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A random sample of 13 hotels in a city had an average nighty room rate of...
Question:The average cost per night of a hotel room in New York City is $472. Assume...
Question:The average cost per night of a hotel room in New York City is $472. Assume this estimate is based on a sample of 75 hotels and that the sample standard deviation is $54. a. At 90% confidence, what is the interval estimate for the population mean? b. At 95% confidence, what is the interval estimate for the population mean? c. At 99% confidence, what is the interval estimate for the population mean? d. Describe the effect of a higher...
The average cost per night of a hotel room in New York City is $273 (SmartMoney,...
The average cost per night of a hotel room in New York City is $273 (SmartMoney, March 2009). Assume this estimate is based on a sample of 45 hotels and that the sample standard deviation is $65. T- statistic is 2.015. a. With 95% confidence, what is the Margin of Error(ME)? b. What is the 95% confidence interval estimate of the population mean? c. Two years ago the average cost of a hotel room in New York City was $229....
The average cost per night of a hotel room in New York City is $263 (SmartMoney,...
The average cost per night of a hotel room in New York City is $263 (SmartMoney, March 2009). Assume this estimate is based on a sample of 45 hotels and that the sample standard deviation is $50. a. With 95% confidence, what is the margin of error? Round your answer to the nearest dollar. b. What is the 95% confidence interval estimate of the population mean? Round your answers to the nearest dollar. c. Two years ago the average cost...
The average cost per night of a hotel room in New York City is $269 (SmartMoney,...
The average cost per night of a hotel room in New York City is $269 (SmartMoney, March 2009). Assume this estimate is based on a sample of 45 hotels and that the sample standard deviation is $55. a. With 95% confidence, what is the margin of error? Round your answer to the nearest dollar. $ b. What is the 95% confidence interval estimate of the population mean? Round your answers to the nearest dollar. to c. Two years ago the...
Problem 2- Confidence Interval The average cost per night of a hotel room in New York...
Problem 2- Confidence Interval The average cost per night of a hotel room in New York City is $273 (SmartMoney, March 2009). Assume this estimate is based on a sample of 45 hotels and that the sample standard deviation is $65. T- statistic is 2.015. a. b, c. With 95% confidence, what is the Margin of Error(ME)? what is the 95% confidence interval estimate of the population mean? Two years ago the average cost of a hotel room in New...
A random sample of 30 hotels and motels on a given Sunday in Monterey found the...
A random sample of 30 hotels and motels on a given Sunday in Monterey found the mean to be $100 with population standard deviation $30. Find a 90% confidence interval for the true mean
A random sample of 109 lightning flashes in a certain region resulted in a sample average...
A random sample of 109 lightning flashes in a certain region resulted in a sample average radar echo duration of 0.82 sec and a sample standard deviation of 0.41 sec. Calculate a 99% (two-sided) confidence interval for the true average echo duration u. (Give answers accurate to 3 decimal places.) (717 923 X) Interpret the resulting interval. o We are 99% confident that this interval does not contain the true population mean. . We are 99% confident that this interval...
A random sample of 13 customers in a Kentucky fast food restaurant revealed an average bill...
A random sample of 13 customers in a Kentucky fast food
restaurant revealed an average bill of OMR 18.75 per person. The
population standard deviation is OMR5.4. Estimate 98%
confidence
interval for the mean bill of all customers.
A random sample of ' n' customers in a Kentucky fast food restaurant revealed an average bill of OMR & per person. The population standard deviation is OMR 6. Estimate 98% confidence interval for the mean bill of all customers. n X...
Suppose that a random sample of ten recently sold houses in a certain city has a...
Suppose that a random sample of ten recently sold houses in a certain city has a mean sales price of $295,000, with a standard deviation of $14,000. Under the assumption that house prices are normally_distributed, find a 99% confidence interval for the mean sales price of all houses in this community. Then complete the table below. Carry your intermediate computations to at least three decimal places. Round your answers to the nearest whole number. (If necessary, consult a list of...
A study of the career paths of hotel general managers sent questionnaires to a random sample...
A study of the career paths of hotel general managers sent questionnaires to a random sample of 97 hotels belonging to major U.S. hotel chains. The average time these 97 general managers had spent with their current company was 12.75 years. (Take it as known that the standard deviation of time with the company for all general managers is 1.8 years.) (a) Find the bound for an 80% confidence interval to estimate the mean time a general manager had spent...
Problem 2- Confidence Interval The average cost per night of a hotel room in New York City is $273 (SmartMoney, March 2009). Assume this estimate is based on a sample of 45 hotels and that the sample standard deviation is $65. T- statistic is 2.015. a. b, c. With 95% confidence, what is the Margin of Error(ME)? what is the 95% confidence interval estimate of the population mean? Two years ago the average cost of a hotel room in New...
A random sample of 109 lightning flashes in a certain region resulted in a sample average radar echo duration of 0.82 sec and a sample standard deviation of 0.41 sec. Calculate a 99% (two-sided) confidence interval for the true average echo duration u. (Give answers accurate to 3 decimal places.) (717 923 X) Interpret the resulting interval. o We are 99% confident that this interval does not contain the true population mean. . We are 99% confident that this interval...
A random sample of 13 customers in a Kentucky fast food
restaurant revealed an average bill of OMR 18.75 per person. The
population standard deviation is OMR5.4. Estimate 98%
confidence
interval for the mean bill of all customers.
A random sample of ' n' customers in a Kentucky fast food restaurant revealed an average bill of OMR & per person. The population standard deviation is OMR 6. Estimate 98% confidence interval for the mean bill of all customers. n X...
Suppose that a random sample of ten recently sold houses in a certain city has a mean sales price of $295,000, with a standard deviation of $14,000. Under the assumption that house prices are normally_distributed, find a 99% confidence interval for the mean sales price of all houses in this community. Then complete the table below. Carry your intermediate computations to at least three decimal places. Round your answers to the nearest whole number. (If necessary, consult a list of...
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