# statistics (7)

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. With 95% confidence, what is the margin of error (to 2 decimals)?

b. What is the 95% confidence interval estimate of the population mean?

NOTE: This is what I got but it is wrong.
a. 18.99
b. [254.008, 291.992]

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1. a) I think the question requires the use of the t-distribution.

standard error=statistic/sqrt(45)
=65/sqrt(45)=9.689
critical value: df=45-1=44, area=.025 so t=2.015

margin of error=t*SE=19.52

b) Confidence interval: Use t-distr: sample mean +- margin of error
=273 +- 19.52
=(253.48, 292.52)

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2. 56

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