Question

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 York City was $2 the change in cost over the two-year period. 29. Discuss

Answer 1

a. Standard deviation (s) =$65

T-statistic (t) =2.015

So, Standard Error (se) =s/√45

=$65/√45=9.689

Margin of Error(me)=t*se

=2.015*9.689=$19.52

b. Confidence interval =average cost +-Margin of Error (me)

=$273+-$19.52

=$292.52 , $253.48

c. At 95% confidence, the population mean is between. $292.52 and $253.48. This interval (the entire span of the interval) is definitely above the $229 level of two years ago. So average hotel room rates have increased.