**PLEASE NOTE** - I am using Excel and it is CRUCIAL that I know the excel funct
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Question
**PLEASE NOTE** - I am using Excel and it is CRUCIAL that I know the excel functions to discover the answers. PLEASE HELP ME UNDERSTAND THIS BY LISTING THEM!!! Finding the answer isn't as hard for me as being able to input the functions via excel. THANK YOU SO MUCH!!!!
**PLEASE NOTE** - I am using Excel and it is CRUCIAL that I know the excel functions to discover the answers. PLEASE HELP ME UNDERSTAND THIS BY LISTING THEM!!! Finding the answer isn't as hard for me as being able to input the functions via excel. THANK YOU SO MUCH!!!!
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. a) Construct a 95% confidence interval of the mean cost of a hotel room. Did you use the z-distribution or the t-distribution Tell me why. b) Give a one-sentence interpretation of the confidence interval c) Two years ago the average cost of a hotel room in New York City was $229. Discuss the change in cost over the two-year period. Be sure to show all calculations. and label all answers appropriately.Explanation / Answer
a) Excel Function to find margin of error,
"=CONFIDENCE.T(0.05, 65, 45)" = 18.53 Note:first argument is the significance level which is 1-0.95 = 0.05, second argument is the sample standard deviation and third is the sample size.
This provides the margin of error. The confidence interval is sample mean +- margin of error
95% CI is (273 - 19.53, 273 + 19.53) = (253.47, 292.53)
T-distribution was used as the population standard deviation is not known.
b) 95% Confident that the population mean of cost per night of a hotel room in New York city falls between $253.47 and $292.53.
c) "=[(273-229)/229]*100" = 19.2%
Over the 2 years, the average cost of hotel rooms per night in New York city has increased by 19.2%
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