Practical 1

Example: Birth Rates in Stirling vs Glasgow

Question 1

Here is the data for two Local Authority Areas (LAA) in Scotland: Stirling and Glasgow City.

The crude birth rate is calculated as the number of births per 1000 people.
\[ \text{Birth Rate} = \frac{\text{Number of Births}}{\text{Number of People}} \cdot 1000 \]

Calculate the birth rate for each LAA

LAA	n_people	n_births	birth_rate
Glasgow City	650300	6150	?
Stirling	94210	659	?

Solution 1.

LAA	n_people	n_births	birth_rate
Glasgow City	650300	6150	9.46
Stirling	94210	659	7.00

Question 2

We can decompose the birth rate into two components: A: the number of births per “woman of child bearing age”(Scotland uses ages 15-44)“, and B: the proportion of the population that are women aged 15-44.

\[ \text{Birth Rate} = A \cdot B \cdot 1000 \]
\[ \text{Birth Rate} = \frac{\text{Nr of Women 15-44}}{\text{Nr of People}} \cdot \frac{\text{Nr of Births}}{\text{Nr of Women 15-44}} \cdot 1000 \]

Calculate these below, and then multiply them to demonstrate to yourselves that this returns the same rates (give or take some rounding!) as the previous question.

LAA	n_people	n_women15to44	n_births	w15-44 / people	births / w15-44	birth_rate
Glasgow City	650300	161105	6150	A1	B1	A1 * B1 * 1000
Stirling	94210	17844	659	A2	B2	A2 * B2 * 1000

Solution 2.

LAA	n_people	n_women15to44	n_births	w15-44 / people	births / w15-44	birth_rate
Glasgow City	650300	161105	6150	0.24774	0.03817	9.46
Stirling	94210	17844	659	0.18941	0.03693	6.99

Question 3

What would Stirling’s birth rate be if:

1. Women aged 15-44 made up the same proportion of its population as they do for Glasgow?

Hints

A1 * B2!

LAA	w15-44 / people	births / w15-44
Glasgow City	A1	B1
Stirling	A2	B2

Solution 3.

LAA	n_people	n_women15to44	n_births	bpw	propw	birth_rate
Glasgow City	650300	161105	6150	0.03817	0.24774	9.456236
Stirling	94210	17844	659	0.03693	0.18941	6.994911

\(0.24774 \times 0.03693 \times 1000 = 9.1490382\)

Question 4

What would each LAA’s birth rates be if:

• The proportion of the population comprised of women aged 15-44 was the same in both populations (i.e., at the average of the two)
• The rates of births in women aged 15-44 were the same in both populations (i.e., at the average of the two)

Hints

B1 * ((A1+A2)/2)
B2 * ((A1+A2)/2)
and
A1 * ((B1+B2)/2)
A2 * ((B1+B2)/2)

LAA	w15-44 / people	births / w15-44
Glasgow City	A1	B1
Stirling	A2	B2

Solution 4.

LAA	n_people	n_women15to44	n_births	bpw	propw	birth_rate
Glasgow City	650300	161105	6150	0.03817	0.24774	9.456236
Stirling	94210	17844	659	0.03693	0.18941	6.994911

if proportion of w15-44 were equal:

• Glasgow: \(0.03817 \times \frac{0.24774+0.18941}{2} \times 1000 = 8.3430078\)
• Stirling: \(0.03693 \times \frac{0.24774+0.18941}{2} \times 1000 = 8.0719747\)

if births per w15-44 were equal:

• Glasgow: \(0.24774 \times \frac{0.03817+0.03693}{2} \times 1000 = 9.302637\)
• Stirling: \(0.18941 \times \frac{0.03817+0.03693}{2} \times 1000 = 7.1123455\)

Question 5

The differences between these two pairs of standardised rates (when using the average of the two populations as the reference) provides the contribution of each factor on the total crude rate difference.

We know from the very first question that Stirling has a crude birth rate that is ~2.46 births per 1000 people fewer than Glasgow.

How much of this is due to Stirling having relatively fewer ‘women of child bearing age’?

