I agree completely with you. This sounds like a binomial distribution situation, where the StDev is give by
(SD) = [n p (1-p)]^0.5
For example, in a poll of 1000 people, if 500 answer one way on a question,
SD = [1000 * 0.5 * 0.5]^0.5 = 15.8
95% falls within 1.96 * SD = 31, or a 3.1% margin of error (assuming we can use a normal distribution approximation for a binomial distribution which works pretty well for n*p > 5). The margin or error goes down toward the tails, so they usually quote the worse-case margin of error at the center.
With two (or more) polls, you could find the total # of responses, and the weighted average for p and find the new uncertainty. The total uncertainty will be larger than either poll separately, but the relative uncertainty will be smaller.
Tim
-----Original Message-----
From: phys-l-bounces@carnot.physics.buffalo.edu on behalf of Bernard Cleyet
Sent: Sun 11/14/2010 12:44 AM
To: Forum for Physics Educators
Subject: [Phys-l] How does one combine uncertainty in multiple sampling?
This is initially directed to JD
I skimmed and read carefully some sections of JD's exposition** of uncertainty (sig. figs, etc.) and didn't find an answer -- suspecting it may be there please point. OTOH, if not, people help.
First, here's my idea [assuming the pols are scientific, i.e. at the same time, random, unbiassed, those poled are likely voters, etc.] The error is due to being a small sample of a large pop. So several polls, when combined, are a larger sample of the same pop. Therefore, the uncertainty of the combined polls is less than the uncertainty of any of the individual ones, no? And again, how does one calculate it?
** Measurements and Uncertainties versus Significant Digits or Significant Figures
bc wishes to reply to a newsletter that claims one can't have a combined error less than the lowest individual one.