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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?

User Nroose
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2 Answers

1 vote

the answer is actually 5,200

source: ap stats quiz and quizlet

User Mluc
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3 votes

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.

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}* (\sigma)/(√(n))


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


1.96* (7200)/(√(n)) < 200\\\\n > (1.96* 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.

User Foredecker
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