Question

Solve the problem. How much money does the average professional football fan spend on food at a single football game? That question was posed to ten randomly selected football fans. The sampled results show that the sample mean and sample standard deviation were $70.00 and $17.50, respectively. Use this information to create a 95 percent confidence interval for the population mean. $$70 \pm 1.833\left(\frac{17.50}{\sqrt{10}}\right)$$ $$70 \pm 2.228\left(\frac{17.50}{\sqrt{10}}\right)$$ $$70 \pm 2.262\left(\frac{17.50}{\sqrt{10}}\right)$$ $$70 \pm 1.960\left(\frac{17.50}{\sqrt{10}}\right)$$

          Solve the problem.

How much money does the average professional football fan spend on food at a single football game? That question was posed to ten randomly selected football fans. The sampled results show that the sample mean and sample standard deviation were $70.00 and $17.50, respectively. Use this information to create a 95 percent confidence interval for the population mean.

$$70 \pm 1.833\left(\frac{17.50}{\sqrt{10}}\right)$$

$$70 \pm 2.228\left(\frac{17.50}{\sqrt{10}}\right)$$

$$70 \pm 2.262\left(\frac{17.50}{\sqrt{10}}\right)$$

$$70 \pm 1.960\left(\frac{17.50}{\sqrt{10}}\right)$$

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solve the problem how much money does the average professional football fan spend on food at a single football game that question was posed to ten randomly selected football fans the sampled 25326

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Elementary Statistics a Step by Step Approach

Allan G. Bluman 9th Edition

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Solved by Expert Madhur L

05/11/2023

Step 1

We are given the following information: Sample size, n = 10 Sample mean, $\bar{x}$ = $70.00 Sample standard deviation, s = $17.50 Confidence level = 95% Since the population standard deviation is unknown and the sample size is small (n < 30), we should use the Show more…

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Solve the problem. How much money does the average professional football fan spend on food at a single football game? That question was posed to ten randomly selected football fans. The sampled results show that the sample mean and sample standard deviation were $70.00 and $17.50, respectively. Use this information to create a 95 percent confidence interval for the population mean. $$70 \pm 1.833\left(\frac{17.50}{\sqrt{10}}\right)$$ $$70 \pm 2.228\left(\frac{17.50}{\sqrt{10}}\right)$$ $$70 \pm 2.262\left(\frac{17.50}{\sqrt{10}}\right)$$ $$70 \pm 1.960\left(\frac{17.50}{\sqrt{10}}\right)$$

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