Markiewicz_Alexandra_Homework_1
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 2.1.4
## ✔ 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
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## ✖ dplyr::filter() masks stats::filter()
## ✖ dplyr::lag() masks stats::lag()
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## [1] "lubridate" "forcats" "stringr" "dplyr" "purrr" "readr"
## [7] "tidyr" "tibble" "ggplot2" "tidyverse" "stats" "graphics"
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## [1] "fst" "lubridate" "forcats" "stringr" "dplyr" "purrr"
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## [13] "graphics" "grDevices" "utils" "datasets" "methods" "base"
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## [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")#estonia_data <- read.fst("estonia_data.fst")#france_data <- read.fst("france_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")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?
norway_happy <- ess %>%
filter(cntry == "NO") %>%
select(happy)norway_happy$y <- norway_happy$happy
table(norway_happy$y)##
## 0 1 2 3 4 5 6 7 8 9 10 77 88
## 15 29 59 163 238 730 817 2617 5235 3796 2344 12 10# need to remove 77, 88, 99 or else will alter results.
# Recode values 77 through 99 to NA
norway_happy$y[norway_happy$y %in% 77:99] <- NA
# checking again
table(norway_happy$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$y, na.rm = TRUE)
cat("Mean of 'y' is:", mean_y, "\n")## Mean of 'y' is: 7.975005belgium_happy <- ess %>%
filter(cntry == "BE") %>%
select(happy)belgium_happy$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# need to remove 77, 88, 99 or else will alter results.
# Recode values 77 through 99 to NA
belgium_happy$y[belgium_happy$y %in% 77:99] <- NA
# checking again
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: 7.737334Based on the ESS data, Norway reports higher happiness levels with an average score of 7.975005, whereas Belgium’s average is slightly lower at 7.737334.
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(wrclmch)
ireland_alcbnge$y <- ireland_alcbnge$wrclmch
table(ireland_alcbnge$y)##
## 1 2 3 4 5 6 7 8
## 386 938 2080 820 217 51 3 32# Recode values 6 through 8 to NA
ireland_alcbnge$y[ireland_alcbnge$y %in% 6:8] <- NA# Converting to categories to get mode as a category instead of a number
df <- ireland_alcbnge %>%
mutate(
y_category = case_when(
y == 1 ~ "Daily or almost daily",
y == 2 ~ "Weekly",
y == 3 ~ "Monthly",
y == 4 ~ "Less than monthly",
y == 5 ~ "Never",
TRUE ~ NA_character_
),
y_category = fct_relevel(factor(y_category),
"Daily or almost daily",
"Weekly",
"Monthly",
"Less than monthly",
"Never")
)
# To confirm the conversion:
table(df$y_category)##
## Daily or almost daily Weekly Monthly
## 386 938 2080
## Less than monthly Never
## 820 217The most common category selected by Irish respondents in terms of frequency of binge drinking is the “Monthly” option, with 2080 responses.
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.
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 :1505When comparing the two countries, Portugal tends to have higher average values for the variable “plnftr,” suggesting that its respondents have a somewhat larger inclination to plan for the future compared to Serbia. Portugal’s values for the 1st quartile (3.000), median (5.000), and mean (6.426) are higher compared to Serbia’s values of 0.000, 4.000, and 4.983, respectively. In addition, Portugal has a larger number of missing values at 14604, whereas Serbia only has 1505.
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.
italy_data <- ess %>%
filter(cntry == "IT")
# Convert gender and stfdem
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) # Convert stfdem values
)# Compute average of stfdem by gender
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 4.78On average, men are more dissatisfied with democracy in Italy, with a mean score of 4.782646, while women have a slightly lower mean score of 4.694652.
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.
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 %>%
# 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()`).
According to ESS data, the median French person is more satisfied with health services than the education system. This can be seen on the boxplot graph, where the median is represented by the line inside both boxes. The median for the education system is 5.0, whereas for health services, it is approximately 7.3.