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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.   :120.00

#Task 1 Provide code and answer.

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"
norway_data <- read.fst("norway_data.fst")
unique(norway_data$cntry)
## [1] "NO"
norway_happy <- norway_data %>%
  filter(cntry == "NO") %>% 
  select(happy)
norway_happy$happy[norway_happy$happy %in% c(77, 88, 99)] <- NA
norway_happy$y <- norway_happy$happy

table(norway_happy$y)
## 
##    0    1    2    3    4    5    6    7    8    9   10 
##   15   29   59  163  238  730  817 2617 5235 3796 2344
mean_y <- mean(norway_happy$y, na.rm = TRUE)
cat("Mean of 'y' is:", mean_y, "\n")
## Mean of 'y' is: 7.975005

#Norways avarge of happiness is higher

#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")
unique(ireland_data$cntry)
## [1] "IE"
ireland_binge <- ireland_data %>%
  select(alcbnge)
ireland_binge <- ireland_data %>%
  filter(cntry == "IE") %>%
  select(alcbnge)

`

table(ireland_binge$alcbnge)
## 
##   1   2   3   4   5   6   7   8 
##  65 650 346 417 239 641  26   6
ireland_binge$alcbnge[ireland_binge$alcbnge %in% 6:8] <- NA
df <- ireland_binge %>%
  mutate( 
    y_category = case_when(
      alcbnge == 1 ~ "Daily or almost daily",
      alcbnge == 2 ~ "Weekly",
      alcbnge == 3 ~ "Monthly",
      alcbnge == 4 ~ "Less than monthly",
      alcbnge == 5 ~ "Never",
      TRUE ~ NA_character_
    ),
    y_category = fct_relevel(factor(y_category), 
                             "Daily or almost daily", 
                             "Weekly", 
                             "Monthly", 
                             "Less than monthly",
                             "Never",)
  )
table(df$y_category)
## 
## Daily or almost daily                Weekly               Monthly 
##                    65                   650                   346 
##     Less than monthly                 Never 
##                   417                   239

#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_data <- read.fst("portugal_data.fst")
portugal_plnftr <- portugal_data %>%
  select(plnftr)
serbia_data <- read.fst("serbia_data.fst")
serbia_plnftr <- serbia_data %>%
  select(plnftr)
serbia_plnftr <- serbia_data %>%
  filter(cntry == "RS") %>%
  select(plnftr)
portugal_plnftr <- portugal_data %>%
  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   :14604
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   :1505
table(portugal_plnftr$plnftr)
## 
##   0   1   2   3   4   5   6   7   8   9  10  88 
## 114 184 313 356 264 481 262 382 345 166 370  40
table(serbia_plnftr$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  17
portugal_plnftr$plnftr[portugal_plnftr$plnftr %in% 88] <- NA
summary(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   :14644
serbia_plnftr$plnftr[serbia_plnftr$plnftr %in% 77:88] <- NA
summary(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   :1526

#some interpertations I can make from the data is that the the 1st Q= 3 in portuglal while for serbia 1st Q= 0, which means that the middle %50 for serbia actually starts at 0, making the data shifted to the left.

##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 <- read.fst("italy_data.fst")
unique(italy_data$cntry)
## [1] "IT"
italy_gndr <-italy_data %>%
  filter(cntry == "IT") %>% 
  select(gndr)
table(italy_gndr)
## gndr
##    1    2    9 
## 4836 5329   13
italy_data$gndr[italy_data$gndr %in% 9] <- NA
italy_stfdem <-italy_data %>%
  filter(cntry == "IT") %>% 
  select(stfdem)
table(italy_data$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    2
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))
italy_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         4.78

women are more dissatisified, but its only a small diffrence

#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 <- 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 = " 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()`).

#based on the graph, the satisfaction of sthlth (health survices) amoung french people is higher.