/*! 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 15 A study first reported in the Jo... [FREE SOLUTION] | 91Ó°ÊÓ

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A study first reported in the Journal of the American Medical Association (7 December 1994) received widespread attention as the first wide-scale study of the use of alcohol on American college campuses and was the subject of an article in Time magazine (19 December 1994, p. 16). The researchers surveyed 17,592 students at 140 four-year colleges in 40 states. One of the results they found was that about \(8.8 \%,\) or about 1550 respondents, were frequent binge drinkers. They defined frequent binge drinking as having had at least four (for women) or five (for men) drinks at a single sitting at least three times during the previous 2 weeks. a. Time magazine (19 December 1994, p. 66) reported that of the approximately 1550 frequent binge drinkers in this study, \(22 \%\) reported having had unprotected sex. Find a \(95 \%\) confidence interval for the population proportion of all frequent binge drinkers who had unprotected sex, and interpret the interval for someone who has no knowledge of statistics. b. Notice that the results quoted in part (a) indicate that about 341 students out of the 17,592 interviewed said they were frequent binge drinkers and had unprotected sex. Compute a \(95 \%\) confidence interval for the proportion of college students who are frequent binge drinkers and who also had unprotected sex. c. Using the results from parts (a) and (b), write two short news articles on the problem of binge drinking and unprotected sex. In one, make the situation sound as disastrous as you can. In the other, try to minimize the problem.

Short Answer

Expert verified
a. CI: (19.86%, 24.14%), b. CI: (1.734%, 2.146%).

Step by step solution

01

Calculate the Confidence Interval for Part (a)

For part (a), we need to find a 95% confidence interval for the proportion of frequent binge drinkers who reported unprotected sex. The given proportion (sample proportion \(\hat{p}\)) is \(0.22\) (22%), and the sample size \(n\) is 1550. To calculate the confidence interval, we use the formula: \[ CI = \hat{p} \pm Z \cdot \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \]Where \(Z\) is the Z-score for a 95% confidence level, which is 1.96.Calculate the standard error: \[ SE = \sqrt{\frac{0.22 \times 0.78}{1550}} \approx 0.0109 \]Now, find the confidence interval: \[ CI = 0.22 \pm 1.96 \times 0.0109 \]\[ CI = 0.22 \pm 0.0214 \]Thus, the 95% confidence interval is approximately (0.1986, 0.2414). This means we are 95% confident that the true proportion of frequent binge drinkers who had unprotected sex is between 19.86% and 24.14%.
02

Calculate the Confidence Interval for Part (b)

For part (b), we need to find a 95% confidence interval for the proportion of all college students who are frequent binge drinkers and also had unprotected sex. The number of such students is mentioned as 341 out of the total surveyed population of 17,592. This gives a sample proportion \(\hat{p}\) of:\[ \hat{p} = \frac{341}{17592} \approx 0.0194 \]We again use the confidence interval formula:\[ CI = \hat{p} \pm Z \cdot \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \]Calculate the standard error:\[ SE = \sqrt{\frac{0.0194 \times 0.9806}{17592}} \approx 0.00105 \]Now, calculate the confidence interval:\[ CI = 0.0194 \pm 1.96 \times 0.00105 \]\[ CI = 0.0194 \pm 0.00206 \]Thus, the 95% confidence interval is approximately (0.01734, 0.02146). This means we are 95% confident that the proportion of all college students who are frequent binge drinkers and had unprotected sex is between 1.734% and 2.146%.
03

Write the News Articles

**Disastrous Angle:** An alarming number of college students are engaging in hazardous behavior involving binge drinking and unprotected sex. New data suggests that nearly one in four frequent binge drinkers could be risking their health and safety by engaging in unprotected sex. This paints a worrying picture of the college social scene, where risky behaviors are seemingly normalized and expected. **Minimized Angle:** The recent study of college student behavior has highlighted an area for improvement. While a minority of frequent binge drinkers admitted to having unprotected sex, it appears to be a contained issue. Less than a quarter of binge drinkers and only about 2% of all surveyed students engage in such risky behavior, indicating that most students act responsibly despite occasional overindulgence.