Hints

(A2 - A1) * ((B1+B2)/2) / (R2 - R1)

LAA	crude rate	w15-44 / people	births / w15-44
Glasgow City	R1	A1	B1
Stirling	R2	A2	B2

Solution 5. From the previous question, these are Glasgow and Stirling’s rates were only the proportion of women aged 15-44 to differ (and not the birth rate per woman aged 15-44):

• Glasgow: \(0.24774 \times \frac{0.03817+0.03693}{2} \times 1000 = 9.302637\)
• Stirling: \(0.18941 \times \frac{0.03817+0.03693}{2} \times 1000 = 7.1123455\)

And the crude rates:

• Glasgow: \(9.4562358\)
• Stirling: \(6.9949113\)

The difference between standardised rates, as a percentage of the difference between crude rates, gives:

\(\frac{7.1123455 - 9.302637}{6.9949113 - 9.4562358} = 88.9883272\%\)

Get The Software

Question 6

Get either R or Stata open, and install the various packages.

In R: DasGuptR
We would recommend installing the development version

library(remotes)
install_github("josiahpjking/DasGuptR@develop")

In Stata, we’re assuming you are familiar with using Stata (if not, we recommend you use R instead). The relevant command is rdecompose

#| eval: false
ssc install rdecompose

Question 7

Here’s the data:

R

p01 <- data.frame(
  LAA = c("Glasgow City","Stirling"),
  n_people = c(650300,94210),
  n_women15to44 = c(161105,17844),
  n_births = c(6150,659)
)

Stata

clear
input str12 LAA long n_people long n_women15to44 int n_births
"Glasgow City" 650300 161105 6150
"Stirling" 94210 17844 659
end

Create the new factors, and use either dgnpop() (in R) or rdecompose (Stata) to conduct the decomposition.

Check that it matches your calculations above.

You might find it easiest to multiply the “births per woman aged 15-44” by 1000 when you create it.

In R, this will look something like this:

library(tidyverse)
library(DasGuptR)
p01 <- p01 |>
  mutate(
    propw =  ????
    bpw =  ???
  )

dres <- dgnpop(x = ???, 
       pop = "???",
       factors = c("???","???"))

dg_table(dres)

Solution 6.

library(tidyverse)
library(DasGuptR)
p01 <- p01 |>
  mutate(
    propw =  n_women15to44/n_people,
    bpw =  n_births/n_women15to44*1000
  )

dres <- dgnpop(x = p01, 
       pop = "LAA",
       factors = c("bpw","propw"))

dg_table(dres)

      Glasgow City Stirling       diff decomp
bpw       8.343778 8.072162 -0.2716165  11.03
propw     9.303243 7.112697 -2.1905460  88.97
crude     9.457174 6.995011 -2.4621625 100.00

Question 8

We can split the birth rate further, into three different factors:

\[ \small \text{Birth Rate} = \frac{\text{Nr of Women}}{\text{Nr of People}} \cdot \frac{\text{Nr of Women 15-44}}{\text{Nr of Women}} \cdot \frac{\text{Nr of Births}}{\text{Nr of Women 15-44}} \cdot 1000 \]

Perform this decomposition now.

R

p01a <- data.frame(
  LAA = c("Glasgow City","Stirling"),
  n_people = c(650300,94210),
  n_women = c(332054, 48731),
  n_women15to44 = c(161105,17844),
  n_births = c(6150,659)
)

Stata

clear
input str12 LAA long n_people long n_women long n_women15to44 int n_births
"Glasgow City" 650300 332054 161105 6150
"Stirling"     94210  48731  17844  659
end

Solution 7.

p01a <- p01a |> mutate(
  propw = n_women/n_people,
  propw1544 = n_women15to44/n_women,
  bpw = n_births/n_women15to44*1000
)

dgnpop(p01a,
       pop="LAA",
       factors=c("propw","propw1544","bpw")) |>
  dg_table()

          Glasgow City Stirling       diff decomp
propw         8.168592 8.274858  0.1062662  -4.32
propw1544     9.363423 7.066774 -2.2966485  93.28
bpw           8.348808 8.077028 -0.2717802  11.04
crude         9.457174 6.995011 -2.4621625 100.00

Choose your own

Birth Rates Across England and Scotland

Just the same as above, but comparing England and Scotland.

