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 sample of 1004 adult drivers, and a bound on the error of estimation of \(3.1 \%\) was reported. Explain how the given bound on the error can be justified.

Short Answer

Expert verified
The error bound of 3.1% can be justified by using the Margin of Error formula with a 95% confidence level. The calculated result closely aligns with the given error bound.

Step by step solution

01

Understanding Margin of Error

The margin of error is an indicator of the estimation error. Assuming a confidence level of 95%, the z-score is approximately 1.96. The formula for margin of error (E) is \( E = Z * \sqrt{ \frac{P*(1-P)}{n} } \) where P is the sample proportion and n the sample size.
02

Apply values to Margin of Error formula

Let's substitute P = 0.36 (the proportion given in the exercise) and n = 1004 (the sample size) into the formula. The Z value is 1.96 for a 95% confidence interval.
03

Calculate the error bound

Perform the calculation: \( E = 1.96 * \sqrt{ \frac{0.36*(1-0.36)}{1004} } \). The calculated result will be the error estimation in decimal, you must convert it to percentage by multiplying the result by 100.
04

Validate the given error bound

After deriving the margin of error from the given data, compare it with the bound error given in the question (3.1%). If they match, it justifies why the error bound is valid.

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!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Sampling Error
When conducting a study, researchers often use a sample—a subset of the population—to draw conclusions about the population as a whole. A sampling error occurs due to the natural variation between the sample and the actual population. It's the difference between the sample statistic (like a sample mean or proportion) and the population parameter it's meant to estimate. The size of the sampling error can fluctuate depending on the size of the sample and the variance within the population.

In the case of the USA Today survey, the sample consisted of 1004 adult drivers. The sampling error in this context reflects the uncertainty that arises because only a portion of all adult drivers were surveyed. Although every adult driver has an equal chance of being included in the sample, each sample can yield a slightly different result. This is why the bound on the error of estimation is crucial—it gives us a way to measure and report this uncertainty.
Confidence Interval
A confidence interval is a range of values that is likely to contain a population parameter with a certain degree of confidence. It’s constructed around a sample statistic to capture the true population parameter. For example, if researchers report a 95% confidence interval for a proportion, it means that if they were to take numerous random samples and calculate a confidence interval for each sample, they expect about 95% of those intervals to contain the true population proportion.

The width of the confidence interval is determined by several factors, including the size of the sample, the level of confidence one desires, and the variability of the data. In the example question provided, the margin of error reported (3.1%) is part of the confidence interval. When we say that 36% of adult drivers admit to using their cell while driving with a margin of error of 3.1%, we are expressing a 95% confidence interval that the true proportion in the entire population would fall between 32.9% (36% - 3.1%) and 39.1% (36% + 3.1%).
Sample Proportion Calculation
Calculating the sample proportion involves taking the number of occurrences of a certain event in a sample and dividing it by the total sample size. It is represented by 'P' in statistical formulas. The formula for the sample proportion is \( P = \frac{x}{n} \) where 'x' is the number of successes and 'n' is the total number of observations in the sample.

The sample proportion calculation in the context of the USA Today survey indicated that 36% of adult drivers often or sometimes talk on a phone while driving, which translates to 0.36 as the sample proportion (P). To verify this with error bounds, statisticians use the margin of error formula: \( E = Z * \sqrt{ \frac{P*(1-P)}{n} } \). By inserting the values of the sample proportion (P=0.36) and sample size (n=1004) into the formula and multiplying by the z-score that corresponds to the desired confidence level, we can calculate the error bound, which helps gauge the precision of the estimated proportion. This calculated bound should ideally be close to the reported bound on the error of estimation, which is a measure of reliability for the reported percentage.

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 article "National Geographic, the Doomsday Machine," which appeared in the March 1976 issue of the Journal of Irreproducible Results (yes, there really is a journal by that name -it's a spoof of technical journals!) predicted dire consequences resulting from a nationwide buildup of National Geographic magazines. The author's predictions are based on the observation that the number of subscriptions for National Geographic is on the rise and that no one ever throws away a copy of National Geographic. A key to the analysis presented in the article is the weight of an issue of the magazine. Suppose that you were assigned the task of estimating the average weight of an issue of National Geographic. How many issues should you sample to estimate the average weight to within \(0.1 \mathrm{oz}\) with \(95 \%\) confidence? Assume that \(\sigma\) is known to be 1 oz.

For each of the following choices, explain which would result in a wider large-sample confidence interval for \(\pi\) : a. \(90 \%\) confidence level or \(95 \%\) confidence level b. \(n=100\) or \(n=400\)

Despite protests from civil libertarians and gay rights activists, many people favor mandatory AIDS testing of certain at-risk groups, and some people even believe that all citizens should be tested. What proportion of the adults in the United States favor mandatory testing for all citizens? To assess public opinion on this issue, researchers conducted a survey of 1014 randomly selected adult U.S. citizens ("Large Majorities Continue to Back AIDS Testing," Gallup Poll Monthly [1991]: 25-28). The article reported that 466 of the 1014 people surveyed believed that all citizens should be tested. Use this information to estimate \(\pi\), the true proportion of all U.S. adults who favor AIDS testing of all citizens.

The eating habits of 12 bats were examined in the article "Foraging Behavior of the Indian False Vampire Bat" (Biotropica \([1991]: 63-67) .\) These bats consume insects and frogs. For these 12 bats, the mean time to consume a frog was \(\bar{x}=21.9 \mathrm{~min}\). Suppose that the standard deviation was \(s=7.7 \mathrm{~min} .\) Construct and interpret a \(90 \%\) confidence interval for the mean suppertime of a vampire bat whose meal consists of a frog. What assumptions must be reasonable for the one-sample \(t\) interval to be appropriate?

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 a \(90 \%\) confidence interval for the proportion of adult Americans who think lying is never justified. b. Construct a \(90 \%\) confidence interval for the proportion of adult American who think that it is often or sometimes OK to lie to avoid hurting someone's feelings. c. Based on the confidence intervals from Parts (a) and (b), comment on the apparent inconsistency in the responses given by the individuals in this sample.

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