wong_choiyingdebbie_Homework_1
2024-01-18
packages <- c("tidyverse", "gapminder", "fst") # 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
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## ✔ 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()
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## ℹ Use the conflicted package (<http://conflicted.r-lib.org/>) to force all conflicts to become errors## [[1]]
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## [13] "stats" "graphics" "grDevices" "utils" "datasets" "methods"
## [19] "base"getwd()## [1] "/Users/debbie/Downloads/SOC222: Rstudio"rm(list=ls()); gc()## used (Mb) gc trigger (Mb) limit (Mb) max used (Mb)
## Ncells 1026635 54.9 2119441 113.2 NA 1292894 69.1
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## Ncells 1026612 54.9 2119441 113.2 NA 1292894 69.1
## Vcells 1774344 13.6 8388608 64.0 16384 2429572 18.6ess <- read_fst("All-ESS-Data.fst")france_data <- read.fst("france_data.fst")R Markdown
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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.
Norway:
Norway_happy <- ess %>% # note: if you work from belgium_data replace "ess" with belgium_data
filter(cntry == "NO") %>%
select(happy)Norway_happy$y <- Norway_happy$happy
table(Norway_happy$y)##
## 0 1 2 3 4 5 6 7 8 9 10 77 88
## 15 29 59 163 238 730 817 2617 5235 3796 2344 12 10# need to remove 77, 88, 99 or else will alter results. See data portal for what they represent (e.g. DK, Refusal, etc.)
# Recode values 77 through 99 to NA
Norway_happy$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 2344Calculating the mean (average)
mean_y <- mean(Norway_happy$y, na.rm = TRUE)
cat("Mean of 'y' is:", mean_y, "\n")## Mean of 'y' is: 7.975005Belgium:
belgium_happy <- ess %>%
filter(cntry == "BE") %>%
select(happy)belgium_happy$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# 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$y)##
## 0 1 2 3 4 5 6 7 8 9 10
## 50 27 104 194 234 830 999 3503 6521 3402 1565Calculating the mean (average)
mean_y <- mean(belgium_happy$y, na.rm = TRUE)
cat("Mean of 'y' is:", mean_y, "\n")## Mean of 'y' is: 7.737334Answer to Task 1: Norway reports a higher level 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_alcbnge <- ess %>%
filter(cntry == "EE") %>%
select(wrclmch)
ireland_alcbnge$y <- ireland_alcbnge$wrclmch
table(ireland_alcbnge$y)##
## 1 2 3 4 5 6 7 8
## 343 852 1612 565 125 56 1 7# Recode values 6 through 8 to NA
ireland_alcbnge$y[ireland_alcbnge$y %in% 6:8] <- NAMean and Median and Interpret
# Compute mean and median
mean_y <- mean(ireland_alcbnge$y, na.rm = TRUE)
median_y <- median(ireland_alcbnge$y, na.rm = TRUE)
cat("Mean of 'y':", mean_y, "\n")## Mean of 'y': 2.793251cat("Median of 'y':", median_y, "\n")## Median of 'y': 3Converting numeric values to categories
# Converting to categories to get mode as a category instead of a number
df <- ireland_alcbnge %>%
mutate(
y_category = case_when(
y == 1 ~ "Daily or almost daily",
y == 2 ~ "Weekly",
y == 3 ~ "Monthly",
y == 4 ~ "Less than monthly",
y == 5 ~ "Never",
TRUE ~ NA_character_
),
y_category = fct_relevel(factor(y_category), ### here you would put the categories in order you want them to appear or else it will appear alphabetically
"Daily or almost daily",
"Weekly",
"Monthly",
"Less than monthly",
"Never")
)
# To confirm the conversion:
table(df$y_category)##
## Daily or almost daily Weekly Monthly
## 343 852 1612
## Less than monthly Never
## 565 125# Let's determine the mode of our newly created category:
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: MonthlyAnswer to Task 2: “Monthly” is the most common category selected, for Irish respondents, for frequency of binge drinking.
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.
Comparing means with groupby
result <- ess %>%
# Step 1: Filter for the countries of interest
filter(cntry %in% c("PT", "SE")) %>%
# Step 2: Recode 77, 88, 99 to NA for 'plnftr'
mutate(plnftr = recode(plnftr, `77` = NA_real_, `88` = NA_real_, `99` = NA_real_)) %>%
# Step 3: Compute mean by country
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 5.02Answer to Task 3: Portugal stands out as higher with a mean of 5.418289(i.e.Closer to taking each day as it comes) than Serbia does with a mean of 5.018229 (i.e Closer to planning for future as much as possible)
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
Average of Y by X
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 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 average of lrscale by gender
means_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: 3 × 2
## gndr stfdem
## <chr> <dbl>
## 1 9 3.25
## 2 Female 4.69
## 3 Male 4.78Answer to Task 4: In this case, Women are slightly more dissatidfied with democracy in Italy than Men do.
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 %>%
# 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()`).
Answer to Task 5: The median French person is more satisfied with the
health services as the boxplot graph indicates, we could tell that the
position of the box and median for health services is generally higher
than that for education.This suggests that the middle 50% of the
satisfaction scores for health services is higher than the education
system.