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Let's do a nonparametric test here. We observe that only two of the ten counties show machine counts favouring Obama. What are the odds of that? This is like tossing a two-headed coin (Clinton on one side, Obama on the other) ten times and getting two or fewer Obamas. The odds of this are 1 + 10 + 10 * 9 /2 divided by 2 to the 10th power, or 56/1024, or 7/128, or about 1/18. This is not quite significant at 95%. Moreover, it would be equally suspicious if machines favoured Clinton in only two counties or less. But for a two-sided alternative the odds of an extreme result are 1/9, so not quite significant even at 90%.

But this suggests repeting the calculation at the ward/municipality level, where there are 300 pairs and the statistical test will be much sharper.

We have met the enemy, and he is us — Pogo

by Carrie (migeru at eurotrib dot com) on Wed Jan 16th, 2008 at 05:58:53 PM EST
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