Some background to
the electoral system:
Each country in the EU has its own voting system for the
European elections. England is divided
into 9 regions and Wales and Scotland are each a region (Northern Ireland has a
different system, so I didn’t involve them).
In each region there are a number of seats, ranging from 10 seat for the
South East region to 3 seats in the North East.
You don’t vote for an individual but for a party. Each party has a ‘list’ of candidates for
each region. Last time around in the
7-seat East of England region most parties had seven candidates. Having worked out that UKIP should get 3
seats, the Conservatives 3 seats and Labour 1 seat (using the D’Hondt method – but that’s
not relevant here), the top three candidates from the UKIP list were elected as
MEPs, the top three from the Conservative list and top Labour candidate.
What we did:
On a shared Excel file, I had each region on a separate
sheet. I divided these up around the
class (some students had a sheet each, others worked as a pair). First they sorted out whether each candidate
was male or female. This wasn’t always
obvious, so they Googled if they weren’t sure.
Then they typed into another sheet the number of female and
male candidates for each party.
They had to deal with some issues. For example, some candidates are standing as
‘independents’, meaning they are not part of a party. We decided to ignore them. Some candidates have withdrawn from the
election; we removed the ones we knew about.
These are the results, for the parties that are standing in
every region:
11% of the UKIP candidates are female, through to Change UK,
for whom 57% of the candidates are female.
Is it interesting that the more pro-Brexit parties tend to
have more male candidates, while the more pro-Remain parties have roughly equal
numbers of men and women?
Better methodology?
But hold on: maybe this doesn’t tell the whole story. What if the LibDems have the same number of
male and female candidates but the top few in each region are all men? It is extremely unlikely that those further
down each list will be elected, so this is important.
Here is how I intend to deal with that.
The students recorded a list of Male and Female candidates
for each party. Here’s part of that
list:
Since the lesson (we will talk about this next week) I have turned that into a set of binary
numbers, where F becomes 1 and M becomes 0.
I then turned the binary into a decimal. This means the first-ranked candidate in each
list is worth double the one that follows, which is worth double the one after
that, etc. If there are 5 candidates in
a list, the 5th one is worth 1 point, the 4th is worth 2 points, the 3rd gets 4 points, the second gets 8 points and the first is 16 points. Add up the number of points for the female
candidates and then divide by the total (which is 1 + 2 + 4 + 8 + 16 = 31). That gives a fractional weighted-score for the female candidates
in each region.
Then I found the average of these.
The second column shows this new figure (with the first
column showing the previous data):
Not much difference (apart from the Green Party)!
Final thoughts:
It was good for the students to see a way to carry out this sort of analysis.
It was good to divide up the work. I couldn't find an easy way of getting hold of the gender of each candidate, so sharing the workload was a good thing.
We had to decide how to deal with problems (withdrawn candidates).
We had to think about how to do the analysis.
Here are links to my original spreadsheet and the one my class filled in.
If you want your class to use the original version then you will need to save it and share it with them (your IT people in College will be able to help if you haven't done this before).
