Question

A large company is considering opening a franchise in St. Louis and wants to estimate the mean household income for the area using simple random sample of households.

Based on information from a pilot study, the company assumes that the standard deviation of household incomes is $\sigma =\$7,200$.

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 with $95$ percent confidence?

Answer 1

Find least number of household:

Given standard deviation $\sigma =\$7,200$,

Margin of error is $E=\$200$, and

From $z$-table we have, $95$ percent confidence represents, $z=1.96$

By $z$- formula we have, number of household is given by $n={\left(\frac{z\times \sigma }{E}\right)}^2$

Putting, $\sigma =\$7,200$, $E=\$200$ and $z=1.96$

$$\begin{gathered} n={\left(\frac{1.96\times 7200}{200}\right)}^2 \\ \Rightarrow n={\left(\frac{14112}{200}\right)}^2 \\ \Rightarrow n=4978.7136 \\ \Rightarrow n\approx 4979 \end{gathered}$$

Hence, the least number of households that should be surveyed is $4979$.