Gender Mix
Harold Nelson
3/11/2022
Topic
If a live birth occurs, what is the probability that the child is male? Most people would say 50%. That is close but not exactly right.
Questions.
What is the proportion of males in a given set of births?
Does this vary by the age of the mother, the geographical location, or the race of the mother.
Setup
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## ✓ readr 2.0.2 ✓ forcats 0.5.1
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Get Data
This data is from CDC Wonder and covers the years 2007 through 2019. The race variable we are using was dropped for 2020.
Gender <- read_delim("Natality, 2007-2020.txt",
delim = "\t", escape_double = FALSE,
col_names = TRUE, trim_ws = TRUE) %>%
select(`Census Region Code`, `Age of Mother 9`,Year,
`Mother's Bridged Race`,Gender,Births) %>%
filter(Year < 2020) %>%
drop_na()
## Rows: 3421 Columns: 12
## ── Column specification ────────────────────────────────────────────────────────
## Delimiter: "\t"
## chr (9): Notes, Census Region, Census Region Code, Mother's Bridged Race, Mo...
## dbl (3): Year, Year Code, Births
##
## ℹ Use `spec()` to retrieve the full column specification for this data.
## ℹ Specify the column types or set `show_col_types = FALSE` to quiet this message.
## Warning: One or more parsing issues, see `problems()` for details
#save(Gender,file = "Gender.Rdata")
Inspect
Do a summary(), glimpse(), and head() of the dataframe.
Solution
## Census Region Code Age of Mother 9 Year Mother's Bridged Race
## Length:3322 Length:3322 Min. :2007 Length:3322
## Class :character Class :character 1st Qu.:2010 Class :character
## Mode :character Mode :character Median :2013 Mode :character
## Mean :2013
## 3rd Qu.:2016
## Max. :2019
## Gender Births
## Length:3322 Min. : 10
## Class :character 1st Qu.: 257
## Mode :character Median : 2142
## Mean : 15605
## 3rd Qu.: 12607
## Max. :172645
## Rows: 3,322
## Columns: 6
## $ `Census Region Code` <chr> "CENS-R1", "CENS-R1", "CENS-R1", "CENS-R1", "C…
## $ `Age of Mother 9` <chr> "15-19 years", "15-19 years", "15-19 years", "…
## $ Year <dbl> 2007, 2007, 2008, 2008, 2009, 2009, 2010, 2010…
## $ `Mother's Bridged Race` <chr> "American Indian or Alaska Native", "American …
## $ Gender <chr> "Female", "Male", "Female", "Male", "Female", …
## $ Births <dbl> 150, 131, 135, 144, 131, 127, 123, 110, 134, 1…
## # A tibble: 6 × 6
## `Census Region Code` `Age of Mother 9` Year `Mother's Bridged … Gender Births
## <chr> <chr> <dbl> <chr> <chr> <dbl>
## 1 CENS-R1 15-19 years 2007 American Indian or… Female 150
## 2 CENS-R1 15-19 years 2007 American Indian or… Male 131
## 3 CENS-R1 15-19 years 2008 American Indian or… Female 135
## 4 CENS-R1 15-19 years 2008 American Indian or… Male 144
## 5 CENS-R1 15-19 years 2009 American Indian or… Female 131
## 6 CENS-R1 15-19 years 2009 American Indian or… Male 127
Pivot
We want to have two separate columns name Male and Female. These columns contain the numbers of births. Use pivot_wider() to accomplish this. Check the results with the three functions above.
Solution
Gender = Gender %>%
pivot_wider(names_from = Gender, values_from = Births) %>%
drop_na()
summary(Gender)
## Census Region Code Age of Mother 9 Year Mother's Bridged Race
## Length:1638 Length:1638 Min. :2007 Length:1638
## Class :character Class :character 1st Qu.:2010 Class :character
## Mode :character Mode :character Median :2013 Mode :character
## Mean :2013
## 3rd Qu.:2016
## Max. :2019
## Female Male
## Min. : 10.0 Min. : 10.0
## 1st Qu.: 266.2 1st Qu.: 278.5
## Median : 2180.5 Median : 2278.5
## Mean : 15454.6 Mean : 16192.4
## 3rd Qu.: 12555.5 3rd Qu.: 13163.5
## Max. :164768.0 Max. :172645.0
## Rows: 1,638
## Columns: 6
## $ `Census Region Code` <chr> "CENS-R1", "CENS-R1", "CENS-R1", "CENS-R1", "C…
## $ `Age of Mother 9` <chr> "15-19 years", "15-19 years", "15-19 years", "…
## $ Year <dbl> 2007, 2008, 2009, 2010, 2011, 2012, 2013, 2014…
## $ `Mother's Bridged Race` <chr> "American Indian or Alaska Native", "American …
## $ Female <dbl> 150, 135, 131, 123, 134, 140, 106, 72, 63, 53,…
## $ Male <dbl> 131, 144, 127, 110, 135, 144, 106, 85, 100, 70…
## # A tibble: 6 × 6
## `Census Region Code` `Age of Mother 9` Year `Mother's Bridged R… Female Male
## <chr> <chr> <dbl> <chr> <dbl> <dbl>
## 1 CENS-R1 15-19 years 2007 American Indian or … 150 131
## 2 CENS-R1 15-19 years 2008 American Indian or … 135 144
## 3 CENS-R1 15-19 years 2009 American Indian or … 131 127
## 4 CENS-R1 15-19 years 2010 American Indian or … 123 110
## 5 CENS-R1 15-19 years 2011 American Indian or … 134 135
## 6 CENS-R1 15-19 years 2012 American Indian or … 140 144
Clean Up
Use the code from Fertility 3 to fix the names and values in this dataframe.
