Wizard
Joined: Oct 14, 2009
• Posts: 22046
October 29th, 2010 at 8:14:14 AM permalink
Let me start with a rule. This thread is meant to discuss election mathematics only. So let's keep political opinions out of this thread, please. If you want to chime in about the pros and cons of either candidate, please make a separate thread for it. I reserve the right to bust any post in this thread taking a position on the race. This warning applies especially to JerryLogan.

The Las' rel='nofollow' target='_blank'>http://www.lvrj.com/news/angle-poll-data-improve-106287803.html]Las Vegas Review Journal just published the latest poll on the race. Here is what they have:

Reid: 45%
Angle: 49%
Margin of error: 4%
Sample size: 625

The questions I would like to bring up are, how is the "margin of error" determined, and what are each candidates odds of winning?

If we multiply the percentage of the vote based on the sample size of 625 we get:

Reid: 281.25
Angle: 306.25
Neither: 37.5

Let's get rid of those pesky indecisive voters and round the others. That would leave us with:

Reid: 281
Angle: 306
Total: 587

In terms of percentages:

Reid: 47.87%
Angle: 52.13%

To determine the probability of Reid winning, I think the question that should be asked is, what is the probability that Angle would get 306 or more in the survey if more than half of actual voters will favor Reid?

I show the variance of the mean in this pole is (.4787)*(.5213)/(587-1) = 0.000426. The standard deviation is the square root of that, or 0.020636.

The LVRJ said the margin of error was 4%. So, is that the standard deviation of the difference between Reid's and Angle's percentages? In other words, two times the variance of the mean (2*2.06%)?

Regardless of how the margin of error is defined and calculated, we see that Angle is 2.13% over 50%. That is 0.0213/.020636 = 1.031917 standard deviations. The probability of Angle falling to the left of the 1.031917 point on the bell curve can be found in any Gaussian curve statistics table, or put normsdist(1.031917) in Excel, and you get 0.848945.

So, if my math is right, and I'm far from sure it is, I say that Angle has an 85% chance of winning. Based on the LVRJ poll only, is my math correct?
It's not whether you win or lose; it's whether or not you had a good bet.
DorothyGale
Joined: Nov 23, 2009
• Posts: 639
October 29th, 2010 at 8:23:17 AM permalink
Quote: Wizard

So, if my math is right, and I'm far from sure it is, I say that Angle has an 85% chance of winning. Based on the LVRJ poll only, is my math correct?

When these polls say "4% margin of error," I always take that to be a 95% confidence interval. Knowing the confidence interval is the key to this, the rest of the calculation is easy from there. Without knowing the confidence interval that the 4% represents, I'm not sure what to do with these results.

From Wikipedia:

Quote:

Like confidence intervals, the margin of error can be defined for any desired confidence level, but usually a level of 90%, 95% or 99% is chosen (typically 95%).

--Ms. D.
"Who would have thought a good little girl like you could destroy my beautiful wickedness!"
JerryLogan
Joined: Jun 28, 2010
• Posts: 1344
October 29th, 2010 at 8:31:48 AM permalink
Isn't the LVRJ involved in some kind of legal issue with LVA for the unauthorized posting of one of their articles on a site without written permission?
Wizard
Joined: Oct 14, 2009
• Posts: 22046
October 29th, 2010 at 8:39:09 AM permalink
Quote: DorothyGale

When these polls say "4% margin of error," I always take that to be a 95% confidence interval. Knowing the confidence interval is the key to this, the rest of the calculation is easy from there. Without knowing the confidence interval that the 4% represents, I'm not sure what to do with these results.

Thanks. So if one standard deviation is 2.06%, then there is a 95% chance that Angle's actual percentage in the election will fall within 2.06%*1.96 = 4.04% of that. So on election day, there is a 95% chance her actual share will be within 48.08% and 56.17%, or 52.13% +/- 1.96*2.06%. For other readers who may be wondering where the 1.96 comes from, there is a 95% chance of falling within 1.96 standard deviations of expectations in any random sampling.

It would be nice of the papers said "The 95% margin of error is 4%," rather than just "The margin of error is 4%." How are we supposed to know they are referring to a 95% confidence interval? Why not 90%, 98%, 99%, or something else?
It's not whether you win or lose; it's whether or not you had a good bet.
crazyiam
Joined: Feb 5, 2010
• Posts: 44
October 29th, 2010 at 8:46:46 AM permalink
Fivethirtyeight might be the best place for election predictions. It uses poll averaging and weighting metrics to come up with predictions. I believe the methodology uses more undecided people to increase the variance of possible results.

