Question

chloe and her family are attending a baseball game at a local stadium. chloe is interested in statistics about the stadium on a \typical\ game day. she asked the stadium manager for estimated means, margins of error and confidence intervals.
step 1: the stadium has concession stands around many of the entrances. the estimated mean for the amount spent at a visit to the concession stand is $8.45 ± $1.25. what is the confidence interval for the amount spent?
step 2: on a typical game day, the concession stand has 26,400 visits. what is the range of how much money the concession stand brings in on a typical game day?
step 3: chloe found out that the confidence interval for the number of people that attend each baseball game is 41,640 to 43,480. what is the estimated mean and margin of error? verify your answer is correct.
step 4: there have been twelve home games so far this season and the manager says that 515,340 tickets to get into the games have been purchased in all. does this statistic seem reasonable based on the information given in step 3? explain.

Explanation:

Step1: Calculate the confidence interval for Step1

The confidence interval is given by the formula \(\text{Mean}\pm\text{Margin of Error}\). Here, the confidence interval is \(\$8.45\pm\$1.25\).

Step2: Calculate the range for Step2

The lower bound of the amount spent per visit is \(8.45 - 1.25=\$7.2\) and the upper bound is \(8.45+ 1.25=\$9.7\).
For \(n = 26400\) visits, the lower - bound of the money brought in is \(26400\times7.2=\$190080\) and the upper - bound is \(26400\times9.7=\$256080\)

Step3: Calculate mean and margin of error for Step3

The formula for the confidence interval is \(\text{Mean}-\text{Margin of Error}\) (lower bound) and \(\text{Mean}+\text{Margin of Error}\) (upper bound). Let the mean be \(\mu\) and margin of error be \(E\).
We have \(\mu - E=41640\) and \(\mu + E=43480\).
Adding these two equations: \((\mu - E)+(\mu + E)=41640 + 43480\), \(2\mu=85120\), \(\mu=\frac{85120}{2}=42560\)
Subtracting the first equation from the second: \((\mu + E)-(\mu - E)=43480 - 41640\), \(2E = 1840\), \(E = 920\)

Step4: Check the reasonability for Step4

The mean number of people per game from Step3 is \(\mu = 42560\). For \(n = 12\) games, the expected number of tickets sold is \(12\times42560=510720\)
The actual number of tickets sold is \(515340\). The difference is \(515340-510720 = 4620\). Since \(4620\) is relatively small compared to \(510720\), the statistic seems reasonable.

Answer:

• Step1: The confidence interval for the amount spent is from \(\$7.2\) to \(\$9.7\)
• Step2: The range of money brought in is from \(\$190080\) to \(\$256080\)
• Step3: The estimated mean is \(42560\) and the margin of error is \(920\)
• Step4: The statistic of \(515340\) tickets sold seems reasonable.