Kaur_Gurjyot_Homework_1
2024-1-23
Set up:
# 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"ess <- read_fst("All-ESS-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:
belgium_happy <- ess %>%
filter(cntry == "BE") %>%
select(happy)
belgium_happy$y <- belgium_happy$happy
belgium_happy$y[belgium_happy$y %in% 77:99] <- NA
table(belgium_happy$y)##
## 0 1 2 3 4 5 6 7 8 9 10
## 50 27 104 194 234 830 999 3503 6521 3402 1565mean_y <- mean(belgium_happy$y, na.rm = TRUE)
cat("Mean of 'y' is:", mean_y, "\n")## Mean of 'y' is: 7.737334Norway:
norway_happy <- ess %>%
filter(cntry == "NO") %>%
select(happy)
norway_happy$y <- norway_happy$happy
norway_happy$y[norway_happy$y %in% 77:99] <- NA
table(norway_happy$y)##
## 0 1 2 3 4 5 6 7 8 9 10
## 15 29 59 163 238 730 817 2617 5235 3796 2344mean_y <- mean(norway_happy$y, na.rm = TRUE)
cat("Mean of 'y' is:", mean_y, "\n")## Mean of 'y' is: 7.975005Therefore, on average, based on the ESS data, Norway reports higher levels of happiness as the mean of happiness in Norway is 7.975005, which is greater than Belgium’s mean of 7.737334.
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_bingedrink <- ess %>%
filter(cntry == "IE") %>%
select(alcbnge)
ireland_bingedrink$y <- ireland_bingedrink$alcbnge
table(ireland_bingedrink$y)##
## 1 2 3 4 5 6 7 8
## 65 650 346 417 239 641 26 6df <- ireland_bingedrink %>%
mutate(
alcbnge = case_when(
y == 1 ~ "Daily or almost daily",
y == 2 ~ "Weekly",
y == 3 ~ "Monthly",
y == 4 ~ "Less than monthly",
y == 5 ~ "Never",
TRUE ~ NA_character_
),
alcbnge = fct_relevel(factor(alcbnge),
"Daily or almost daily",
"Weekly",
"Monthly",
"Less than monthly",
"Never")
)
# To confirm the conversion:
table(df$alcbnge)##
## Daily or almost daily Weekly Monthly
## 65 650 346
## Less than monthly Never
## 417 239This code when operated shows that the most common category selected by Irish respondents in terms of the frequency of binge drinking is “weekly” as its value is 650, greater than the rest.
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.
portugal_plnftr <- ess %>%
filter(cntry == "PT") %>%
select(plnftr)
summary(portugal_plnftr)## plnftr
## Min. : 0.000
## 1st Qu.: 3.000
## Median : 5.000
## Mean : 6.426
## 3rd Qu.: 8.000
## Max. :88.000
## NA's :14604serbia_plnftr <- ess %>%
filter(cntry == "RS") %>%
select(plnftr)
summary(serbia_plnftr)## plnftr
## Min. : 0.000
## 1st Qu.: 0.000
## Median : 4.000
## Mean : 4.983
## 3rd Qu.: 8.000
## Max. :88.000
## NA's :1505Interpretation: There are a couple of factors that stand out as different when comparing data on the variable plnftr for Portugal and Serbia. The first is that the mean of Portugal is higher than the mean of Serbia. Looking at plnftr’s information on the ESS website, individuals can choose from 0 to 10 to express if they plan for their futures or just take each day as it comes; 0 as in ‘I plan for my future as much as possible’ and 10 as in ‘I just take each day as it comes.’ This indicates that Portugal’s mean of 6.426 represents that their population is, on average, more likely to take each day as it comes compared to Serbia’s population whose mean is 4.983. Secondly, the median of Portugal is also higher than the median of Serbia and each country’s respective medians are close to their means, demonstrating the data is not skewed and, thus, has a more or less of an even distribution. Lastly, the 1st quartile of Portugal is greater than Serbia’s, which means Portugal has more data points that are below the 25th percentile compared to Serbia.
Task 4
Provide code and answer.
Prompt and question: using the variables stfdem and gndr, answer the following: on average, who is more dissatisfied 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_dissatdem <- ess %>%
filter(cntry == "IT")
# Convert gender and stfdem
italy_data <- italy_dissatdem %>%
mutate(
gndr = case_when(
gndr == 1 ~ "Male",
gndr == 2 ~ "Female",
TRUE ~ as.character(gndr)
),
stfdem = ifelse(stfdem %in% c(77, 88), 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 4.782646# Compute mean for male
mean_female_stfdem <- italy_data %>%
filter(gndr == "Female") %>%
summarize(mean_stfdem_female = mean(stfdem, na.rm = TRUE))
print(mean_female_stfdem)## mean_stfdem_female
## 1 4.694652Therefore, on average, men are more dissatisfied with democracy in Italy because the output of the male’s mean is 4.782646, which is greater than the female’s mean of 4.694652.
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 <- ess %>%
filter(cntry == "FR")
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()`).
Therefore, according to ESS data, the median French person is more satisfied with the health services because the average of stfhlth (health services), which represents the middle line inside the box, is sitting greater than 6.25, indicating it is greater than stfedu’s (education system) median of 5.0.