Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

USA Today (October 14,2002 ) reported that \(36 \%\) of adult drivers admit that they often or sometimes talk on a cell phone when driving. This estimate was based on data from a representative sample of 1,004 adult drivers. A margin of error of \(3.1 \%\) was also reported. Is this margin of error correct? Explain.

Short Answer

Expert verified
To find out if the 3.1% margin of error quoted in the exercise is correct, we need to use the formula for the margin of error and compare the result to the value provided. It's essential to show how to apply the margin of error formula meticulously.

Step by step solution

01

Define the Variables

Firstly, we identify the variables from the problem. The sample proportion \(p\) is 0.36 (36%). The sample size \(n\), equals 1,004.
02

Apply Margin of Error Formula

We apply the margin of error formula \(E = Z \sqrt{\frac{{p(1-p)}}{n}}\). Here, we use the standard \(Z\) value of 1.96 for a 95% confidence interval, \(p=0.36\), and \(n=1004\). Then calculate the value.
03

Compare Calculated Margin of Error with Given Margin of Error

After calculating the margin of error, we compare it with the given margin of error of 3.1%. If they are close or approximately equal, the stated margin of error would be assumed correct. Otherwise, it isn't correct.

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

One thousand randomly selected adult Americans participated in a survey conducted by the Associated Press (June, 2006). When asked "Do you think it is sometimes justified to lie, or do you think lying is never justified?" \(52 \%\) responded that lying was never justified. When asked about lying to avoid hurting someone's feelings, 650 responded that this was often or sometimes OK. a. Construct and interpret a \(90 \%\) confidence interval for the proportion of adult Americans who would say that lying is never justified. b. Construct and interpret a \(90 \%\) confidence interval for the proportion of adult Americans who think that it is often or sometimes OK to lie to avoid hurting someone's feelings. c. Using the confidence intervals from Parts (a) and (b), comment on the apparent inconsistency in the responses.

Suppose that a campus bookstore manager wants to know the proportion of students at the college who purchase some or all of their textbooks online. Two different people independently selected random samples of students at the college and used their sample data to construct the following confidence intervals for the population proportion: Interval 1:(0.54,0.57) Interval 2:(0.46,0.62) a. Explain how it is possible that the two confidence intervals are not centered in the same place. b. Which of the two intervals conveys more precise information about the value of the population proportion? c. If both confidence intervals have a \(95 \%\) confidence level, which confidence interval was based on the smaller sample size? How can you tell? d. If both confidence intervals were based on the same sample size, which interval has the higher confidence level? How can you tell?

Suppose that 935 smokers each received a nicotine patch, which delivers nicotine to the bloodstream at a much slower rate than cigarettes do. Dosage was decreased to 0 over a 12 -week period. Of these 935 people, 245 were still not smoking 6 months after treatment. Assume this sample is representative of all smokers. a. Use the given information to estimate the proportion of all smokers who, when given this treatment, would refrain from smoking for at least 6 months. b. Verify that the conditions needed in order for the margin of error formula to be appropriate are met. c. Calculate the margin of error. d. Interpret the margin of error in the context of this problem.

Consider taking a random sample from a population with \(p=0.25\) a. What is the standard error of \(\hat{p}\) for random samples of size \(400 ?\) b. Would the standard error of \(\hat{p}\) be smaller for random samples of size 200 or samples of size \(400 ?\) c. Does cutting the sample size in half from 400 to 200 double the standard error of \(\hat{p} ?\)

Consumption of fast food is a topic of interest to researchers in the field of nutrition. The article "Effects of Fast-Food Consumption on Energy Intake and Diet Quality Among Children" (Pediatrics [2004]: 112-118) reported that 1,720 of those in a random sample of 6,212 American children indicated that they ate fast food on a typical day. a. Use the given information to estimate the proportion of American children who eat fast food on a typical day. b. Verify that the conditions needed in order for the margin of error formula to be appropriate are met. c. Calculate the margin of error. d. Interpret the margin of error in the context of this problem.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

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