Nagpal_Angela_Makeup
2024-02-22
R Markdown
This is an R Markdown document. Markdown is a simple formatting syntax for authoring HTML, PDF, and MS Word documents. For more details on using R Markdown see http://rmarkdown.rstudio.com.
When you click the Knit button a document will be generated that includes both content as well as the output of any embedded R code chunks within the document. You can embed an R code chunk like this:
summary(cars)## speed dist
## Min. : 4.0 Min. : 2.00
## 1st Qu.:12.0 1st Qu.: 26.00
## Median :15.0 Median : 36.00
## Mean :15.4 Mean : 42.98
## 3rd Qu.:19.0 3rd Qu.: 56.00
## Max. :25.0 Max. :120.00Including Plots
You can also embed plots, for example:
Note that the echo = FALSE parameter was added to the
code chunk to prevent printing of the R code that generated the
plot.
# 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"rm(list=ls()); gc() ## used (Mb) gc trigger (Mb) limit (Mb) max used (Mb)
## Ncells 1223726 65.4 2169680 115.9 NA 2169680 115.9
## Vcells 2095014 16.0 8388608 64.0 16384 3079498 23.5ess <- 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_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$y)##
## 0 1 2 3 4 5 6 7 8 9 10 77 88 99
## 50 27 104 194 234 830 999 3503 6521 3402 1565 3 16 3# need to remove 77, 88, 99 or else will alter results. See data portal for what they represent (e.g. DK, Refusal, etc.)
# Recode values 77 through 99 to NA
belgium_happy$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_belgium <- mean(belgium_happy$y, na.rm = TRUE)
cat("Mean of 'y' is:", mean_belgium, "\n")## Mean of 'y' is: 7.737334norway_data <- read.fst("norway_data.fst")norway_happy <- norway_data %>%
select(happy)
# Recode invalid values to NA
norway_happy$happy[norway_happy$happy %in% c(77, 88, 99)] <- NA
# Calculate mean
mean_norway <- mean(norway_happy$happy, na.rm = TRUE)
cat("Mean of 'y' is:", mean_norway, "\n")## Mean of 'y' is: 7.975005Belgium = 7.74, Norway = 7.98, thur norway reports slightly higher happiness
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_data <- read.fst("ireland_data.fst")ireland_binge <- ireland_data %>%
select(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"Most common category selected for frequency of binge drinking is weekly.
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.
# 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 :15261st quartile and mean stand out the most
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.
italy_data <- read.fst("italy_data.fst")table(italy_data$gndr)##
## 1 2 9
## 4836 5329 13italy_data$gndr[italy_data$gndr %in% 9] <- NAtable(italy_data$stfdem)##
## 0 1 2 3 4 5 6 7 8 9 10 77 88 99
## 712 365 733 962 1221 1785 1724 1300 733 138 102 57 344 2italy_data$stfdem[italy_data$stfdem %in% 77:99] <- NAitaly_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))table(italy_data$gndr)##
## Female Male
## 5084 4679table(italy_data$stfdem)##
## 0 1 2 3 4 5 6 7 8 9 10
## 710 365 731 958 1220 1785 1722 1299 733 138 102italy_data %>%
group_by(gndr) %>%
summarize(meanstfdem = mean(stfdem, na.rm = TRUE))## # A tibble: 2 × 2
## gndr meanstfdem
## <chr> <dbl>
## 1 Female 4.66
## 2 Male 4.78women are more dissatidfied, although very little difference
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 <- 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 = "Y-axis", x = "X-axis", title = "Boxplot of stfedu vs. stfhlth") +
theme_minimal()## Warning: Removed 364 rows containing non-finite values (`stat_boxplot()`).
median(france_data$stfhlth)## [1] 7median(france_data$stfedu)## [1] 5france_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)) %>%
ggplot(aes(x = variable, y = value)) +
geom_boxplot() +
labs(title = "Boxplot of satisfaction with the state of education vs. health services",
x = "", # Removing x-axis label
y = "Satisfaction (0-10)") +
theme_minimal()## Warning: Removed 364 rows containing non-finite values (`stat_boxplot()`).