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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