## Related questions with answers

The U.S. Department of Transportation provides the number of miles that residents of the 75 largest metropolitan areas travel per day in a car. Suppose that for a simple random sample of 50 Buffalo residents the mean is $22.5$ miles a day and the standard deviation is $8.4$ miles a day, and for an independent simple random sample of 40 Boston residents the mean is $18.6$ miles a day and the standard deviation is $7.4$ miles a day.

What is the point estimate of the difference between the mean number of miles that Buffalo residents travel per day and the mean number of miles that Boston residents travel per day?

What is the $95 \%$ confidence interval for the difference between the two population means?

Solution

Verified**(a)** This is the first part of a two-part exercise, in which we are given details about samples drawn from residents of two cities. The data were about per day miles traveled in a car. Organized into a table, the given details are:

$\begin{array}{lll} & \rm Buffalo & \rm Boston\\ \rm Sample\,size & n_{1}=50 & n_{2}=40\\ \rm Sample\,mean & \bar x_{1}=22.5 & \bar x_{1}=18.6\\ \rm Sample\,standard\,deviation & s_{1}=8.4 & s_{2}=7.4 \end{array}$

Our job here is to compute the point estimate of the difference between the two population means.

*What is a point estimate?*

## Create an account to view solutions

## Create an account to view solutions

## Recommended textbook solutions

#### Business Statistics: Communicating with Numbers

2nd Edition•ISBN: 9780078020551Alison Kelly, Sanjiv Jaggia#### Basic Business Statistics: Concepts and Applications

12th Edition•ISBN: 9780132168380 (15 more)David M. Levine, Mark L. Berenson, Timothy C. Krehbiel#### Modern Business Statistics with Microsoft Excel

5th Edition•ISBN: 9781285433301David R. Anderson, Dennis J. Sweeney, Thomas A. Williams## More related questions

1/4

1/7