Assignment 5

Human Mortality Database

This is an initial exploration of the Human Mortality Database, which is at https://www.mortality.org/.

Download the entire database and place it in your current working directory.

Setup

library(tidyverse)

## Warning: package 'tidyverse' was built under R version 4.4.1

## Warning: package 'ggplot2' was built under R version 4.4.1

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library(plotly)

## Warning: package 'plotly' was built under R version 4.4.1

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USAM Data

Load the data for USA males. Add a variable country and set it to “USA”.

Select country, Year, Age and qx.

Make Age numeric.

Eliminate any missing data.

Solution

USAM <- read_table("USA_mltper_1x1.txt", skip = 2) %>% 
mutate(country = "USA") %>% 
select(country, Year, Age, qx) %>% 
mutate(Age = as.numeric(Age)) %>% 
filter(Age < 85) %>% 
rename(male_prob_death = qx) %>% 
drop_na()

## 
## ── Column specification ────────────────────────────────────────────────────────
## cols(
##   Year = col_double(),
##   Age = col_character(),
##   mx = col_double(),
##   qx = col_double(),
##   ax = col_double(),
##   lx = col_double(),
##   dx = col_double(),
##   Lx = col_double(),
##   Tx = col_double(),
##   ex = col_double()
## )

## Warning: There was 1 warning in `mutate()`.
## ℹ In argument: `Age = as.numeric(Age)`.
## Caused by warning:
## ! NAs introduced by coercion

summary(USAM)

##    country               Year           Age     male_prob_death   
##  Length:7650        Min.   :1933   Min.   : 0   Min.   :0.000100  
##  Class :character   1st Qu.:1955   1st Qu.:21   1st Qu.:0.001662  
##  Mode  :character   Median :1978   Median :42   Median :0.004540  
##                     Mean   :1978   Mean   :42   Mean   :0.019836  
##                     3rd Qu.:2000   3rd Qu.:63   3rd Qu.:0.024307  
##                     Max.   :2022   Max.   :84   Max.   :0.172840

Canada

Do the same for Canada.

Solution

CANM <- read_table("CAN_mltper_1x1.txt", skip = 2) %>% 
mutate(country = "Canada") %>% 
select(country, Year, Age, qx) %>% 
mutate(Age = as.numeric(Age)) %>% 
filter(Age < 85) %>% 
rename(male_prob_death = qx) %>%
drop_na()

## 
## ── Column specification ────────────────────────────────────────────────────────
## cols(
##   Year = col_double(),
##   Age = col_character(),
##   mx = col_double(),
##   qx = col_double(),
##   ax = col_double(),
##   lx = col_double(),
##   dx = col_double(),
##   Lx = col_double(),
##   Tx = col_double(),
##   ex = col_double()
## )

## Warning: There was 1 warning in `mutate()`.
## ℹ In argument: `Age = as.numeric(Age)`.
## Caused by warning:
## ! NAs introduced by coercion

summary(CANM)

##    country               Year           Age     male_prob_death  
##  Length:8670        Min.   :1921   Min.   : 0   Min.   :0.00005  
##  Class :character   1st Qu.:1946   1st Qu.:21   1st Qu.:0.00141  
##  Mode  :character   Median :1972   Median :42   Median :0.00386  
##                     Mean   :1972   Mean   :42   Mean   :0.01870  
##                     3rd Qu.:1997   3rd Qu.:63   3rd Qu.:0.02082  
##                     Max.   :2022   Max.   :84   Max.   :0.18114

Combine

Combine the two dataframes into USA_CANM using rbind().

