A research firm wants to estimate the true population proportion of families having annual incomes that exceed $100,000 in a particular city. If it costs $20 to obtain each sample value, how large must the sampling budget be?"

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A research firm wants to estimate the true population proportion of families having annual incomes that exceed $100,000 in a particular city. If it costs $20 to obtain each sample value, how large must the sampling budget be?"

The firm would like to have 98% confidence, and a margin of error no larger than 3 percentage points, but currently has no information regarding how large the true population proportion might be.

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Answer 1 · 2017-07-23T17:41:05

one approach is to use the confidence interval formula and determine n
$n = \frac{z_{\frac{α}{2}}^{2} \cdot p \cdot (1 - p)}{m^{2}}$ $n=(z_(\alpha/2)^2*p*(1-p))/m^2$

we are given the following constraints $z_{\frac{α}{2}} = 2.33, m = .03$ $z_(\alpha/2) =2.33, m = .03$

We don't know the variance in order to solve for $n$ $n$ but we can always take the most conservative estimate of $p = .5$ $p=.5$

$n = \frac{{(2.33)}^{2} \cdot .25}{{.03}^{2}} = 1508$ $n = ((2.33)^2*.25)/.03^2 = 1508$

so we know we shouldn't need to select any more than 1508 users at worst case, the budget can be reduced to 1508*20 = $30,160

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A research firm wants to estimate the true population proportion of families having annual incomes that exceed $100,000 in a particular city. If it costs $20 to obtain each sample value, how large must the sampling budget be?"

The firm would like to have 98% confidence, and a margin of error no larger than 3 percentage points, but currently has no information regarding how large the true population proportion might be.

1 Answer

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