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One hundred randomly chosen adult subjects in Jacksonville, FL, were asked their annual income.

The data reported from the sample showed a mean of $\$46{,}203$ with a standard deviation of $\$10{,}201$.

A $95\%$ confidence interval would show, approximately, that we are $95 \% $ confident the average annual income for United States residents is between

A

$\$36{,}002$ and $\$56{,}404$

B

$\$44{,}204$ and $\$48{,}202$

C

$\$25{,}801$ and $\$66{,}605$

D

$\$44{,}179$ and $\$48{,}227$

E

Assumptions are not met in this situation to compute an appropriate $95\%$ confidence interval.

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