Basis for Margin of Error in Opinion Polls?

[Bacle]

Hi,

Just curious as to what is the basis of the margin of error given in polls, e.g.,
in statements of the form:" 30% of people are in favor of candidate x. The poll
has a margin of error of +/- 5 %"

Thanks.

 

[mathman]

The error margin is obtained from an assumption of a binomial distribution. The net result is the estimated error is proportional to the square root of the sample size. For example if 10000 people are asked, the error estimate is around 100, which if the answers are split 5000 yes and 5000 no, the error estimate would be about 2%.

 

[Bacle]

Thanks, M:

Would the result extend to a multinomial if those polled were asked to choose between
more than two alternatives?

Thanks.

 

[mathman]

Thanks, M:

Would the result extend to a multinomial if those polled were asked to choose between
more than two alternatives?

Thanks.

Essentially yes. The proportion to the square root of the sample size remains,

 

[mathman]

To get back to basics. Binomial distribution with sample size n and probability of success p, the mean number of successes is np and the variance (square of the standard deviation) is np(1-p). When doing a statistical analysis, p is unknown, so it is usually estimated as the number of successes divided by the sample size.

When used to analyze poling data, the simplest way is to use the binomial for each choice. For example if there are 3 choices with probabilities x, y, z, then the variances can be estimated as nx(1-x), ny(1-y), and nz(1-z).

 

[Bacle]

Thanks Again.

When you talk about error, is it related to the variance/standard deviation?
Please ignore my ignorance--and the question-- if this is too basic/obvious.

 

[Bacle]

Never mind the last question and thanks; sorry for being so slow.

 

[moonman239]

The margin of error given is actually a confidence interval, I think. The assumption for a confidence interval is that the data follows a normal (not binomial) distribution, which is an assumption that should be checked using chi-square first.

The standard error (we'll call it SE) equals the standard deviation (the square root of the variance, which (in this case) = sample size * frequency of x) divided by the square root of the sample size.

The confidence interval is z * the standard error, where z is the Z-score that corresponds to the confidence level of your choice. For example, if your confidence level is 95%, then z = 1.96.

 

[Gokul43201]

The margin of error given is actually a confidence interval, I think.

Is it really? I think it is the spread corresponding to a specific confidence level. So, if the distribution is roughly Gaussian, then for a 95% confidence, the MoE is about 2 standard deviations.

 

[mathman]

The margin of error given is actually a confidence interval, I think. The assumption for a confidence interval is that the data follows a normal (not binomial) distribution, which is an assumption that should be checked using chi-square first.

The standard error (we'll call it SE) equals the standard deviation (the square root of the variance, which (in this case) = sample size * frequency of x) divided by the square root of the sample size.

The confidence interval is z * the standard error, where z is the Z-score that corresponds to the confidence level of your choice. For example, if your confidence level is 95%, then z = 1.96.

Gaussian is assumed because for large sample size, binomial is much harder to handle. In practice the average and the estimated deviation are used. The binomial gives the value for the parameters while the Gaussian is used to estimate distribution values.