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Key Concepts

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

Confidence Interval
When analyzing data in statistics, a confidence interval provides a range within which we expect the true value of a parameter to lie.For example, in our study, we were interested in the proportion of frequent binge drinkers involved in unprotected sex.A confidence interval helps to express how much uncertainty there is when estimating a population parameter based on a sample statistic.
  • The **sample proportion** \(\hat{p}\) gives us an estimate from our data. Here, \(\hat{p} = 0.22\), or 22% of frequent binge drinkers reported unprotected sex.
  • The formula to calculate a confidence interval is: \[ CI = \hat{p} \pm Z \cdot \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \]
  • The **Z-score** for a 95% confidence interval is typically 1.96, which implies a 5% chance that the true parameter is outside this interval.
By applying the formula, we derive a confidence interval of (0.1986, 0.2414) for our example.This indicates we can be 95% confident that the true proportion of frequent binge drinkers having unprotected sex lies between 19.86% and 24.14%.Confidence intervals are powerful because they provide more insight than a single estimate, showing both the estimate and the uncertainty around it.
Proportion
A proportion is a statistical term that reflects a part of the whole.In this context, it deals with the part of college students who engage in both binge drinking and unprotected sex.Understanding proportions helps us assess how widespread certain behaviors might be.
  • The proportion of a specific group is calculated by dividing the number of individuals in that group by the total number of individuals considered.
  • For instance, if 341 students out of 17,592 were frequent binge drinkers who had unprotected sex, the proportion is \( \frac{341}{17592} = 0.0194 \) or 1.94%.
  • Proportions allow us to interpret data in relative terms and make comparisons across different groups or time periods.
In our analysis, understanding these proportions provides a clearer picture of behavioral patterns among college students.By determining these figures, stakeholders can decide whether intervention or education is needed to address the issue.
Binge Drinking
Binge drinking is defined as consuming an excessive amount of alcohol in a short period. For women, this involves having four drinks, and for men, five drinks in one sitting. The study shows that about 8.8% of college students engage in binge drinking frequently.
  • Binge drinking poses numerous health risks including alcohol poisoning, impaired judgment, and potential for addiction.
  • It is a significant concern on college campuses due to the combination of freedom, peer pressure, and lack of supervision.
Addressing binge drinking requires understanding its causes and promoting responsible behavior. Initiatives could include educational programs that highlight the risks and the impact of excessive drinking on one's health and academic performance.
Unprotected Sex
Unprotected sex refers to sexual activity without the use of contraceptives or protective barriers, such as condoms. This behavior increases the risk of sexually transmitted infections (STIs) and unintended pregnancies. In the study, it was found that 22% of frequent binge drinkers reported having had unprotected sex.
  • Engaging in unprotected sex while under the influence of alcohol can further elevate the risks due to impaired decision-making.
  • Educational measures are vital in raising awareness about the importance of protection and safe practices.
  • Programs that address both drinking habits and safe sex practices may be most effective in reducing the incidence of these behaviors on college campuses.
Through increased awareness and targeted interventions, the goal is to promote healthier decision-making among young adults.

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Most popular questions from this chapter

In a 2008 Scripps Howard News/Ohio University survey of 1003 randomly selected adults, 56% said that it is either very or somewhat likely that there is intelligent life on other planets (Source: www.scrippsnews.com/node/34758). For each part, use the desired confidence level to find a confidence interval for the proportion of all adults in 2008 who thought it was either very or somewhat likely that there is intelligent life on other planets. a. \(90 \%\) b. \(95 \%\) c. \(99 \%\) d. Using the 99% confidence interval for part (c), would you conclude with 99% confidence that a majority of adults (more than \(50 \%\) ) in 2008 thought it was either very or somewhat likely that there is intelligent life on other planets? Explain.

In a special double issue of Time magazine, the cover story featured Pope John Paul II as "Man of the Year" (26 December 1994-2 January 1995, pp. 74-76). As part of the story, Time reported on the results of a survey of 507 adult American Catholics, taken by telephone on December \(7-8 .\) It was also reported that "sampling error is \(\pm 4.4 \% . "\) a. One question asked was, "Do you favor allowing women to be priests?" to which 59\% of the respondents answered yes. Using the reported margin of error of 4.4\%, calculate a 95\% confidence interval for the response to this question. Write a sentence interpreting the interval that could be understood by someone who knows nothing about statistics. Be careful about specifying the correct population. b. Calculate a \(95 \%\) confidence interval for the question in part (a), using the formula in this chapter rather than the reported margin of error. Compare your answer to the answer in part (a). c. Another question in the survey was, "Is it possible to disagree with the Pope and still be a good Catholic?" to which 89\% of respondents said yes. Using the formula in this chapter, compute a \(95 \%\) confidence interval for the true percentage who would answer yes to the question. Now compute a \(95 \%\) confidence interval using the reported margin of error of \(4.4 \%\) Compare your two intervals. d. If you computed your intervals correctly, you would have found that the two intervals in parts (a) and (b) were quite similar to each other, whereas the two intervals in part (c) were not. In part (c), the interval computed using the reported margin of error was wider than the one computed using the formula. Explain why the two methods for computing the intervals agreed more closely for the survey question in parts (a) and (b) than for the survey question in part (c).

Use the Empirical Rule to specify what level of confidence would accompany each of the following confidence intervals. a. Sample proportion \(\pm 1(S E P)\) b. Sample proportion \(\pm 2(S E P)\) c. Sample proportion \(\pm 3(S E P)\)

On September \(10,1998,\) the "Starr Report," alleging impeachable offenses by President Bill Clinton, was released to Congress. That evening, the Gallup Organization conducted a poll of 645 adults nationwide to assess initial reaction (reported at www.gallup.com). One of the questions asked was: "Based on what you know at this point, do you think that Bill Clinton should or should not be impeached and removed from office?" The response "Yes, should" was selected by 31\% of the respondents. a. The Gallup webpage said, "For results based on the total sample of adults nationwide, one can say with \(95 \%\) confidence that the margin of sampling error is no greater than ±4 percentage points." Explain what this means and verify that the statement is accurate. b. Give a \(95 \%\) confidence interval for the percentage of all adults who would have said President Clinton should be impeached had they been asked that evening. c.A similar Gallup Poll taken a few months earlier, in June 1998 , found that \(19 \%\) responded that President Clinton should be impeached. Do you think the difference between the results of the two polls can be attributed to chance variation in the samples taken, or does it represent a real difference of opinion in the population in June versus mid-September? Explain.

Does a confidence interval for a proportion estimate a sample proportion, or a population proportion?

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