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
Jul 23, 2017

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