Posted on 10/07/2004 7:48:54 PM PDT by Republican Extremist
If there are 5 polls, each having a margin of error of 3%, and each showing the same candidate leading by 3% (Bush), what is the cmulative MOE for all 5 polls, if they are combined?
I am maintaining that it is far lower than 3%, but am rusty on my statistics.
Thanks
I'm fairly math-phobic, but I thought the MOE on multiple polls was computed as an average; i.e. If there are 5 polls, each with a MOE of 3, you multiply 3 x 5 then divide by 5 (the number of polls). But please don't take this as gospel - I'm pretty miserable with statistics.
W = Percentage of voters for Bush
K = percentage of voters for sKerry
n = sample size
MoE = sqrt( (W X K)/(n-1))
If you have three polls, the best way would to treat all five polls as a single weighted item.
You know they get their MOE by assuming they have a perfect random sample representative of the group and each poll has it's own special sauce on what it assumes will be the makeup of likely voters. The actual MOE for all these polls is actually much higher then they quote.
lol, yes you are.
When you combine polls, you increase the sample size and thereby decrease the margin of error of the composite result.
But remember, polls don't measure "truth!" They only predict what another sample set might say, be it true or false.
The MOE should remain constant if all of the polls have the same MOE and Average. so if 5 polls show bush up by 3 and the MOE in all five polls is 3, then the final average moe will be
3.
That's what I thought, but I have to prove it somehow, as I made the claim. LOL!
Actually, it's a little more complicated than simply adding them together to create a larger sample size.
The problem is that each poll is likely to have been stratified and weighted. The specific details are often not published, making consolidation of the poll results much more complex than it appears.
Could you define your variables please?
Then may I assume that you can answer Republican Extremist's question? Please do so.
No, the MOE drops as the sample size (e.g., number of polls) increases. See above posts.
You can't necessarily compare polls. A poll of likely voters is different than a poll of registered voters. And way different than a poll of eligible voters.
And you also have to be careful about how the poll is weighted between the parties and different age groups.
Averaging polls makes no sense. The best you can do with polls is look at the trends within the same poll. If all of the polls show one candidate ahead of the other, there's some sense there. But that's all you can tell.
Increasing the sample size (with similar polls and results) should always decrease the MOE.
AAAAAAAAAAAAAUUUUUUUUUUUUUUGGGGGGGGGGGGHHHHHHHHHHHH!!!!!!!!
Sorry, that gave me a flashback.
If the polls are compatible in terms of asking the same question and using the same methodology, the margin of error varies inversely with the square root of the sample size.
In other words, the same poll, same methodology, that asks 2x as many people will have about 70% of the first poll's margin of error (1 divided by the square root of 2). So in a perfect world, with 5 proper "combine-able" polls, the margin of error would decrease by a factor of the square root of 5, or about 2.23, that is, about 44% of the original MOE.
Sure but you would have to consider all polls equally weighted with the same or mostly similar questioning for it to drop, but one thing is for certain though the MOE would not increase.
Sorry, I just thought his answer was funny. If you multiply by the number of polls then divide by the number of polls, you have just multiplied by one.
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