/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 35 In a recent National Survey of D... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

In a recent National Survey of Drug Use and Health, 2312 of 5914 randomly selected full-time U.S. college students were classified as binge drinkers. \({ }^{13}\) (a) Calculate and interpret a \(99 \%\) confidence interval for the population proportion \(p\) that are binge drinkers. (b) A newspaper article claims that \(45 \%\) of full-time U.S. college students are binge drinkers. Use your result from part (a) to comment on this claim.

Short Answer

Expert verified
The 99% confidence interval is (37.46%, 40.72%), contradicting the 45% claim.

Step by step solution

01

Calculate Sample Proportion

To begin, calculate the sample proportion of binge drinkers. The sample proportion \( \hat{p} \) is given by \( \hat{p} = \frac{x}{n} \), where \( x \) is the number of binge drinkers and \( n \) is the total number of surveyed students. Here, \( x = 2312 \) and \( n = 5914 \). Thus, \( \hat{p} = \frac{2312}{5914} \approx 0.3909 \).
02

Determine Standard Error

Next, calculate the standard error (SE) for the sample proportion. The formula for the standard error is \( SE = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \). Substituting the values, \( SE = \sqrt{\frac{0.3909 \times (1-0.3909)}{5914}} \approx 0.00632 \).
03

Find Critical Value for 99% Confidence Interval

For a 99% confidence interval, find the critical value (z*) by looking it up in the standard normal distribution table. The critical value for 99% confidence is approximately 2.576.
04

Calculate Confidence Interval

Now, calculate the confidence interval using the formula: \( \hat{p} \pm z^* \times SE \). This gives us the interval \( 0.3909 \pm 2.576 \times 0.00632 \), which is \( 0.3909 \pm 0.0163 \). Thus, the 99% confidence interval is approximately \( (0.3746, 0.4072) \).
05

Interpret Confidence Interval

The 99% confidence interval for the population proportion of binge drinkers is between 37.46% and 40.72%. This means we are 99% confident that the true proportion of full-time U.S. college students who are binge drinkers falls within this range.
06

Comment on the Newspaper Claim

The newspaper article claims that 45% of students are binge drinkers. Since 45% is outside of our calculated confidence interval of (37.46%, 40.72%), the data does not support the newspaper's claim. It suggests that the actual proportion is likely lower than the reported 45%.

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 91Ó°ÊÓ!

Key Concepts

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

Sample Proportion
In statistics, the sample proportion is a way of estimating the proportion of a specific outcome in a population based on a sample. For example, if we want to understand the proportion of full-time U.S. college students who are binge drinkers, we can conduct a survey to get a sample. To calculate the sample proportion \( \hat{p} \), we use the formula:\[\hat{p} = \frac{x}{n}\]where \( x \) is the number of students identified as binge drinkers, and \( n \) is the total number of students surveyed. In the given exercise, \( x = 2312 \) and \( n = 5914 \), so:\[\hat{p} = \frac{2312}{5914} \approx 0.3909\]This means that approximately 39.09% of the sampled students were binge drinkers. The sample proportion gives us a snapshot of the larger population's characteristics, but with some level of uncertainty.
Standard Error
The standard error is a statistical measure that provides an estimate of the variability or "standard error" of a sample statistic. In simpler words, it tells us how much the sample proportion might vary from the true population proportion. The formula for the standard error of the sample proportion is:\[SE = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\]Where:- \( \hat{p} \) is the sample proportion- \( n \) is the sample sizeFor our exercise, substituting the known values:\[SE = \sqrt{\frac{0.3909 \times (1-0.3909)}{5914}} \approx 0.00632\]This small standard error suggests that our sample proportion is a reasonably good estimate of the actual population proportion, and the smaller the standard error, the more precise the estimate.
Critical Value
The critical value is an essential component in creating a confidence interval. It represents the number of standard errors a sample proportion is away from the population proportion under the standard normal curve.When constructing a confidence interval, we multiply the critical value by the standard error to determine the range in which the true population proportion should fall with a certain level of confidence.For a 99% confidence interval, the critical value \( z^* \), is approximately 2.576. This value is chosen from a standard normal distribution (z-table) and indicates that our interval covers the middle 99% of possible sample proportions.This value is pivotal in calculating the confidence interval as it scales up the standard error to ensure the interval is wide enough to "capture" the true population proportion with the desired level of certainty.
Statistical Interpretation
Statistical interpretation is the process of making sense of calculated statistical measures, such as confidence intervals. Once we have our confidence interval, it provides a range of plausible values for the population parameter we're estimating. In this case, we calculated a 99% confidence interval for the proportion of binge drinkers among full-time U.S. college students to be approximately \((0.3746, 0.4072)\). This means we can say with 99% confidence that the true proportion of binge drinking students falls within this range.This confidence interval plays a crucial role in evaluating claims. For example, a newspaper claims that 45% of students are binge drinkers. However, since 45% lies outside the computed confidence interval, our statistical analysis suggests that the newspaper's claim is not supported by the survey data.In essence, statistical interpretation helps translate numbers into insights, enabling us to draw meaningful conclusions and make informed decisions based on data.

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

Find \(z^{\circ}\) for a \(93 \%\) confidence interval using Table A or your calculator. Show your method

Computers in some vehicles calculate various quantities related to performance. One of these is fuel efficiency, or gas mileage, usually expressed as miles per gallon (mpg). For one vehicle equipped in this way, the miles per gallon were recorded each time the gas tank was filled and the computer was then reset. \(^{27}\) Here are the mpg values for a random sample of 20 of these records: $$\begin{array}{llllllllll}15.8 & 13.6 & 15.6 & 19.1 & 22.4 & 15.6 & 22.5 & 17.2 & 19.4 & 22.6 \\\19.4 & 18.0 & 14.6 & 18.7 & 21.0 & 14.8 & 22.6 & 21.5 & 14.3 & 20.9\end{array}$$ Construct and interpret a \(95 \%\) confidence interval for the mean fuel efficiency \(\mu\) for this vehicle.

A $$ 95 \%$$ confidence interval for the mean body mass index (BMI) of young American women is \(26.8 \pm 0.6\). Discuss whether each of the following explanations is correct. (a) We are confident that \(95 \%\) of all young women have BMI between 26.2 and 27.4 . (b) We are \(95 \%\) confident that future samples of young women will have mean BMI between 26.2 and 27.4 . (c) Any value from 26.2 to 27.4 is believable as the true mean BMI of young American women. (d) If we take many samples, the population mean BMI will be between 26.2 and 27.4 in about \(95 \%\) of those samples. (e) The mean BMI of young American women cannot be 28 .

Gloria Chavez and Ronald Flynn are the candidates for mayor in a large city. We want to estimate the proportion \(p\) of all registered voters in the city who plan to vote for Chavez with \(95 \%\) confidence and a margin of error no greater than 0.03 . How large a random sample do we need? Show your work.

As Gallup indicates, the 3 percentage point margin of error for this poll includes only sampling variability (what they call "sampling error"). What other potential sources of error (Gallup calls these "nonsampling errors") could affect the accuracy of the \(95 \%\) confidence interval?

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.