Solution
Gender = Gender %>%
rename(Region = `Census Region Code`,
Race = `Mother's Bridged Race`,
Age = `Age of Mother 9`) %>%
mutate(Region = ifelse(Region == "CENS-R1","NE",Region),
Region = ifelse(Region == "CENS-R2","MW",Region),
Region = ifelse(Region == "CENS-R3","SO",Region),
Region = ifelse(Region == "CENS-R4","WE",Region),
Race = ifelse(Race == "American Indian or Alaska Native","AmInd",Race),
Race = ifelse(Race == "Asian or Pacific Islander","API",Race),
Race = ifelse(Race == "Black or African American","Black",Race))
# View(Gender)
The Ratio
Use all of the data to compute the proportion of males in live births.
Solution
Gender %>%
summarize(Male = sum(Male),
Female = sum(Female)) %>%
mutate(p = Male/(Male + Female))
## # A tibble: 1 × 3
## Male Female p
## <dbl> <dbl> <dbl>
## 1 26523138 25314578 0.512
Let’s take .5116 as the best estimate. Use the spreadsheet to look at the implications of this and the different probabilities of death for males and females at different ages.
For Wednesday
Use this data to ask if the Male ratio differs by Age, Race, or Region.
Solution for Age
Gender %>%
group_by(Age) %>%
summarize(n = sum(Male + Female),
p = sum(Male)/(n)) %>%
mutate(se = sqrt(p*(1-p)/n),
ub = p + 1.96 * se,
lb = p - 1.96 * se) %>%
ggplot(aes(y = Age ) ) +
geom_point(aes(x = p), color = "black") +
geom_point(aes(x = lb), color = "red") +
geom_point(aes(x = ub), color = "blue") +
ggtitle("95% Confidence Intervals for Male Ratio") +
geom_vline(aes(xintercept =.5116))
Solution for Race
Solution
Gender %>%
group_by(Race) %>%
summarize(n = sum(Male + Female),
p = sum(Male)/(n)) %>%
mutate(se = sqrt(p*(1-p)/n),
ub = p + 1.96 * se,
lb = p - 1.96 * se) %>%
ggplot(aes(y = Race ) ) +
geom_point(aes(x = p), color = "black") +
geom_point(aes(x = lb), color = "red") +
geom_point(aes(x = ub), color = "blue") +
ggtitle("95% Confidence Intervals for Male Ratio") +
geom_vline(aes(xintercept =.5116))
Solution for Year
Solution
Gender %>%
group_by(Year) %>%
summarize(n = sum(Male + Female),
p = sum(Male)/(n)) %>%
mutate(se = sqrt(p*(1-p)/n),
ub = p + 1.96 * se,
lb = p - 1.96 * se) %>%
ggplot(aes(y = Year ) ) +
geom_point(aes(x = p), color = "black") +
geom_point(aes(x = lb), color = "red") +
geom_point(aes(x = ub), color = "blue") +
ggtitle("95% Confidence Intervals for Male Ratio") +
geom_vline(aes(xintercept =.5116))
Solution for Region
Solution
Gender %>%
group_by(Region) %>%
summarize(n = sum(Male + Female),
p = sum(Male)/(n)) %>%
mutate(se = sqrt(p*(1-p)/n),
ub = p + 1.96 * se,
lb = p - 1.96 * se) %>%
ggplot(aes(y = Region ) ) +
geom_point(aes(x = p), color = "black") +
geom_point(aes(x = lb), color = "red") +
geom_point(aes(x = ub), color = "blue") +
ggtitle("95% Confidence Intervals for Male Ratio") +
geom_vline(aes(xintercept =.5116))
Transition to Mortality
Use the SSA life table to compute the ratio of the male probability of death to the female probability of death at every age.
Solution
ProbDeath <- read_csv("ProbDeath.csv")
## Rows: 120 Columns: 3
## ── Column specification ────────────────────────────────────────────────────────
## Delimiter: ","
## dbl (3): Age, MProbDeath, FProbDeath
##
## ℹ Use `spec()` to retrieve the full column specification for this data.
## ℹ Specify the column types or set `show_col_types = FALSE` to quiet this message.
ProbDeath %>%
mutate(Ratio = MProbDeath/FProbDeath) %>%
ggplot(aes(x = Age, y = Ratio)) +
geom_point() +
ggtitle("Ratio of Male Porbability of Death to Female")
