How people elect parliaments
One of the final polls before Iowa’s caucuses tonight has support among Republican voters for Donald Trump at 28%, Ted Cruz at 23% and Marco Rubio at 15%. It’s been pretty much at that level all week. (All others must total 34%.)
So that’s a substantial lead for Trump, then? But – what if the rates of turnout of these supporters differ markedly?
Suppose that these polling rates reflect 280 Trump voters, 230 Cruz voters, and 150 Rubio voters (and to add up to 1,000, there must be 340 other voters).
But not every one polled actually bothers to turn up to the time-demanding caucuses. Suppose that 70% of Trump’s voters actually show up, but 90% of Cruz’s show up (and say 80% of the others show up).
That means 196 Trump voters show up, 207 Cruz voters, 120 Rubio voters and 152 others.
After such an attendance, everyone would recalculate the numbers and conclude that the vote went 29% Trump, 31% Cruz, 18% Rubio and 22% all others together. The media would rave about how Cruz beat Trump with an impressive support surge of 8% in the last day, as many of the ‘others’ came off the fence in big numbers for Cruz – even though no such thing had happened at all. Such is the power of turnout rates in voluntary elections.
Presidential nomination politics is highly local in Iowa (image: Phil Reader, Wikimedia commons)
That’s why US pollsters need to pay so much attention to whether respondents are ‘likely voters’ before they do their calculations.
The better polling agencies attempt to take this problem into account before they publish their estimates, but doing so is very difficult. It’s very hard to get reliable answers when asking poll interviewees whether they intend to turn out.
The lesson is – differences between pre-poll estimates and calculated actual results do NOT necessarily mean that there was any change in public opinion in the final hours of a campaign – it may simply be caused by people failing to show up to vote.