Kashvi Shah | 1009413706
# 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 ──
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## ✔ 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()
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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"rm(list=ls()); gc() ## used (Mb) gc trigger (Mb) limit (Mb) max used (Mb)
## Ncells 1208565 64.6 2196208 117.3 NA 2196208 117.3
## Vcells 2060295 15.8 8388608 64.0 16384 2882391 22.0ess <- read_fst("All-ESS-Data.fst")unique(ess$cntry)## [1] "AT" "BE" "CH" "CZ" "DE" "DK" "ES" "FI" "FR" "GB" "GR" "HU" "IE" "IL" "IT"
## [16] "LU" "NL" "NO" "PL" "PT" "SE" "SI" "EE" "IS" "SK" "TR" "UA" "BG" "CY" "RU"
## [31] "HR" "LV" "RO" "LT" "AL" "XK" "ME" "RS" "MK"Task 1: Question and Answer Question: Provide code and answer.
Prompt and question: calculate the average for the variable ‘happy’ for the country of Norway. On average, based on the ESS data, who reports higher levels of happiness: Norway or Belgium?
Note: we already did it for Belgium. You just need to compare to Norway’s average, making sure to provide the code for both.
belgium_happy <- ess %>% # note: if you work from belgium_data replace "ess" with belgium_data
filter(cntry == "BE") %>%
select(happy)belgium_happy$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("Mean of 'y' is:", mean_belgium, "\n")## Mean of 'y' is: 7.737334#Norway Data
norway_happy <- ess %>%
filter(cntry == "NO") %>%
select(happy)norway_happy$y <- norway_happy$happy
table(norway_happy$y)##
## 0 1 2 3 4 5 6 7 8 9 10 77 88
## 15 29 59 163 238 730 817 2617 5235 3796 2344 12 10norway_happy$happy[norway_happy$happy %in% c(77, 88, 99)] <- NA
table(norway_happy$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 10mean_norway <- mean(norway_happy$happy, na.rm = TRUE)
cat("Mean of 'y' is:", mean_norway, "\n")## Mean of 'y' is: 7.975005mean_belgium## [1] 7.737334mean_norway## [1] 7.975005#In comparison to Belgium (7.73), Norway (7.97) reports “0.24” higher levels of happiness on average.
Task 2 Provide code and answer.
Question Prompt and question: what is the most common category selected, for Irish respondents, for frequency of binge drinking? The variable of interest is: alcbnge.
ireland_binge <- ess %>%
filter(cntry == "IE") %>%
select(alcbnge)table(ireland_binge$alcbnge)##
## 1 2 3 4 5 6 7 8
## 65 650 346 417 239 641 26 6ireland_binge$alcbnge[ireland_binge$alcbnge %in% 6:8] <- NAtable(ireland_binge$alcbnge)##
## 1 2 3 4 5
## 65 650 346 417 239ireland_binge$alcbnge_category <- factor(ireland_binge$alcbnge, labels = c("Daily or almost daily", "Weekly", "Monthly", "Less than monthly", "Never"))
table(ireland_binge$alcbnge_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_ireland_binge <- get_mode(ireland_binge$alcbnge_category)
mode_ireland_binge## [1] "Weekly"mode2 <- ireland_binge%>%
mutate(
y_category = case_when(
alcbnge == 1 ~ "Daily or almost daily",
alcbnge == 2 ~ "Weekly",
alcbnge == 3 ~ "Monthly",
alcbnge == 4 ~ "Less than monthly",
alcbnge == 5 ~ "Never",
TRUE ~ NA_character_
),
y_category = fct_relevel(factor(y_category),
"Daily or almost daily",
"Weekly",
"Monthly",
"Less than monthly",
"Never")
)
table(mode2$y_category)##
## Daily or almost daily Weekly Monthly
## 65 650 346
## Less than monthly Never
## 417 239mode_ireland_binge2 <- get_mode(mode2$y_category)
mode_ireland_binge2## [1] "Weekly"Task 3 Question Provide code and answer.
