Question

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.

Answer 1

one approach is to use the confidence interval formula and determine n
`n=(z_(\alpha/2)^2*p*(1-p))/m^2`

we are given the following constraints `z_(\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 = ((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