Academic Honesty Statement (fill your name in the blank)

I, keyan______, hereby state that I have not gained information in any way not allowed by the exam rules during this exam, and that all work is my own.

Load packages

library(tidyverse)
library(openintro)
library(nycflights13)
library(forcats)

1. Data Tidying and Relational Data

The following questions shall be answered by working with the world_bank_pop and who data sets from the openinto library.

  1. The data set world_bank_pop is not clean. Clean the data set such that the after data tidying you have six columns: country, year, SP.URB.TOTL, SP.URB.GROW, SP.POP.TOTL, SP.POP.GROW. Give your code and show the first 10 rows of the data set after being tidied. Then explain the meaning of each column.
data(world_bank_pop)

tidy_data <- world_bank_pop %>%
  pivot_longer(
    cols = -c(country, indicator),
    names_to = "year",
    values_to = "value"
  ) %>%
  pivot_wider(
    names_from = indicator,
    values_from = value
  )

head(tidy_data, 10)

Answer:
country: three-letter country code｜ year: the year｜SP.URB.TOTL: total urban population｜SP.URB.GROW: urban population growth rate｜SP.POP.TOTL: total population｜SP.POP.GROW: population growth rate

  1. Replace the country column of the tided data set in step a) with full names of the country (for example, replace USA with United States of America) by checking the data frame who, which contains the full name of each country corresponding to the three-digit country code. Give your code and show the updated data set in a manner to illustrate that the task is correctly fulfilled.
data(world_bank_pop)
data(who)

tidy_data <- world_bank_pop %>%
  pivot_longer(
    cols = -c(country, indicator),
    names_to = "year",
    values_to = "value"
  ) %>%
  pivot_wider(
    names_from = indicator,
    values_from = value
  )

updated_data <- tidy_data %>%
  rename(iso3 = country) %>%
  left_join(who %>% select(iso3, country), by = "iso3") %>%
  select(country, everything(), -iso3)
## Warning in left_join(., who %>% select(iso3, country), by = "iso3"): Detected an unexpected many-to-many relationship between `x` and `y`.
## ℹ Row 1 of `x` matches multiple rows in `y`.
## ℹ Row 341 of `y` matches multiple rows in `x`.
## ℹ If a many-to-many relationship is expected, set `relationship =
##   "many-to-many"` to silence this warning.
head(updated_data, 10)

Answer:
country column changed to full country name｜join works by matching same country code｜who dataset is like a dictionary

  1. With the data set obtained in step b), answer which countries had undergone significant urbanization between 2000 and 2017. You need to show the code and the results (either graphs or tables) to support your answer.
result <- updated_data %>%
  filter(year %in% c("2000", "2017")) %>%
  group_by(country, year) %>%
  summarise(SP.URB.TOTL = mean(SP.URB.TOTL, na.rm = TRUE), .groups = "drop") %>%
  pivot_wider(names_from = year, values_from = SP.URB.TOTL) %>%
  mutate(change = `2017` - `2000`) %>%
  arrange(desc(change))

head(result, 10)

Answer:
These countries had significant urbanization because their urban population increased the most between 2000 and 2017

2. Factors and Relational Data

For the following tasks, use data set planes and flights from the nycflights13 package.

  1. For the planes data set, only keep planes from manufacturers that have more than 10 samples in the data set. Then convert manufacturer column into a factor. Then combine AIRBUS and AIRBUS INDUSTRIE as a single category AIRBUS; combine MCDONNELL DOUGLAS, MCDONNELL DOUGLAS AIRCRAFT CO and MCDONNELL DOUGLAS CORPORATION into a single category MCDONNELL. Save your data frame as a new one. Show your code and the first 10 rows of the updated data frame.
data(planes)

planes_updated <- planes %>%
  group_by(manufacturer) %>%
  filter(n() > 10) %>%
  ungroup() %>%
  mutate(manufacturer = as.factor(manufacturer)) %>%
  mutate(manufacturer = fct_collapse(manufacturer,
    "AIRBUS" = c("AIRBUS", "AIRBUS INDUSTRIE"),
    "MCDONNELL" = c("MCDONNELL DOUGLAS", "MCDONNELL DOUGLAS AIRCRAFT CO", "MCDONNELL DOUGLAS CORPORATION")
  ))

head(planes_updated, 10)
  1. Join the flights data set with the planes data set, study how plane models correlate with the flight distance with proper data visualizations or summary tables. You are required to summarize your findings concisely in your own words.
data(flights)
data(planes)

joined <- flights %>%
  left_join(planes, by = "tailnum")

result <- joined %>%
  filter(!is.na(model), !is.na(distance)) %>%
  group_by(model) %>%
  summarise(avg_distance = mean(distance, na.rm = TRUE), n = n()) %>%
  arrange(desc(avg_distance))

head(result, 10)

Answer:
Long-range or larger aircraft models generally fly longer distances, while smaller or short-range models fly shorter routes overall.

3. Datetime and Data Transformation

For the following tasks, use the data set weather, flights or planes from the nycflights13 package.

