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.

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

The following sample data are from a normal population: 10,8,12,15,13,11,6,5 a. What is the point estimate of the population mean? b. What is the point estimate of the population standard deviation? c. With \(95 \%\) confidence, what is the margin of error for the estimation of the population mean? d. What is the \(95 \%\) confidence interval for the population mean?

A survey conducted by the American Automobile Association showed that a family of four spends an average of \(\$ 215.60\) per day while on vacation. Suppose a sample of 64 families of four vacationing at Niagara Falls resulted in a sample mean of \(\$ 252.45\) per day and a sample standard deviation of \(\$ 74.50\) a. Develop a \(95 \%\) confidence interval estimate of the mean amount spent per day by a family of four visiting Niagara Falls. b. Based on the confidence interval from part (a), does it appear that the population mean amount spent per day by families visiting Niagara Falls differs from the mean reported by the American Automobile Association? Explain.

A simple random sample with \(n=54\) provided a sample mean of 22.5 and a sample standard deviation of 4.4 a. Develop a \(90 \%\) confidence interval for the population mean. b. Develop a \(95 \%\) confidence interval for the population mean. c. Develop a \(99 \%\) confidence interval for the population mean. d. What happens to the margin of error and the confidence interval as the confidence level is increased?

The National Center for Education Statistics reported that \(47 \%\) of college students work to pay for tuition and living expenses. Assume that a sample of 450 college students was used in the study. a. Provide a \(95 \%\) confidence interval for the population proportion of college students who work to pay for tuition and living expenses. b. Provide a \(99 \%\) confidence interval for the population proportion of college students who work to pay for tuition and living expenses. c. What happens to the margin of error as the confidence is increased from \(95 \%\) to \(99 \% ?\)

For a \(t\) distribution with 16 degrees of freedom, find the area, or probability, in each region. a. To the right of 2.120 b. To the left of 1.337 c. To the left of -1.746 d. To the right of 2.583 e. Between -2.120 and 2.120 f. Between -1.746 and 1.746

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