Amin_Mustafa_Homework1
2024-02-10
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 2.1.5
## ✔ forcats 1.0.0 ✔ stringr 1.5.1
## ✔ ggplot2 3.4.4 ✔ tibble 3.2.1
## ✔ lubridate 1.9.3 ✔ tidyr 1.3.1
## ✔ 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"
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## [[2]]
## [1] "fst" "lubridate" "forcats" "stringr" "dplyr" "purrr"
## [7] "readr" "tidyr" "tibble" "ggplot2" "tidyverse" "stats"
## [13] "graphics" "grDevices" "utils" "datasets" "methods" "base"
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## [[3]]
## [1] "modelsummary" "fst" "lubridate" "forcats" "stringr"
## [6] "dplyr" "purrr" "readr" "tidyr" "tibble"
## [11] "ggplot2" "tidyverse" "stats" "graphics" "grDevices"
## [16] "utils" "datasets" "methods" "base"getwd()## [1] "C:/Users/Mustafa/Documents/SOC222"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.
unique(ess$cntry)## [1] "AT" "BE" "CH" "CZ" "DE" "DK" "ES" "FI" "FR" "GB" "GR" "HU" "IE" "IL" "IT"
## [16] "LU" "NL" "NO" "PL" "PT" "SE" "SI" "EE" "IS" "SK" "TR" "UA" "BG" "CY" "RU"
## [31] "HR" "LV" "RO" "LT" "AL" "XK" "ME" "RS" "MK"belgium_happy <- ess %>% # note: if you work from belgium_data replace "ess" with belgium_data
filter(cntry == "BE") %>%
select(happy)
belgium_happy$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
# 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: 7.737334norway_happy <- ess %>%
filter(cntry == "NO") %>%
select(happy)
norway_happy$y <- norway_happy$happy
table(norway_happy$y)##
## 0 1 2 3 4 5 6 7 8 9 10 77 88
## 15 29 59 163 238 730 817 2617 5235 3796 2344 12 10norway_happy$y[norway_happy$y %in% 77:88] <- NA
# checking again
table(norway_happy$y)##
## 0 1 2 3 4 5 6 7 8 9 10
## 15 29 59 163 238 730 817 2617 5235 3796 2344meanNO_Y <- mean(norway_happy$y, na.rm = TRUE)
cat("Mean of 'meanNO_Y' is:", meanNO_Y, "\n")## Mean of 'meanNO_Y' is: 7.975005The data indicates Norway reports higher rates of happiness
Task 2
Provide code and answer.
Prompt and question: what is the most common category selected, for Irish respondents, for frequency of binge drinking? The variable of interest is: alcbnge.
More info here: https://ess-search.nsd.no/en/variable/0c65116e-7481-4ca6-b1d9-f237db99a694.
Hint: need to convert numeric value entries to categories as specified in the variable information link. We did similar steps for Estonia and the climate change attitude variable.
ireland_alcbnge <- ess %>% # note: if you work from belgium_data replace "ess" with belgium_data
filter(cntry == "IE") %>%
select(alcbnge)
ireland_alcbnge$y <- ireland_alcbnge$alcbnge
table(ireland_alcbnge$y)##
## 1 2 3 4 5 6 7 8
## 65 650 346 417 239 641 26 6ireland_alcbnge$y[ireland_alcbnge$y %in% 6:8] <- NA
# checking again
table(ireland_alcbnge$y)##
## 1 2 3 4 5
## 65 650 346 417 239# Converting to categories to get mode as a category instead of a number
df <- ireland_alcbnge %>%
mutate(
y_category = case_when(
y == 1 ~ "Daily or almost daily",
y == 2 ~ "Weekly",
y == 3 ~ "Monthly",
y == 4 ~ "Less than monthly",
y == 5 ~ "Never",
TRUE ~ NA_character_
),
y_category = fct_relevel(factor(y_category), ### here you would put the categories in order you want them to appear or else it will appear alphabetically
"Daily or almost daily",
"Weekly",
"Monthly",
"Less than monthly",
"Never")
)
# To confirm the conversion:
table(df$y_category)##
## Daily or almost daily Weekly Monthly
## 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_values <- get_mode(df$y_category)
cat("Mode of y category:", paste(mode_values, collapse = ", "), "\n")## Mode of y category: WeeklyThe most common category was weekly drinking
Task 3
Provide code and answer.
Prompt and question: when you use the summary() function for the variable plnftr (about planning for future or taking every each day as it comes from 0-10) for both the countries of Portugal and Serbia, what do you notice? What stands out as different when you compare the two countries (note: look up the variable information on the ESS website to help with interpretation)? Explain while referring to the output generated.
portugal_plnftr <- ess %>%
filter(cntry == "PT") %>%
select(happy)
portugal_plnftr$y <- portugal_plnftr$happy
table(portugal_plnftr$y)##
## 0 1 2 3 4 5 6 7 8 9 10 77 88 99
## 139 127 349 647 805 2589 2687 3664 3835 1610 1328 26 70 5portugal_plnftr$y[portugal_plnftr$y %in% 77:99] <- NA
# checking again
table(portugal_plnftr$y)##
## 0 1 2 3 4 5 6 7 8 9 10
## 139 127 349 647 805 2589 2687 3664 3835 1610 1328summary(portugal_plnftr)## happy y
## Min. : 0.000 Min. : 0.000
## 1st Qu.: 5.000 1st Qu.: 5.000
## Median : 7.000 Median : 7.000
## Mean : 7.148 Mean : 6.701
## 3rd Qu.: 8.000 3rd Qu.: 8.000
## Max. :99.000 Max. :10.000
## NA's :101serbia_plnftr <- ess %>%
filter(cntry == "RS") %>%
select(plnftr)
serbia_plnftr$y <- serbia_plnftr$plnftr
table(serbia_plnftr$y)##
## 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$y[serbia_plnftr$y %in% 77:99] <- NA
# checking again
table(serbia_plnftr$y)##
## 0 1 2 3 4 5 6 7 8 9 10
## 587 133 152 138 95 246 70 87 103 47 364summary(serbia_plnftr)## plnftr y
## Min. : 0.000 Min. : 0.000
## 1st Qu.: 0.000 1st Qu.: 0.000
## Median : 4.000 Median : 4.000
## Mean : 4.983 Mean : 4.143
## 3rd Qu.: 8.000 3rd Qu.: 8.000
## Max. :88.000 Max. :10.000
## NA's :1505 NA's :1526Interpretation:
Both Portugal and Serbia range from 0-10 indicating that respondents are providing responses across the full spectrum of the range. However, there are less extreme values compared to Portugal, with no outliers. Additionally, the mean values for Serbia :4.983: is slightly higher than the mean in Portugal :4.143:, indicating to a higher level of planning for the future compared to Portugal’s. Overall, both countries contain similar range of responses however the distribution and means establish key distinctions into how both countries approach planning for the future.
Task 4
Provide code and answer.
Prompt and question: using the variables stfdem and gndr, answer the following: on average, who is more dissastified with democracy in Italy, men or women? Explain while referring to the output generated.
Info on variable here: https://ess.sikt.no/en/variable/query/stfdem/page/1
italy_data <- ess %>%
filter(cntry == "IT")
italy_data$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 4.78Females are more dissatisfied with democracy in Italy
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()`).
The median line within the boxplot of satisfaction with health services is higher, therefore satisfaction with healthcare is higher in France.