Poll of Polls

Who is they?

 

Because "they" refers in this case to people who don't know what they are talking about.

 

That doesn't make it right.

 

I interpret that as "due to constraints error can't be estimated" not "we're perfect."

 

I agree with Student100: what they seem to be saying is that they are not estimating the margin of error, not that they are estimating it and that it is zero.

 

I interpreted the OP's original statement "why do they say poll of polls has no margin of error" as meaning there was no error. If they're just saying they can't figure out what it is, that's a different story.

 

So did I, which is why I wanted him to tell us where he got this from.

 

How does Chebyshev help at all??

 

"Poll of polls?" Isn't that on a par with combining the forecast tracks for hurricanes?

 

It is possible to combine the results of individual polls to obtain a meta-poll. It's also possible to calculate a margin of error for that meta-poll, and that margin of error will have a expected, general trend to be [itex] \frac{1}{\sqrt{N}} [/itex] of that of an individual poll, where [itex] N [/itex] is the number of individual polls (this assumes that the individual polls are polling the same thing [i.e., apples-to-apples] and that their individual margins of error are comparable).

As a simplified example, suppose we have [itex] N [/itex] polls for comparison. For the sake of simplicity, suppose all polls have identical margins of error, which is proportional to the the poll's standard deviation which I'll call [itex] \sigma [/itex]. (This is essentially saying that each poll is an equally valid predictor, even though each poll might give a unique prediction.)

We'll treat each poll as a random variable [itex] x_n [/itex] with a mean [itex] \mu_n [/itex] and a standard deviation [itex] \sigma_n [/itex], where [itex] \sigma_n = \sigma [/itex]: the same value for all polls.

Summing the results of all the polls into the random variable [itex] y [/itex],

[itex] y = x_1 + x_2 + x_3 + \dots + x_{N-1} + x_N [/itex]

gives the mean,

[itex] \mu_y = \mu_1 + \mu_2 + \mu_3 + \dots + \mu_{N-1} + \mu_N [/itex]

and variance

[itex] \sigma^2_y = \sigma^2_1 + \sigma^2_2 + \sigma^2_3 + \dots + \sigma^2_{N-1} + \sigma^2_N [/itex]

but since the individual variances are all the same in this simple example, we can say,

[itex] \sigma^2_y = N \sigma^2. [/itex]

and standard deviation

[itex] \sigma_y = \sqrt{N} \sigma. [/itex]

But in the end, we're not really interested in the sum but rather the average. So we scale the mean of the sum and the standard deviation of the sum by [itex] N [/itex].

[itex] \mu_{ave} = \frac{\mu_y}{N} [/itex]

[itex] \sigma_{ave} = \frac{\sigma_{ave}}{N} = \frac{\sqrt{N} \sigma}{N} = \frac{\sigma}{\sqrt{N}} [/itex]

And that last one is the kicker. It shows that when you combine multiple random variables, the average tends to reduce the "noise" by an amount [itex] \frac{1}{\sqrt{N}} [/itex].

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The simple example above has a glaring limitation that it assumes that all the individual polls are created equal. In the real world that is not the case. Still, statisticians have mathematical tools to weigh the individual polls before combining, but that is getting out of the scope of this thread.

My point was just to say that it is possible to combine the results of polls into a meta-poll and still obtain statically significant results. The claim in the original post that the meta-poll has "no margin of error" is not true.

(This of course assumes that the individual polls are comparing apples-to-apples. It doesn't make any sense to combine a poll for the 2016, US presidential election with another poll regarding favorite ice-cream flavors, for example.)

 

IIRC, only in the case that the polls are i) statistically independent, ii) the errors Gaussian. A big IF.

 

Seems like the easiest way is to take the raw data from each sample, apply your own methodology, and recalculate error. Your sample size is larger than any one poll, but still realitivity small compared to the entire voting population. You could then look at one polls methodology, apply it to the sample, and your MOE should then be improved compared to that poll.

 

Also, what do you do if you get one poll that measures 40 +/- 1% and the other that measures 60 +/- 1%? I don't think 50 +/- 0.7% is the right answer.

 

The errors do not have to be Gaussian. They can be of any probability distribution that you can think up (assuming the standard deviations are finite). The sum will approach being Gaussian, but that is not a requirement of the individual trials. That's part of the beauty of the Central Limit Theorem.

Of course we're assuming a certain level of statistical independence in the individual polls -- such as we can assume that a given individual poll is not going to base its results on that of another individual poll. Each individual poll is conducted individually. That's not an unrealistic assumption.

 

Is it not? Assuming that we're careful on how percentage and percentage points are used here, I think your method of combining is correct.

What I mean by that is assuming that the "40" and "60" are in units of percent, and the margin of error is +/- 1 percentage point then I think the method of combination is okay. Otherwise just get rid of the percent sign. Combine 40 units +/- 1 unit with 60 units +/- 1 unit to make 50 units +/- 0.7 units.

If something seems wrong here it is because you know right away that one or both of the original data points (original polls, if you will) represents a statistical fluke. Assuming the methodology that went into determining the margin of error to be +/- 1 unit is correct, it is very, very improbable that the initial data points will be different by a whopping 20 units. It's not necessarily impossible, but very improbable.

Initial data points like that would make me question the validity of the original 40 unit and 60 unit values, and the claim that their margin of error is +/- 1 unit. It's a hint that there might be something wrong there. Is it possible that these values are actually correct and are merely the result of random chance? Yes, it is possible, yet very improbable.