge.
More info here: https://ess-search.nsd.no/en/variable/0c65116e-7481-4ca6-b1d9-f237db99a694.
Hint: need to convert numeric value entries to categories as specified in the variable information link. We did similar steps for Estonia and the climate change attitude variable.
ireland_ccdrinking <- ess %>%
filter(cntry == "IL") %>%
select(alcbnge)
ireland_ccdrinking$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 survey participants commonly identify their binge drinking habits with the descriptor “Extreme Binge Drinking” as the prevailing category for their consumption patterns
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: Portugal exhibits a subtle inclination toward an “planning for the future” outlook, slightly surpassing Serbia with average scores of 5.41 and 5.01, respectively.
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), 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 analysis of the data, it is evident that in Italy, male respondents express a higher level of contentment with democracy compared to their female counterparts, as indicated by their superior average score of 4.78.
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 %>%
# 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()`).