How much money does the average professional football fan spend on food at a single football game? That question was posed to 60 randomly selected football fans. The sampled results show
That the sample mean was $70.00 and prior sampling indicated that the population standard
Deviation was $17.50. Use this information to create a 95 percent confidence interval for the
Population mean.
A) $70 \pm 1.671 \left( \frac { 17.50 } { \sqrt { 60 } } \right)$
B) $70 \pm 1.833 \left( \frac { 17.50 } { \sqrt { 60 } } \right)$
C) $70 \pm 1.645 \left( \frac { 17.50 } { \sqrt { 60 } } \right)$
D) $70 \pm 1.960 \left( \frac { 17.50 } { \sqrt { 60 } } \right)$

Question

How much money does the average professional football fan spend on food at a single football game? That question was posed to 60 randomly selected football fans. The sampled results show
That the sample mean was $70.00 and prior sampling indicated that the population standard
Deviation was $17.50. Use this information to create a 95 percent confidence interval for the
Population mean. 
A) $70 \pm 1.671 \left( \frac { 17.50 } { \sqrt { 60 } } \right)$
B) $70 \pm 1.833 \left( \frac { 17.50 } { \sqrt { 60 } } \right)$
C) $70 \pm 1.645 \left( \frac { 17.50 } { \sqrt { 60 } } \right)$
D) $70 \pm 1.960 \left( \frac { 17.50 } { \sqrt { 60 } } \right)$

Quizplus · Accepted Answer

To construct a 95% confidence interval for the population mean when the population standard deviation is known, we use the Z-distribution. The Z-score corresponding to a 95% confidence level is 1.960. The formula for the confidence interval is $\mu \pm Z \left( \frac{\sigma}{\sqrt{n}} \right)$, where $\mu$ is the sample mean, $\sigma$ is the population standard deviation, and $n$ is the sample size. Therefore, option D correctly applies this formula with a Z-score of 1.960.

How Much Money Does the Average Professional Football Fan Spend 70±1.671(17.5060)70 \pm 1.671 \left( \frac { 17.50 } { \sqrt { 60 } } \right)70±1.671(60​17.50​)

How Much Money Does the Average Professional Football Fan Spend $70 \pm 1.671 \left( \frac { 17.50 } { \sqrt { 60 } } \right)$