jackman.stanford.edu/blog
• 114th U.S. Senate
• ideal point estimates svg pdf csv 8/3/15
• scatterplot against 2012 Obama vote share svg pdf
• roll call object: RData
• 114th U.S. House
• ideal point estimates svg pdf csv 7/29/15
• scatterplot vs Obama vote share svg pdf
• roll call object: RData
• Bayesian Analysis for the Social Sciences Wiley; Amazon; errata as of 3/6/15

## Sunday September 24, 2006

Filed under: general — jackman @ 9:38 pm

Blogging from the homeland (Brisbane, Queensland, Australia) with some observations that I wanted to commit to a quasi-permanent form.

Comments Off on Australian Observations

## Thursday September 7, 2006

Filed under: general — jackman @ 5:23 pm

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.

Comments Off on Bob Pape knows something you don’t

## Monday September 4, 2006

Filed under: politics — jackman @ 1:28 pm

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).

Comments Off on Queensland election

## Friday September 1, 2006

Filed under: politics,statistics — jackman @ 8:08 am

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 Indicatorâ€™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-Laborâ€™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