  1. Create a plot of the temperature change across the whole year of 2013 at the JFK airport. (Hint: You need to first create a datetime variable for each hour.)
weather_jfk <- weather %>%
  filter(origin == "JFK") %>%
  mutate(datetime = make_datetime(year, month, day, hour))

ggplot(weather_jfk, aes(x = datetime, y = temp)) +
  geom_line(color = "steelblue", alpha = 0.3, linewidth = 0.3) +
  geom_smooth(method = "loess", span = 0.2, color = "red", se = FALSE, linewidth = 1) +
  labs(
    title = "Temperature Change at JFK in 2013",
    x = "Time",
    y = "Temperature (F)"
  ) +
  theme_minimal()

Answer:
Temperature changes over time show seasonal patterns across the year.From an overall perspective, temperature increases.From the beginning of the year, reaches a peak in summer, and then decreases toward the end of the year.

  1. Find out which day of the year has the largest temperature difference (defined as the difference between the highest and the lowest temperature) across the day (0am - 11pm).
data(weather)

result <- weather %>%
  filter(origin == "JFK") %>%
  mutate(date = as.Date(make_datetime(year, month, day, hour))) %>%
  group_by(date) %>%
  summarise(temp_diff = max(temp, na.rm = TRUE) - min(temp, na.rm = TRUE)) %>%
  arrange(desc(temp_diff))

head(result, 1)

Answer:
the day with the largest temperature difference is the one with the highest daily variation between max and min temperature

  1. Find a way to select all overnight flights (also called “Red Eye Flights” that depart at late night and arrive in the early morning) from the flights data set. Here overnight flights are defined as flights that departed between 10pm and 1am, and having an air time of over 4 hours . Create a categorical variable overnight_flag with YES or NO as the possible values. Show your code and the updated data frame.
data(flights)

flights_new <- flights %>%
  mutate(dep_hour = dep_time %/% 100) %>%
  mutate(
    overnight_flag = case_when(
      (dep_hour >= 22 | dep_hour <= 1) & air_time > 240 ~ "YES",
      TRUE ~ "NO"
    )
  )

head(flights_new, 10)

Answer:
overnight flights are those departing late night with long air time

  1. Someone says that most overnight flights use relatively small planes. Verify whether this is true with the data frame obtained in c) and the planes data set.
data(flights)
data(planes)

flights_new <- flights %>%
  mutate(dep_hour = dep_time %/% 100) %>%
  mutate(
    overnight_flag = case_when(
      (dep_hour >= 22 | dep_hour <= 1) & air_time > 240 ~ "YES",
      TRUE ~ "NO"
    )
  )

joined <- flights_new %>%
  left_join(planes, by = "tailnum")

result <- joined %>%
  filter(!is.na(seats)) %>%
  group_by(overnight_flag) %>%
  summarise(avg_seats = mean(seats, na.rm = TRUE), n = n())

result

Answer:
overnight flights tend to use larger planes on average, so the statement is not true.

4. General Analysis and Statistical Tests

Answer the following questions with data visualization or summary. You are required to summarize your findings concisely in your own words and support your conclusion with proper graphs or tables.

  1. From the gss_cat data set, find factors that are significantly correlated with the reported income.
gss_clean <- gss_cat %>%
  filter(!rincome %in% c("No answer", "Don't know", "Refused", "Not applicable")) %>%
  mutate(rincome = fct_drop(rincome))

ggplot(gss_clean, aes(x = rincome, y = age)) +
  geom_boxplot(fill = "skyblue", color = "darkgreen", outlier.alpha = 0.3) +
  coord_flip() +
  labs(title = "Age Distribution by Income",
       x = "Income",
       y = "Age") +
  theme_minimal() +
  theme(plot.title = element_text(hjust = 0.5),
        axis.text.y = element_text(size = 9))
## Warning: Removed 25 rows containing non-finite outside the scale range
## (`stat_boxplot()`).

ggplot(gss_clean, aes(x = marital, fill = rincome)) +
  geom_bar(position = "fill", width = 0.7) +
  coord_flip() +
  scale_fill_brewer(palette = "Blues", direction = 1) +
  labs(title = "Income Distribution by Marital Status",
       x = "Marital Status",
       y = "Proportion",
       fill = "Income") +
  theme_minimal() +
  theme(plot.title = element_text(hjust = 0.5))
## Warning in RColorBrewer::brewer.pal(n, pal): n too large, allowed maximum for palette Blues is 9
## Returning the palette you asked for with that many colors

Answer:
in summary, age and marital status are related to income. older and married people are more likely to have higher income. younger and single people are more likely to have lower income.

  1. From the smoking data set of the openintro package, find find factors that are significantly correlated with the smoking status and the number of cigarettes smoked per day.
ggplot(smoking, aes(x = gender, fill = smoke)) +
  geom_bar(position = "fill", width = 0.7) +
  scale_fill_manual(values = c("Yes" = "orange", "No" = "gray")) +
  labs(title = "Smoking Status by Gender",
       x = "Gender",
       y = "Proportion",
       fill = "Smoking") +
  theme_minimal() +
  theme(plot.title = element_text(hjust = 0.5))

smoking_active <- smoking %>% filter(smoke == "Yes")

ggplot(smoking_active, aes(x = age, y = amt_weekends)) +
  geom_point(alpha = 0.4, color = "blue") +
  geom_smooth(method = "lm", se = TRUE, color = "red") +
  labs(title = "Age vs Cigarettes Smoked (Weekends)",
       x = "Age",
       y = "Cigarettes (Weekends)") +
  theme_minimal() +
  theme(plot.title = element_text(hjust = 0.5))

Answer:
gender is related to smoking status, and males and females show different proportions of smokers
age is related to smoking intensity, and some age groups smoke more than others
overall, gender and age both affect smoking behavior, including whether people smoke and how much they smoke