My Math Forum sample size for confidence interval of proportion

 November 28th, 2015, 07:36 AM #1 Newbie   Joined: Nov 2015 From: HK Posts: 8 Thanks: 0 sample size for confidence interval of proportion Question: A state legistlator wishes to survey residents of her district to see what proportion of the electorate is aware of her position on using state funds to pay for abortions. What sample size is necessary if the 95% confidence interval for p is to have width at most 0.10 irrespective of p? (p bar) + Z0.025 sigma/sqrt(n) - (p bar - Z0.025 sigma / sqrt(n)) <= 0.1 2(Z0.025 sigma/ sqrt(n) ) <= 0.1 but sigma is not known
 November 28th, 2015, 07:44 AM #2 Senior Member   Joined: Dec 2012 From: Hong Kong Posts: 853 Thanks: 311 Math Focus: Stochastic processes, statistical inference, data mining, computational linguistics That's not the correct formula for sample proportions. You should refer to your textbook/lecture notes again... The formula is like this: $\displaystyle \hat{p} \pm z_{\alpha / 2} \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$ since the variance of $\displaystyle \hat{p}$ is approximated by $\displaystyle \frac{\hat{p}(1-\hat{p})}{n}$ (true variance is $\displaystyle \frac{p(1-p)}{n}$). Out of curiosity, is this HKDSE M1 or tertiary-level? Last edited by 123qwerty; November 28th, 2015 at 07:47 AM.
November 28th, 2015, 08:04 AM   #3
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 Originally Posted by 123qwerty That's not the correct formula for sample proportions. You should refer to your textbook/lecture notes again... The formula is like this: $\displaystyle \hat{p} \pm z_{\alpha / 2} \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$ since the variance of $\displaystyle \hat{p}$ is approximated by $\displaystyle \frac{\hat{p}(1-\hat{p})}{n}$ (true variance is $\displaystyle \frac{p(1-p)}{n}$). Out of curiosity, is this HKDSE M1 or tertiary-level?
thanks!
It is tertiary level

November 28th, 2015, 05:55 PM   #4
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Quote:
 Originally Posted by 123qwerty That's not the correct formula for sample proportions. You should refer to your textbook/lecture notes again... The formula is like this: $\displaystyle \hat{p} \pm z_{\alpha / 2} \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$ since the variance of $\displaystyle \hat{p}$ is approximated by $\displaystyle \frac{\hat{p}(1-\hat{p})}{n}$ (true variance is $\displaystyle \frac{p(1-p)}{n}$). Out of curiosity, is this HKDSE M1 or tertiary-level?
but how to get p?

November 28th, 2015, 06:02 PM   #5
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 Originally Posted by tttim but how to get p?
You can't, which is why you use $\displaystyle \hat{p}$ to approximate p.

November 28th, 2015, 11:54 PM   #6
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 Originally Posted by 123qwerty You can't, which is why you use $\displaystyle \hat{p}$ to approximate p.
p hat = 0.9n / n = 0.9?

November 29th, 2015, 07:22 AM   #7
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Quote:
 Originally Posted by tttim p hat = 0.9n / n = 0.9?
We take the largest possible value of $\displaystyle \hat{p}(1-\hat{p})$. Let f(x) = $\displaystyle \hat{p}(1-\hat{p})$. Then f'(x) = $\displaystyle 2\hat{p} - 1$. $\displaystyle 2\hat{p} - 1 = 0$ at $\displaystyle \hat{p} = 0.5$, so we assume this value.

November 29th, 2015, 11:08 PM   #8
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 Originally Posted by 123qwerty We take the largest possible value of $\displaystyle \hat{p}(1-\hat{p})$. Let f(x) = $\displaystyle \hat{p}(1-\hat{p})$. Then f'(x) = $\displaystyle 2\hat{p} - 1$. $\displaystyle 2\hat{p} - 1 = 0$ at $\displaystyle \hat{p} = 0.5$, so we assume this value.
where does " f'(x) = 2p̂ −1. 2p̂ −1=0" come from?

November 30th, 2015, 05:20 AM   #9
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 Originally Posted by tttim where does " f'(x) = 2p̂ −1. 2p̂ −1=0" come from?
I expanded and differentiated $\displaystyle \hat{p}(1-\hat{p})$.

November 30th, 2015, 06:16 AM   #10
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 Originally Posted by 123qwerty I expanded and differentiated $\displaystyle \hat{p}(1-\hat{p})$.
I haven't learnt differentiation

 Tags confidence, interval, proportion, sample, size

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