I have an expression
[tex]\sigma = \sqrt{p(1-p)/n}[/tex]
floating around in my head; let's see if it makes sense.
The variance of binary valued random variable b that comes up 1 with probability p (so has mean p) is
[tex]E( (b-p)^2 ) = (-p)^2(1-p) + (1-p)^2p = p(1-p)[/tex]
That's for an individual sample. For the observed average we divide by n, and for the standard deviation we take the square root:
[tex]\sigma = \sqrt{p(1-p)/n}[/tex]
Plugging in the numbers,
[tex]\sigma = \sqrt{.24(1-.24)/490} = 0.019 = 1.9\%[/tex]
One standard deviation of the average is almost 2% so a 27% outcome was 3/1.9 = 1.6 standard deviations from the mean, corresponding to a two sided probability of a bit bigger than 10% of happening by chance.
So this is borderline suspect; most surveys will include a two sigma margin of error, say plus or minus 4 percent here, and the results were within those bounds.
