Question

A large company is considering opening a franchise in St. Louis and wants to estimate the mean household income for the area using a simple random sample of households. Based on information from a pilot study, the company assumes that the standard deviation of household incomes is σ = $7,200. Of the following, which is the least number of households that should be surveyed to obtain an estimate that is within $200 of the true mean household income with 95 percent confidence?

Answer 1

Answer:

4979  is the least number of households that should be surveyed to obtain an estimate that is within $200 of the true mean household income with 95 percent confidence.      

Step-by-step explanation:

We are given the following in the question:

Standard Deviation, σ = $7,200

Margin of error = $\pm\ 200$

We have to find the least number of households that should be surveyed to obtain an estimate that is within $200.

Formula:

$\text{Margin of error} = \text{Test Statistic}\times \dfrac{\sigma}{\sqrt{n}}$

$z_{critical}\text{ at}~\alpha_{0.05} = 1.96$

$1.96\times \dfrac{7200}{\sqrt{n}} < 200\\\\n > \dfrac{1.96\times 7200}{200} = 4978.7 \approx 4979$

4979  is the least number of households that should be surveyed to obtain an estimate that is within $200 of the true mean household income with 95 percent confidence.

Answer 2

the answer is actually 5,200

source: ap stats quiz and quizlet