# 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.

Jul 23, 2017

one approach is to use the confidence interval formula and determine n
$n = \frac{{z}_{\setminus \frac{\alpha}{2}}^{2} \cdot p \cdot \left(1 - p\right)}{m} ^ 2$

we are given the following constraints ${z}_{\setminus \frac{\alpha}{2}} = 2.33 , m = .03$

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

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

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