A travel agent wants to gather information on the per-night cost at hotels in Caribbean countries. She took a random sample of 52 rooms from various hotels in those countries. The sample produced a mean cost for the 52 rooms to be $208.35 per night. If the population standard deviation of costs for a one-night stay in Caribbean hotels is $47.45, find a 99% confidence interval for the average cost per night in Caribbean hotels.
Answer to relevant QuestionsA city planner wants to estimate the average monthly residential water usage in the city. He selected a random sample of 40 households from the city, which gave the mean water usage to be 3415.70 gallons over a 1-month ...A bank manager wants to know the mean amount of mortgage paid per month by homeowners in an area. A random sample of 120 homeowners selected from this area showed that they pay an average of $1575 per month for their ...Briefly explain the similarities and the differences between the standard normal distribution and the t distribution. For each of the following, find the area in the appropriate tail of the t distribution. a. t = –1.302 and df = 42 b. t = 2.797 and n = 25 c. t = 1.397 and n = 9 d. t = –2.383 and df = 67 a. A sample of 400 observations taken from a population produced a sample mean equal to 92.45 and a standard deviation equal to 12.20. Make a 98% confidence interval for µ b. Another sample of 400 observations taken from ...
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