Question

A Census Bureau report on the income of Americans says that with 90% confidence the median income of all U.S. households in a recent year was \(57,005 with a margin of error of $\pm$\)742. This means that

(a) 90% of all households had incomes in the range \(57,005 $\pm$\)742.

(b) we can be sure that the median income for all households in the country lies in the range \(57,005 $\pm$\)742.

(c) 90% of the households in the sample interviewed by the Census Bureau had incomes in the range \(57,005 $\pm$\)742.

(d) the Census Bureau got the result \(57,005 $\pm$\)742 using a method that will cover the true median income 90% of the time when used repeatedly.

(e) 90% of all possible samples of this same size would result in a sample median that falls within \(742 of \)57,005.

Short answer

This means that (d) the Census Bureau got the result $\mathrm{\$}57,005\pm \mathrm{\$}742$ using a method that will cover the true median income$90\%$ of the time when used repeatedly.

Step-by-step solution

introduction

Given,

The income of Americans says that with $90\%$confidence the median income of all U.S. households in a recent year was $\$575005$ with a margin of error of$\$742$.

explanation

The confidence interval says assuming the many examples of the same size from a similar populace are taken out then there is assurance that genuine population mean will lie between the scope of two qualities. Here, one might say that acquired the outcome $\mathrm{\$}57,005\pm \mathrm{\$}742$by applying a strategy that is probably going to cover the median pay $90\%$of the time assuming it is done commonly.