\[ \small \text{Birth Rate} = \frac{\text{Nr of Women}}{\text{Nr of People}} \cdot \frac{\text{Nr of Women 15-44}}{\text{Nr of Women}} \cdot \frac{\text{Nr of Births}}{\text{Nr of Women 15-44}} \cdot 1000 \]

Data:

R

scot_v_eng <- 
  tribble(
    ~country, ~n_people, ~n_women, ~n_women1544, ~n_births,
    "Scotland", 5546900, 2851740, 1064441, 45763,
    "England", 58620101, 29895762, 11583779, 567708
  )

Stata

clear
input str8 country long n_people long n_women long n_women1544 long n_births
"Scotland"  5546900  2851740  1064441  45763
"England"   58620101 29895762 11583779 567708
end

Solution 8.

scot_v_eng <- 
  tribble(
    ~country, ~n_people, ~n_women, ~n_women1544, ~n_births,
    "Scotland", 5546900, 2851740, 1064441, 45763,
    "England", 58620101, 29895762, 11583779, 567708
  )

# crude rates
scot_v_eng |>
  transmute(
    country, 
    cr = n_births/n_people*1e3
  )

# A tibble: 2 × 2
  country     cr
  <chr>    <dbl>
1 Scotland  8.25
2 England   9.68

# create factors:
scot_v_eng <- 
  scot_v_eng |>
  mutate(
    propw = n_women/n_people,
    propw1544 = n_women1544/n_women,
    bpw = n_births/n_women1544
  )

# decomposition
dgnpop(scot_v_eng, 
       pop = "country",
       factors = c("propw","propw1544","bpw")) |>
  dg_table()

              England    Scotland          diff decomp
propw     0.008927015 0.008999177  7.216193e-05  -5.03
propw1544 0.009126028 0.008791294 -3.347335e-04  23.34
bpw       0.009545104 0.008373342 -1.171762e-03  81.69
crude     0.009684528 0.008250194 -1.434334e-03 100.00

The Gender Earnings-Gap

In 2025 in the UK, the estimated weekly earnings per working aged Males was £179. The weekly earnings per working aged Females was only £7.

These numbers can be calculated as simply:

\[ \text{weekly £ per working aged person} = \frac{\text{total weekly earnings}}{\text{number of working aged people}} \]

This difference could be attributable to:

• a greater number of males in the “labour force” (people who are employed or actively seeking employment. this excludes students etc).
• higher rates of employment for males in the labour force compared to females in the labour force (employment bias)
• males working more hours per week compared to females (i.e., fewer part time roles)
• a higher average hourly wage for males compared to females (a pay bias - this is the typically “gender pay gap”, unadjusted for specific roles)

So we can decompose the total weekly earnings per person as:

\[ \text{weekly £ per working aged person} = A \cdot B \cdot C \cdot D \]

Where

• A = proportion of working aged population in the labour force
• B = the proportion of the labour force that are employed
• C = the average hours worked per employee
• D = the average earnings per hour

You can find this data for a whole load of different countries at https://ilostat.ilo.org/data/?cat_mode=subject, if you would like to do this for somewhere else.

R

work25 <- data.frame(
  pop = c("males","females"),
  working_age = c(23718900, 24613110),
  labour_force = c(18442860, 5767405),
  employees = c(4823169, 280557),
  avg_hours_pp = c(37.9, 31.7),
  total_weekly_earnings = c(4240916038, 178762504)
)

Stata

clear
input str7 pop long working_age long labour_force long employees float avg_hours_pp double total_weekly_earnings
"males"   23718900 18442860 4823169 37.9 4240916038
"females" 24613110 5767405  280557  31.7 178762504
end

Solution 9.

