Question

A sample of 35 luxury hotels in the US had an average room rate of $252.80 per night. Assume the population standard deviation is $41 per night.

1.Calculate the margin of error if the confidence level is 90%.

2. Calculate the 90% confidence interval lower bound.

3.Calculate the 90% confidence interval upper bound

Answer 1

Given that, sample size ( n ) = 35

sample mean = $252.80

population standard deviation = $41

1) A 90% confidence level has significance level of 0.10 and critical value is,

Margin of error ( E ) is,

The margin of error is 11.4003

2) The 90% confidence interval is,

Where, lower bound = 241.3997 and upper bound = 264.2003

Therefore the 90% confidence interval lower bound = $241.3997 or $241.4 (rounded to 1 decimal place)

3) The 90% confidence interval upper bound = $264.2003

or $264.2 (rounded to 1 decimal place)