Solution

USAM_CANM = rbind(USAM, CANM)
summary(USAM_CANM)

##    country               Year           Age     male_prob_death  
##  Length:16320       Min.   :1921   Min.   : 0   Min.   :0.00005  
##  Class :character   1st Qu.:1951   1st Qu.:21   1st Qu.:0.00152  
##  Mode  :character   Median :1974   Median :42   Median :0.00411  
##                     Mean   :1974   Mean   :42   Mean   :0.01923  
##                     3rd Qu.:1998   3rd Qu.:63   3rd Qu.:0.02257  
##                     Max.   :2022   Max.   :84   Max.   :0.18114

Male Infant Mortality USA and Canada

Produce a graph showing the probability of male death at age 0 for the USA and Canada. Use color to see two time-series plots. Create this graph beginning in 1940.

Solutiom

USAM_CANM %>% 
  filter(Age == 0 & Year > 1940) %>% 
  ggplot(aes(x = Year, y = male_prob_death, color = country)) +
  geom_point() +
  ggtitle("Male Infant Mortality - USA and Canada")

USA/Canada 2

Create a graph comparing USA and Canadian male mortality at age 79.

Solution

USAM_CANM %>% 
  filter(Age == 79 & Year > 1940) %>% 
  ggplot(aes(x = Year, y = male_prob_death, color = country)) +
  geom_point() +
  ggtitle("Age 79 Male Mortality - USA and Canada")

Task 1

Copy and modify the code above to produce USAF, CANF and USA_CANF. Do summaries to verify your work.

USAF <- read_table("USA_fltper_1x1.txt", skip = 2) %>% 
mutate(country = "USA") %>% 
select(country, Year, Age, qx) %>% 
mutate(Age = as.numeric(Age)) %>% 
filter(Age < 85) %>% 
rename(female_prob_death = qx) %>% 
drop_na()

## 
## ── Column specification ────────────────────────────────────────────────────────
## cols(
##   Year = col_double(),
##   Age = col_character(),
##   mx = col_double(),
##   qx = col_double(),
##   ax = col_double(),
##   lx = col_double(),
##   dx = col_double(),
##   Lx = col_double(),
##   Tx = col_double(),
##   ex = col_double()
## )

## Warning: There was 1 warning in `mutate()`.
## ℹ In argument: `Age = as.numeric(Age)`.
## Caused by warning:
## ! NAs introduced by coercion

summary(USAF)

##    country               Year           Age     female_prob_death
##  Length:7650        Min.   :1933   Min.   : 0   Min.   :0.00008  
##  Class :character   1st Qu.:1955   1st Qu.:21   1st Qu.:0.00073  
##  Mode  :character   Median :1978   Median :42   Median :0.00297  
##                     Mean   :1978   Mean   :42   Mean   :0.01343  
##                     3rd Qu.:2000   3rd Qu.:63   3rd Qu.:0.01431  
##                     Max.   :2022   Max.   :84   Max.   :0.15084

CANF <- read_table("CAN_fltper_1x1.txt", skip = 2) %>% 
mutate(country = "Canada") %>% 
select(country, Year, Age, qx) %>% 
mutate(Age = as.numeric(Age)) %>% 
filter(Age < 85) %>% 
rename(female_prob_death = qx) %>%
drop_na()

## 
## ── Column specification ────────────────────────────────────────────────────────
## cols(
##   Year = col_double(),
##   Age = col_character(),
##   mx = col_double(),
##   qx = col_double(),
##   ax = col_double(),
##   lx = col_double(),
##   dx = col_double(),
##   Lx = col_double(),
##   Tx = col_double(),
##   ex = col_double()
## )

## Warning: There was 1 warning in `mutate()`.
## ℹ In argument: `Age = as.numeric(Age)`.
## Caused by warning:
## ! NAs introduced by coercion

summary(CANF)

##    country               Year           Age     female_prob_death 
##  Length:8670        Min.   :1921   Min.   : 0   Min.   :0.000030  
##  Class :character   1st Qu.:1946   1st Qu.:21   1st Qu.:0.000650  
##  Mode  :character   Median :1972   Median :42   Median :0.003125  
##                     Mean   :1972   Mean   :42   Mean   :0.013531  
##                     3rd Qu.:1997   3rd Qu.:63   3rd Qu.:0.013325  
##                     Max.   :2022   Max.   :84   Max.   :0.159520