# crude rates
work25 |> transmute(
  pop,
  weeklyearnings_perperson = total_weekly_earnings/working_age
)

      pop weeklyearnings_perperson
1   males               178.799018
2 females                 7.262898

# create factors
work25 <- 
  work25 |> 
  mutate(
    participation = labour_force / working_age,
    employment = employees / labour_force,
    hours = avg_hours_pp,
    earnings = total_weekly_earnings/hours/employees
  )

# decompose
dgnpop(work25, 
       pop="pop",
       factors=c("participation","employment","hours","earnings")) |>
  dg_table()

                females     males       diff decomp
participation 28.443354  94.38448  65.941123  38.44
employment    19.115488 102.76588  83.650392  48.77
hours         62.034915  74.16793  12.133012   7.07
earnings      63.617110  73.42870   9.811594   5.72
crude          7.262898 178.79902 171.536121 100.00

I guess it’s good that a pay-bias doesn’t account for much of this difference (only ~5%).
It’s worrying that the employment-bias account for almost half!

This is a good example to note that these methods are not going to give us evidence of a bias being present (or absent).

All we are saying is that the different employment rates for men vs women contributes 49% of the total earnings disparity. That is not evidence of a bias against employing women - such a bias happens at the unit level, not at the level of the population. In this way, Das Gupta’s decomposition is essentially an act of book-keeping. The counterfactuals we get from this method (A, below) are not the same as those that we need to think about to evidence an gender employment bias (B, below).

• A: “if Fs and Ms had same proportions of people in the labour force, worked the same hours, and had the same hourly wage, but differed in their employment rates in the way that they do, what would their weekly earnings be?”
• B: “if a given individual F was perceived to be M, does their probability of employment change (all else being equal)?”

Scotland’s Recycling: 2011 and 2022

The Scottish Environment Protection Agency (SEPA) provides data on the weights of waste that is generated, recycled, landfilled etc.

One indicator tracked by many countries and organisations is the amount of recycled waste per citizen, i.e.:
\[ \text{Recycled Tonnes Per Person} = \frac{\text{Recycled Waste}}{\text{Nr People}} \]

In Scotland, this indicator has actually decreased from 195kg per person in 2011 to 186kg per person in 2022. On face value this could be interpreted as an indication that we are, as a nation, recycling less!

But we can decompose the indicator into the amount of waste generated per person, and the proportion of waste that is recycled.

This decomposition will allow us to see how much of the decrease in the population indicator can be attributable to changes in the amounts of waste generated, or in the proportion of waste that is recycled.

\[ \text{Recycled Tonnes Per Person} = \frac{\text{Total Waste}}{\text{Nr People}} \cdot \frac{\text{Recycled Waste}}{\text{Total Waste}} \]

Data are available at https://josiahpjking.github.io/sgsss_std_decomp/data/recycling_2011_2022.csv

Solution 10.

df <- read_csv("https://josiahpjking.github.io/sgsss_std_decomp/data/recycling_2011_2022.csv")

# crude rates and factors
df <- df |>
  mutate(
    # crude rate
    rpp = recycled/pop*1e3,
    # factors
    wpp = totalwaste/pop*1e3,
    recp = recycled/totalwaste,
  )

df

# A tibble: 2 × 9
   year     pop recycled landfilled diverted totalwaste   rpp   wpp  recp
  <dbl>   <dbl>    <dbl>      <dbl>    <dbl>      <dbl> <dbl> <dbl> <dbl>
1  2011 5299900  1032156    1453789   120688    2606633  195.  492. 0.396
2  2022 5447000  1012045     585461   736277    2333783  186.  428. 0.434

dgnpop(df,
       pop="year",
       factors=c("wpp","recp")) |> 
  dg_table()

          2011     2022       diff  decomp
wpp   204.0154 177.7272 -26.288226  293.67
recp  182.2029 199.5397  17.336741 -193.67
crude 194.7501 185.7986  -8.951485  100.00

This example is a good one for figuring out directions etc.

We get from this:

1. the crude indicator of “kg recycled per person” goes down
2. if the waste generated per person stayed the same, but we only change the proportion of waste that is recycled, then “kg recycled per person” would have gone up.
   • this is good! it means we’re recycling a greater proportion of waste!
3. if the proportion of waste that is recycled stays the same, but we only change the amount of waste generated per person, then “kg recycled per person” would have gone down about ~3 times as much as it did.
   • this is also good! it suggests we’re creating a lot less waste!

together, we have this antagonistic relationship between these two factors and the eventual outcome. A population can make the indicator go up either by generating loads more waste (a bad thing) or by recycling a greater proportion of the waste (a good thing)!