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?

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.Quote:WizardSo, 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?

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.

Quote:DorothyGaleWhen 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?

http://elections.nytimes.com/2010/forecasts/senate/nevada

Quote:JerryLoganIsn'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?

Yes. Please visit the copyrighted material thread.

Also, everybody, please don't quote entire articles in this forum, especially from the LVRJ. Just small quotes, and properly attribute them.

Quote:crazyiamFivethirtyeight 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.

Quote:WizardThanks. 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.

Quote:matildaYour 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?

Here are the crosstabs for this poll.