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A survey organization draws a simple random sample of 1,000 registered voters in a certain town. In the sample, 32% approve of the Mayor. The organization estimates that 32% of all 50,000 registered voters in the town approve of the Mayor. How to figure the SE? The organization realizes that the number in the sample who approve __ 1,000 draws __box . Fill in each blank (33 words or less). Then work out the SE.

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One possible 'filling in' of the blanks is as follows:

"The organization realizes that the number in the sample who approve of the Mayor is equivalent to calculating the sum of 1000 draws from 50,000 tickets, of which 32% (16000) are marked 1 (representing approval of the Mayor) and the remaining 68% (34000) are marked 0, taken randomly and without replacement, from a box."

If we let X be the random variable 'sum of 1000 tickets taken randomly from the box', then obviously the expected value of X is 320 (32% of 1000). However, we could expect some variance in X due to the fact that our sample size is only a small percentage of the total population.

Formally, we can say that if the probability of drawing a 1 is $p$, then the expected sum of $n$ draws is $E(X) = np$.

The variance of X, defined in terms of expectation, is given by:

$V(X) = E[(X-E(X))^{2}]$

$= E(X^{2})-[E(X)]^{2}$

$= p - p^{2}$

$= p(1-p)$

So the standard error is:

$s = \sqrt{\frac{V(X)}{n}}$

$= \sqrt{\frac{p(1-p)}{n}}$

$= \sqrt{\frac{np(1-p)}{n^{2}}}$

$= \frac{\sqrt{np(1-p)}}{n}$

The standard error for the sum of n draws (with replacement) is therefore:

$SE = ns = \sqrt{np(1-p)}$

$\sqrt{1000\times0.32(1-0.32)} = 14.75$

Note: Calculating the standard error of our survey using this formula (which is based on the sum of n draws with replacement) will result in a slight bias (error) in the result, since this means we could be surveying the same person more than once. However, if the population is quite large compared to the number of draws, $n$ (our sample size) this is still gives a good approximation for calculating the standard error of our survey.

The exact standard error for the sum of n draws taken from a population N without replacement is:

$SE = \sqrt{np(1-p)}\sqrt{\frac{N-n}{N-1}}$

$= 14.75\times \sqrt{\frac{50000-1000}{50000-1}}$

$= 14.60$