Question

1. A 2009 sample of New York hotel room prices had an average of $273. Assume this was from a sample of 48 hotels with a sample standard deviation of $72. Construct a 95% confidence interval and report the lower bound for the interval. (Answer to one decimal place and do not include a $ sign).

2.A 2009 sample of New York hotel room prices had an average of $273. Assume this was from a sample of 48 hotels with a sample standard deviation of $72. Construct a 95% confidence interval and report the upper bound for the interval. Your number should be larger than for the previous problem. (Answer to one decimal place and do not include a $ sign).

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 273

sample standard deviation = s = 72

sample size = n = 48

Degrees of freedom = df = n - 1 = 48 - 1 = 47

At 95% confidence level

= 1 - 95%

=1 - 0.95 =0.05

/2 = 0.025

t/2,df = t0.025,47 = 2.012

Margin of error = E = t/2,df * (s /n)

= 2.012 * ( 72 / 48)

Margin of error = E = 20.9

1) The 95% lower bound confidence interval estimate of the population mean is,

- E

= 273 - 20.9 = 252.1

lower bound = 252.1

2) The 95% upper bound confidence interval estimate of the population mean is,

+ E

= 273 + 20.9 = 293.9

upper bound = 293.9