# If Ralph Nader has 2% of the vote with a +/- 3% margin of error, what will that mean?

#### sbs12

##### New Member
What should we conclude from that?

Let's say if Ralph Nader has 2% of the votes of those polled, with a 3% margin of error [assuming 1000 people polled, so 1/(1000^1/2) = 3%].

With a 95% confidence level and a 3% margin of error, should we then conclude that on 95 out of 100 polls, Ralph Nader who has 2% of the votes will receive between 0 to 5% of the votes? (of course, ignoring the negative percentage points since that's not possible.)

Am I on the right path, or is there something more to it?

I have a feeling I'm doing something wrong. Any ideas?

#### Xenu

##### New Member
First of all, you wouldn't get a 3% margin of error with 1000 people polled. Where do you get the 1/(1000^1/2) from?

The confidence interval means, that if you repeat the survey, your interval would cover the (unknown) true value in 95% of the cases.

#### sbs12

##### New Member
Hi Xenu,

Thanks for your quick response. I got the 3% margin of error from the formula 1/(sample size squared), which in this case means 1/(1000^[1/2]).

I've checked with other sources, and they got the same 3% as I did (e.g. http://www.isixsigma.com/library/content/c040607a.asp).

Maybe I'm misinterpreting you somehow? My apologies about that.

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#### Xenu

##### New Member
The error of margin is approximately 2 times the standard deviations divided by the square root of size of the sample.

So the formula you use would be correct if the true proportion is about 50%, because then the standard deviation would be SQRT(0.5*0.5)=0.5. However, in this case, your standard deviation is much lower because the proportion is lower. So it would be better to approximate the standard deviation using the estimated proportion. So you have: STD = SQRT(0.98*0.02) = 0.14, wich gives an approximate error of margin of 2*0.14/SQRT(1000)= 0.009.

#### sbs12

##### New Member
Xenu, that makes a lot of sense now. Thank you so much for the help. I really appreciate it