Hotels want to gather information about their rooms, they take a random sample of 52 rooms to produce a mean cost of $208.35 a night. If the population standard deviation of costs for a one-night stay is $47.45. 1. Construct a 99% confidence interval for the average cost per night in a hotel. 2. What is the value of your point estimate for part 1? 3. Construct a 95% confidence interval for the average cost per night in a hotel. 4. What is the value of your margin of error for part 3?
Hotels want to gather information about their rooms, they take a random sample of 52 rooms to produce a mean cost of $208.35 a night. If the population standard deviation of costs for a one-night stay is $47.45.
1. Construct a 99% confidence interval for the average cost per night in a hotel.
2. What is the value of your point estimate for part 1?
3. Construct a 95% confidence interval for the average cost per night in a hotel.
4. What is the value of your margin of error for part 3?
Solution:
1.
Let X be the Cost for a one-night stay.
From the given information, x-bar=$208.35, n=52 and σ=$47.45.
The confidence level is 0.99.
Then, a 99% confidence interval for the average cost per night in a hotel is
2.
The point is the sample mean.
Thus, the value of the point estimate for part 1 is $208.35.
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