[SOLVED] population proportion

**noob mathematician** · August 30th 2009, 12:22 AM

2 populations are independently surveyed using s.r.s of size n, and 2 proportions, p1 and p2, are estimated. It is expected that both population proportions are close to 0.5. What should the sample size be so that the standard error of the difference, $\hat p_1-\hat p_2$ will be less than 0.02?

I can find $\sigma_{\hat p}=\sqrt{\frac{p(1-p)}{n}(\frac{N-n}{N-1})}$ for the individual component, but I don't know how to apply for $\hat p_1-\hat p_2$ .
Answer is 1250.

**matheagle** · August 30th 2009, 06:36 AM

I would set

$\sigma_{\hat p_1-\hat p_2}=\sqrt{ {p_1(1-p_1)\over n_1 }+{p_2(1-p_2)\over n_2 } }\le .02$

Then let all the p's equal .5 and $n_1=n_2=n$

This leads to

$\biggl({1\over .02}\biggr)^2\le 2n$

I also get 1250 or LARGER for BOTH of the n's.

**matheagle** · August 30th 2009, 12:11 PM

you had the st deviation for a hypergeometric, but for only ONE parameter.

So $n_1\ge 1250$ and $n_2\ge 1250$

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