Question

This is non graded practices question but i would like to know how to do it step by step.

a. The MGM Great Hotel in Las Vegas wants to estimate the mean time taken by its operators to make reservations. In a random sample of 100 reservation telephone calls to the reservation center (out of tens of thousands received during that time period), the mean time taken to complete a reservation is six minutes with a standard deviation of two minutes. Provide a 90% confidence interval for the mean time its operators take to complete a reservation. b. The owner of the hotel, Mr Gambel, after taking a look at your confidence interval in part (a), says "You know, I am really impressed with how precisely you can estimate the mean time to make a reservation based on a sample of only 100 I think we should use this same sample size for our hotel in Ailantic City, New Jersey. I've been concerned because while telephone operators at the hotel more or less have the same experience and have been with the hotel the same length of time, the Atlantic City hotel employs a broad mix of telephone operators. Some are experienced and efficient while others are new on the job and often have to ask questions of their fellow operators in the middle of a conversation with a customer. Let's get moving on the sampling in New Jersey. I can't wait to see the results." Do you think Mr. Gambel will be as happy with the precision of the results in New Jersey as he has been in Vegas? Please explain in no more than three sentences. Las Vegas

Answer 1

a) The confidence interval for mean is given by

$\left ( \bar{x}-{\sigma\over \sqrt{n}}z_{\alpha/2}\,,\,\bar{x}+{\sigma\over \sqrt{n}}z_{\alpha/2} \right )$

Where $\bar{x}=\text{sample mean}=6$

$\sigma=\text{population standard deviation}$ which can be approximated by sample standard deviation which is 2,

$n=\text{sample size}=100$ , $z_{\alpha/2}=\text{critical value for }~(1-\alpha)100\%~\text{confidence interval}=1.645\text{ for 90\%}$

So, the interval is

$\left ( 6-{2\over\sqrt{100}}(1.645)\,,\, 6+{2\over\sqrt{100}}(1.645)\right )=(5.671\,,\,6.329)$

b) The data will have outliers as some operators are experts so will take less time while others are freshers. Because of the outliers, the standard deviation will be large and hence the interval will be wider.