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. 17.50 V60 17.50) B) 70 + 1.960- V60 A) 70 + 1.833 17.50 C) 70 ± 1.671 17.50 D) 70 + 1.645

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Chapter1: Combinatorial Analysis
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**Transcription of Text:**

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) \( 70 \pm 1.833 \left(\frac{17.50}{\sqrt{60}}\right) \)

B) \( 70 \pm 1.960 \left(\frac{17.50}{\sqrt{60}}\right) \)

C) \( 70 \pm 1.671 \left(\frac{17.50}{\sqrt{60}}\right) \)

D) \( 70 \pm 1.645 \left(\frac{17.50}{\sqrt{60}}\right) \)

**Explanation:**

This problem provides data to calculate a 95% confidence interval for the average amount of money a typical professional football fan spends on food at a game. The given options involve different critical values and use the sample mean ($70.00), standard deviation ($17.50), and sample size (60) in calculating the confidence interval. Each option represents a potential confidence interval using a different z-value multiplier.
Transcribed Image Text:**Transcription of Text:** 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) \( 70 \pm 1.833 \left(\frac{17.50}{\sqrt{60}}\right) \) B) \( 70 \pm 1.960 \left(\frac{17.50}{\sqrt{60}}\right) \) C) \( 70 \pm 1.671 \left(\frac{17.50}{\sqrt{60}}\right) \) D) \( 70 \pm 1.645 \left(\frac{17.50}{\sqrt{60}}\right) \) **Explanation:** This problem provides data to calculate a 95% confidence interval for the average amount of money a typical professional football fan spends on food at a game. The given options involve different critical values and use the sample mean ($70.00), standard deviation ($17.50), and sample size (60) in calculating the confidence interval. Each option represents a potential confidence interval using a different z-value multiplier.
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