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Suppose 500 individuals were asked one question having possible answers ”yes” or “no”. Denote by p the proportion of “yes” in the whole population. Estimate the probability that the proportion of “yes” in the sample (out of 500) is greater than in the whole population (p) by more than 5 %.

I was thinking of using the normal approx of the binomial where..
n = 500
p = 0.5
q = 0.5

Formula: \begin{align*} P\left(\frac{N - np}{\sqrt{npq}}\right)\\ \end{align*} However, I'm not sure what the N value would be in this case.

Andy
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  • What did you get from the answer there? And what exactly do you call $$P\left(\frac{N - np}{\sqrt{npq}}\right)\ ?$$ – Did Apr 09 '15 at 05:19
  • Didn't get any answer from there. And that's the normal approx of the binomial rv except it's poorly written as I didn't include P( N < X). – Andy Apr 09 '15 at 05:27
  • You got TWO answers "from there", actually. – Did Apr 09 '15 at 05:31
  • Not sure why you marked it as a duplicate as the questions are completely different. And I have sent an email to my prof already to ask why my answer was incorrect. I have already tried out all the given answers from that question and all of them were marked as incorrect. Until I get a reply back from my prof, I will leave it as unanswered for now. – Andy Apr 09 '15 at 05:35
  • Also duplicate of this... and probably of countless others. – Did Apr 09 '15 at 05:38
  • That question was not posted yet when I posted mine. There was a 20 minute difference. Even if those two questions are duplicates, there was no reason to mark this question as a duplicate. – Andy Apr 09 '15 at 05:41
  • Who cares? It was posted way before the present one. – Did Apr 09 '15 at 05:44
  • There is a complication in this one since $p$ is unspecified, and if $p(1-p)$ is small the variance will be smallish. – André Nicolas Apr 09 '15 at 05:50

1 Answers1

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In the population the porpotion is $0.5=50\%$. Now you have to add $5\%$. This is 55%.

To calculate N you have to calculate $55\% \cdot 500$.

The formula is like you have posted.

$P(X\leq N)=\ \Phi\left(\frac{N - np}{\sqrt{npq}}\right)$

You need the formula for the converse probability, because of "... by more than 5 %"

$P(X> N)=\ 1-\Phi\left(\frac{N - np}{\sqrt{npq}}\right)$

callculus42
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  • There is no reason to think $p=1/2$. If $p$ is reasonable, then $\sqrt{npq}$ is not far from $\sqrt{500(1/2)(1/2)}$, and we can come up with an answer. – André Nicolas Apr 09 '15 at 05:46