Blogging from the homeland (Brisbane, Queensland, Australia) with some observations that I wanted to commit to a quasi-permanent form.
Wow… According to the University of Chicago’s Bob Pape, Melbourne is Australia’s most likely terrorist target over the next 12 months…. Read more here.
Pollbludger is doing another outstanding job, this time tracking developments in the upcoming election for the Queensland Legislative Assembly (September 9, 2006). I was a little surprised to see a report of a poll showing a 13 percentage point dive in the incumbent Premier’s approval rating; polbludger reminded us that the poll had a sample size of only 268. The same poll has Labor on 52.5 of the two-party preferred vote, which is the lowest Labor number I think I’ve seen all campaign. Beattie should cruise home, and as pollbludger also picked up on, the untimely death of local legend Steve Irwin will drown out any noise by the opposition for a news cycle or two.
For political scientists, the interesting facet of this election will be the same as the last one: the huge disjuncture between state-level voting and Federal-level voting in Australia at the moment, with Queensland perhaps the most aberrant state of all on this score (thumpingly massive victories for Labor in state elections, with the almost opposite outcome in Federal elections, sometimes less than 12 months apart). We’re about to get another nice set of data to further document and analyze this phenomenon. This disjuncture is especially interesting given that party identification is considered the dominant explanation of voting in (Australian) electoral behavior; time to think again, or at the very least, do some work on this. It is a real shame there is nothing like the AES for state level elections in Australia (or at least not that I know of).
Richard Farmer on crikey.com.au writes:
Newspoll this week had a two-party preferred prediction of Labor 58% to 42% for non-Labor. The Crikey indicator, based on the prices of the major internet bookmakers, has Labor a 91.1% chance of winning. Translate the Indicator’s probability into the same format as the opinion polls and the market is predicting a final outcome of Labor with around 53% of the two party preferred vote to non-Labor’s 47%.
Huh? How does that work? Given a probability p, that a binomial proportion x, exceeds .5, solve for x. The answer depends on n, the “sample size” of the “poll” being considered. In this case we’re told that p = .911. What levels of x generate p = .911? The answer depends on n. Details appear below the fold, but if I’ve got n = 1,000,000 (one million), I can find the probability of a Labor victory to be .911 with x as small as 50.07%. If n is only 100, then to find a Labor victory with probability .911 we’d need to see x = .567. So, when Richard Farmer says a Labor probability of victory of .911 means Labor has 53% of the 2PP, the implicit assumption is that we’re talking about a poll of 500 people. But the point here is that the comparison isn’t well defined, until you pin down the size of the “poll” being used. The “problem” here is that the mapping from a betting market’s implied “probability of victory” to a vote proportion is not well defined, and nor is the inverse mapping; hence, assumptions are required. I continue to wonder if an opinion poll’s aggregate estimate of “who do you think will win the election” is a better way to compare betting markets and opinion polls…
The following table summarizes the situation in this case, assuming the conditions given in Farmer’s crikey post:
p n [1,] 0.5667428 100 [2,] 0.5474069 200 [3,] 0.5300640 500 [4,] 0.5212777 1000 [5,] 0.5173891 1500 [6,] 0.5122653 3000 [7,] 0.5095060 5000 [8,] 0.5067255 10000 [9,] 0.5042557 25000 [10,] 0.5030100 50000 [11,] 0.5021288 100000 [12,] 0.5015055 200000 [13,] 0.5009522 500000 [14,] 0.5006734 1000000 [15,] 0.5004259 2500000
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