Hoping to attract more shoppers,a city builds a new public parking garage downtown.The city plans to pay for the structure through parking fees.The consultant,who advised the city on this project,randomly selected 44 weekdays.Daily fees collected averaged $126.Based on data from other parking structures,the consultant will assume parking fees at this parking garage are Normally distributed with a standard deviation of $15. What is a 90% confidence interval for the mean daily income this parking garage will generate from the parking fees?
A)($107.31,$144.69)
B)($121.57,$130.43)
C)($122.28,$129.72)
D)($125.44,$126.56)

Question

Quizplus · Accepted Answer

First, we need to calculate the margin of error:
Margin of Error = z * (standard deviation / sqrt(sample size))
We are looking for a 90% confidence interval, so the z-value is 1.645.
Standard deviation is given as $15.
Sample size is 44.

Margin of Error = 1.645 * ($15 / sqrt(44)) ≈ $4.72

Next, we need to calculate the mean of the sample:
Sample mean = $126

Finally, we can construct the confidence interval:
Confidence interval = Sample mean ± Margin of Error
Confidence interval = $126 ± $4.72
Confidence interval ≈ ($122.28, $129.72)

Therefore, the best choice is C as it provides the correct 90% confidence interval for the mean daily income this parking garage will generate from the parking fees.

Hoping to Attract More Shoppers,a City Builds a New Public