Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 8: Problem 29

The travel-to-work time for residents of the 15 largest cities in the United States is reported in the 2003 Information Please Almanac. Suppose that a preliminary simple random sample of residents of San Francisco is used to develop a planning value of 6.25 minutes for the population standard deviation. a. If we want to estimate the population mean travel-to-work time for San Francisco residents with a margin of error of 2 minutes, what sample size should be used? Assume \(95 \%\) confidence b. If we want to estimate the population mean travel-to-work time for San Francisco residents with a margin of error of 1 minute, what sample size should be used? Assume \(95 \%\) confidence

Short Answer

Expert verified
a. To estimate the population mean travel-to-work time for San Francisco residents with a margin of error of 2 minutes, a sample size of \(37\) should be used. b. To estimate the population mean travel-to-work time for San Francisco residents with a margin of error of 1 minute, a sample size of \(145\) should be used.

Step by step solution

01

Find the Z-score

For a \(95\%\) confidence interval, the Z-score is \(Z_\frac{α}{2} = Z_{0.975}\), which is approximately \(1.96\). You can find this value using a standard normal table or a calculator.
02

Determine the Margin of Error

For part a, the margin of error is 2 minutes. For part b, the margin of error is 1 minute.
03

Solve for the Sample Size

Rearrange the formula for the margin of error to solve for the sample size, n: \(n = (\frac{Z_\frac{α}{2} * σ}{Margin\ of\ Error})^2\)
04

Calculate the Sample Size for Part A

Use the given values to calculate the sample size for a margin of error of 2 minutes: \(n = (\frac{1.96 * 6.25}{2})^2\) \(n \approx 36.03\) Round up to the nearest whole number because you can't have a fraction of a sample. So, the required sample size for part a is 37.
05

Calculate the Sample Size for Part B

Use the given values to calculate the sample size for a margin of error of 1 minute: \(n = (\frac{1.96 * 6.25}{1})^2\) \(n \approx 144.12\) Round up to the nearest whole number because you can't have a fraction of a sample. So, the required sample size for part b is 145.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Disney's Hannah Montana: The Movie opened on Easter weekend in April 2009. Over the three-day weekend, the movie became the number-one box office attraction (The Wall Street Journal, April 13,2009 ). The ticket sales revenue in dollars for a sample of 25 theaters is as follows. $$\begin{array}{rrrrr} 20,200 & 10,150 & 13,000 & 11,320 & 9,700 \\ 8,350 & 7,300 & 14,000 & 9,940 & 11,200 \\ 10,750 & 6,240 & 12,700 & 7,430 & 13,500 \\ 13,900 & 4,200 & 6,750 & 6,700 & 9,330 \\ 13,185 & 9,200 & 21,400 & 11,380 & 10,800 \end{array}$$ a. What is the \(95 \%\) confidence interval estimate for the mean ticket sales revenue per theater? Interpret this result. b. Using the movie ticket price of \(\$ 7.16\) per ticket, what is the estimate of the mean number of customers per theater? c. The movie was shown in 3118 theaters. Estimate the total number of customers who saw Hannah Montana: The Movie and the total box office ticket sales for the threeday weekend.

If a perfectly competitive firm sells 100 units of output at a market price of \(\$ 100\) per unit, its marginal revenue per unit is a. \(\$ 1\) b. \(\$ 100\) c. more than \(\$ 1,\) but less than \(\$ 100\). d. less than \(\$ 100\).

The 2003 Statistical Abstract of the United States reported the percentage of people 18 years of age and older who smoke. Suppose that a study designed to collect new data on smokers and nonsmokers uses a preliminary estimate of the proportion who smoke of .30 a. How large a sample should be taken to estimate the proportion of smokers in the population with a margin of error of \(.02 ?\) Use \(95 \%\) confidence. b. Assume that the study uses your sample size recommendation in part (a) and finds 520 smokers. What is the point estimate of the proportion of smokers in the population? c. What is the \(95 \%\) confidence interval for the proportion of smokers in the population?

The average cost per night of a hotel room in New York City is \(\$ 273\) (SmartMoney, March 2009 . Assume this estimate is based on a sample of 45 hotels and that the sample standard deviation is \(\$ 65\) a. With \(95 \%\) confidence, what is the margin of error? b. What is the \(95 \%\) confidence interval estimate of the population mean? c. Two years ago the average cost of a hotel room in New York City was \(\$ 229 .\) Discuss the change in cost over the two-year period.

Find the \(t\) value(s) for each of the following cases. a. Upper tail area of .025 with 12 degrees of freedom b. Lower tail area of .05 with 50 degrees of freedom c. Upper tail area of .01 with 30 degrees of freedom d. Where \(90 \%\) of the area falls between these two \(t\) values with 25 degrees of freedom e. Where \(95 \%\) of the area falls between these two \(t\) values with 45 degrees of freedom
See all solutions

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free