Question

A random sample of 17 hotels in Orlando had an average nightly room rate of $165.40 with a sample standard deviation of $21.70. The margin of error for a 95% confidence interval around this sample mean is

Answer 1

Margin of Error = 11.16

Explanation:

A random sample of 17 hotels in Orlando had an average nightly room rate of $165.40 with a sample standard deviation of $21.70. The margin of error for a 95% confidence interval around this sample mean is

The formula for Margin of Error =

z × Standard deviation/√n

z = z score of 95% confidence Interval = 1.96

Standard deviation = $21.70

Random sample = 17

1.96 × 21.70/√17

= 11.16

Margin of Error = 11.16