Prompt and question: when you use the summary() function for the variable plnftr (about planning for future or taking every each day as it comes from 0-10) for both the countries of Portugal and Serbia, what do you notice? What stands out as different when you compare the two countries (note: look up the variable information on the ESS website to help with interpretation)? Explain while referring to the output generated.
# For Portugal
portugal_plnftr <- ess %>%
filter(cntry == "PT") %>%
select(plnftr)
# For Serbia
serbia_plnftr <- ess %>%
filter(cntry == "RS") %>%
select(plnftr)summary(portugal_plnftr)## plnftr
## Min. : 0.000
## 1st Qu.: 3.000
## Median : 5.000
## Mean : 6.426
## 3rd Qu.: 8.000
## Max. :88.000
## NA's :14604summary(serbia_plnftr)## plnftr
## Min. : 0.000
## 1st Qu.: 0.000
## Median : 4.000
## Mean : 4.983
## 3rd Qu.: 8.000
## Max. :88.000
## NA's :1505table(portugal_plnftr$plnftr)##
## 0 1 2 3 4 5 6 7 8 9 10 88
## 114 184 313 356 264 481 262 382 345 166 370 40portugal_plnftr$plnftr[portugal_plnftr$plnftr %in% 88] <- NAtable(serbia_plnftr$plnftr)##
## 0 1 2 3 4 5 6 7 8 9 10 77 88
## 587 133 152 138 95 246 70 87 103 47 364 4 17serbia_plnftr$plnftr[serbia_plnftr$plnftr %in% 77:88] <- NAsummary(portugal_plnftr)## plnftr
## Min. : 0.000
## 1st Qu.: 3.000
## Median : 5.000
## Mean : 5.418
## 3rd Qu.: 8.000
## Max. :10.000
## NA's :14644summary(serbia_plnftr)## plnftr
## Min. : 0.000
## 1st Qu.: 0.000
## Median : 4.000
## Mean : 4.143
## 3rd Qu.: 8.000
## Max. :10.000
## NA's :1526ggplot(serbia_plnftr, aes(x = plnftr)) +
geom_histogram(bins = 10, fill = "purple", color = "hotpink") +
theme_minimal() +
labs(title = "Distribution for Serbia",
x = "Scale",
y = "Count")## Warning: Removed 1526 rows containing non-finite values (`stat_bin()`).
ggplot(portugal_plnftr, aes(x = plnftr)) +
geom_histogram(bins = 10, fill = "lightblue", color = "darkgreen") +
theme_minimal() +
labs(title = "Distribution for Portugal",
x = "Scale",
y = "Count")## Warning: Removed 14644 rows containing non-finite values (`stat_bin()`).
Task 4 Question Provide code and answer.
Prompt and question: using the variables stfdem and gndr, answer the following: on average, who is more dissastified with democracy in Italy, men or women? Explain while referring to the output generated.
italy_data <- ess %>%
filter(cntry == "IT")
italy_data$gndr[italy_data$gndr %in% 9] <- NA
italy_data$stfdem[italy_data$stfdem %in% 77:99] <- NA
italy_data <- italy_data %>%
mutate(
gndr = case_when(
gndr == 1 ~ "Male",
gndr == 2 ~ "Female",
TRUE ~ as.character(gndr)
)
)
italy_data <- italy_data %>% filter(!is.na(gndr))
italy_data <- italy_data %>% filter(!is.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) %>% # here you are "grouping by" your second variable
summarize(stfdem = mean(stfdem, na.rm = TRUE)) # here you are summarizing your variable of interest
print(means_by_gender)## # A tibble: 2 × 2
## gndr stfdem
## <chr> <dbl>
## 1 Female 4.66
## 2 Male 4.78#Females are more dissatisfied with democracy in Italy
Task 5 Question 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).
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()`).
#The median line within the boxplot of satisfaction with health services
is higher, therefore satisfaction with healthcare is higher in
France.