Answer:
99% CI: [45.60; 58.00]min
Step-by-step explanation:
Hello!
Your study variable is:
X: Time a customer stays in a certain restaurant. (min)
X~N(μ; σ²)
The population standard distribution is σ= 17 min
Sample n= 50
Sample mean X[bar]= 51.8 min
Sample standard deviation S= 27.68
You are asked to construct a 99% Confidence Interval. Since the variable has a normal distribution and the population variance is known, the statistic to use is the standard normal Z. The formula to construct the interval is:
X[bar] ± *(σ/√n)
Upper level: 51.8 - 2.58*(17/√50) = 45.5972 ≅ 45.60 min
Lower level: 51.8 + 2.58*(17/√50) = 58.0027 ≅58.00 min
With a confidence level of 99%, you'd expect that the interval [45.60; 58.00]min will contain the true value of the average time customers spend in a certain restaurant.
I hope you have a SUPER day!
PS: Missing Data in the attached files.