USAF_CANF = rbind(USAF, CANF)
summary(USAF_CANF)

##    country               Year           Age     female_prob_death
##  Length:16320       Min.   :1921   Min.   : 0   Min.   :0.00003  
##  Class :character   1st Qu.:1951   1st Qu.:21   1st Qu.:0.00069  
##  Mode  :character   Median :1974   Median :42   Median :0.00306  
##                     Mean   :1974   Mean   :42   Mean   :0.01348  
##                     3rd Qu.:1998   3rd Qu.:63   3rd Qu.:0.01371  
##                     Max.   :2022   Max.   :84   Max.   :0.15952

Task 2

Redo the graphs you produced above for females in the USA and Canada. Do you see the same patterns?

USAF_CANF %>% 
  filter(Age == 0 & Year > 1940) %>% 
  ggplot(aes(x = Year, y = female_prob_death, color = country)) +
  geom_point() +
  ggtitle("Female Infant Mortality - USA and Canada")

USAF_CANF %>% 
  filter(Age == 79 & Year > 1940) %>% 
  ggplot(aes(x = Year, y = female_prob_death, color = country)) +
  geom_point() +
  ggtitle("Age 79 Female Mortality - USA and Canada")

Task 3: Male + Female

Combine USAM and USAF into USA. This new dataframe will have both male and female probabilities of death. Run a summary to verify your work.

USA <- merge(
  USAM[, c("country", "Year", "Age", "male_prob_death")], 
  USAF[, c("country", "Year", "Age", "female_prob_death")], 
  by = c("country", "Year", "Age"), 
  suffixes = c("_M", "_F"), # Add suffixes instead of renaming
  all = TRUE
)
summary(USA)

##    country               Year           Age     male_prob_death   
##  Length:7650        Min.   :1933   Min.   : 0   Min.   :0.000100  
##  Class :character   1st Qu.:1955   1st Qu.:21   1st Qu.:0.001662  
##  Mode  :character   Median :1978   Median :42   Median :0.004540  
##                     Mean   :1978   Mean   :42   Mean   :0.019836  
##                     3rd Qu.:2000   3rd Qu.:63   3rd Qu.:0.024307  
##                     Max.   :2022   Max.   :84   Max.   :0.172840  
##  female_prob_death
##  Min.   :0.00008  
##  1st Qu.:0.00073  
##  Median :0.00297  
##  Mean   :0.01343  
##  3rd Qu.:0.01431  
##  Max.   :0.15084

Task 4: The Ratio

Compute a new variable ratio. It is the ratio of the male probability of death to the female probability. For the year 2019, plot this ratio with Age on the horizontal axis. Use geom_point().

USA$ratio <- USA$male_prob_death / USA$female_prob_death
USA_2019 <- subset(USA, Year == 2019)
ggplot(USA_2019, aes(x = Age, y = ratio)) +
  geom_point() +
  labs(title = "Male to Female Probability of Death Ratio in 2019", 
       x = "Age", 
       y = "Male to Female Ratio") +
  theme_minimal()

summary(USA$ratio)

##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
##   1.000   1.407   1.652   1.740   1.907   3.523

Task 5: Comments

Describe what you saw in Task 4. How would you explain this? I saw an incline between 15 and 20 years but also noticed that as they got older it started declining. The oldest drops between 1 and 1.5.Females generally have a longer life.

Assignment 5

Miguel Seaton

10/28/2024

Human Mortality Database

Setup

USAM Data

Solution

Canada

Solution

Combine

Solution

Male Infant Mortality USA and Canada

Solutiom

USA/Canada 2

Solution

Task 1

Task 2

Task 3: Male + Female

Task 4: The Ratio

Task 5: Comments