Wizard
Joined: Oct 14, 2009
• Posts: 22046
October 29th, 2010 at 8:47:35 AM permalink
Quote: JerryLogan

Isn't the LVRJ involved in some kind of legal issue with LVA for the unauthorized posting of one of their articles on a site without written permission?

Also, everybody, please don't quote entire articles in this forum, especially from the LVRJ. Just small quotes, and properly attribute them.
It's not whether you win or lose; it's whether or not you had a good bet.
Wizard
Joined: Oct 14, 2009
• Posts: 22046
October 29th, 2010 at 9:00:10 AM permalink
Quote: crazyiam

Fivethirtyeight might be the best place for election predictions. It uses poll averaging and weighting metrics to come up with predictions. I believe the methodology uses more undecided people to increase the variance of possible results.

That is why I described those other 38 people in the poll is "pesky." It would be one thing if they were wasting their votes on a third party candidate. However, it does add more variance if they are still undecided. I wish the LVRJ would have made that clear.

Nice to see my election odds are close to those of the New York Times (77.2% Angle, 22.8% Reid). I would not expect them to match exactly, since they used a different survey.
It's not whether you win or lose; it's whether or not you had a good bet.
matilda
Joined: Feb 4, 2010
• Posts: 317
October 29th, 2010 at 9:02:38 AM permalink
Quote: Wizard

Thanks. So if one standard deviation is 2.06%, then there is a 95% chance that Angle's actual percentage in the election will fall within 2.06%*1.96 = 4.04% of that. So on election day, there is a 95% chance her actual share will be within 48.08% and 56.17%, or 52.13% +/- 1.96*2.06%. For other readers who may be wondering where the 1.96 comes from, there is a 95% chance of falling within 1.96 standard deviations of expectations in any random sampling.

It would be nice of the papers said "The 95% margin of error is 4%," rather than just "The margin of error is 4%." How are we supposed to know they are referring to a 95% confidence interval? Why not 90%, 98%, 99%, or something else?

Your interpretation of the probability of a confidence interval is incorrect. The reason is that the population parameter that the sample is used to estimate is a constant. It is the sample estimate that varies according to a distribution such as the normal. Thus the correct statement is that if a large number of samples were taken, then 95% of the intervals constructed would contain the parameter being estimated. As for the probability of the parameter being contained in a single interval, it is zero or 1 depending on whether it is in the interval or not.
Wizard
Joined: Oct 14, 2009
• Posts: 22046
October 29th, 2010 at 9:13:05 AM permalink
Quote: matilda

Your interpretation of the probability of a confidence interval is incorrect. The reason is that the population parameter that the sample is used to estimate is a constant. It is the sample estimate that varies according to a distribution such as the normal. Thus the correct statement is that if a large number of samples were taken, then 95% of the intervals constructed would contain the parameter being estimated. As for the probability of the parameter being contained in a single interval, it is zero or 1 depending on whether it is in the interval or not.

I'm not following you. What is the 4% margin of error telling us in this poll?
It's not whether you win or lose; it's whether or not you had a good bet.
rdw4potus
Joined: Mar 11, 2010
• Posts: 7056
October 29th, 2010 at 9:18:21 AM permalink
I agree with your math. Usually, there is an additional out-clause that is included in the press release for political polls. I don't see it in the LVJR article. There are two ways for error to be introduced into political polls. One is essentially statistical variance, which is accounted for by the 4% MOE that is stated on the poll. The other is methodological error. Political pollsters, including Mason-Dixon, randomly call phone numbers off of a list to get their survey sample. Then they use a combination of census data, exit-polling info, voter registration info, and intuition to manipulate the data. For example, any telephone poll will under-sample young voters and black voters and over-sample older voters and whites. So the pollster is left adjusting the results to more closely match the expected electorate. That usually results in a line at the bottom of the release that reads something like "in addition to the stated MOE for this poll, there is a second separate source of potential error that is less easy to quantify. This poll may be statistically sound and still vary from the election results by more than the stated MOE."

Here are the crosstabs for this poll.
"So as the clock ticked and the day passed, opportunity met preparation, and luck happened." - Maurice Clarett