a.
Least amount of money can be spent, when we have the minimum sample size to have margin of error 1.5% with 90% confidence interval.
Margin of error = 1.5% = 0.015
z value for 90% confidence interval is 1.645
As, we do not know the estimates of p, we take conservative
approach to have p = 0.5 which will give the required sample size
to have margin of error 1.5% with 90% confidence interval.
Minimum sample size = (z / E)2 * p(1-p) = (1.645 / 0.015)2 * 0.5 * (1 - 0.5) = 3007 (Rounded to nearest integer)
Least amount of money = 3007 * $5 = $15035
b.
Assuming that the upper limit of true proportion is 1/5 = 0.2
So, the true proportion would be estimated as difference between upper limit and margin of error = 0.2 - 0.015 = 0.185
Now,
Minimum sample size = (z / E)2 * p(1-p) = (1.645 / 0.015)2 * 0.185 * (1 - 0.185) = 1813 (Rounded to nearest integer)
New minimal cost estimate = 1813 * $5 = $9065
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