/*! 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 54 In \(2010,\) the National Footba... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

In \(2010,\) the National Football League adopted new rules designed to limit head injuries. In a survey conducted in 2015 by the Harris Poll, 1216 of 2096 adults indicated that they were football fans and followed professional football. Of these football fans, 692 said they thought that the new rules were effective in limiting head injuries (December 21 , \(2015,\) www.theharrispoll.com/sports/Football-Injuries.html, retrieved May 6,2017 ). a. Assuming that the sample is representative of adults in the United States, construct and interpret a \(95 \%\) confidence interval for the proportion of U.S. adults who consider themselves to be football fans. b. Construct and interpret a \(95 \%\) confidence interval for the proportion of football fans who think that the new rules have been effective in limiting head injuries. c. Explain why the confidence intervals in Parts (a) and (b) are not the same width even though they both have a confidence level of \(95 \%\)

Short Answer

Expert verified
In summary: a. The 95% confidence interval for the proportion of U.S. adults who consider themselves to be football fans is \(\hat{p}_1 \pm 1.96 \sqrt{\frac{\hat{p}_1(1 - \hat{p}_1)}{2096}}\). We are 95% confident that the true proportion of adults in the U.S. who are football fans lies within this range. b. The 95% confidence interval for the proportion of football fans who think that the new rules have been effective in limiting head injuries is \(\hat{p}_2 \pm 1.96 \sqrt{\frac{\hat{p}_2(1 - \hat{p}_2)}{1216}}\). We are 95% confident that the true proportion of football fans who think the new rules are effective lies within this range. c. The confidence intervals in Parts (a) and (b) are not the same width because they have different sample proportions and sample sizes. Even though both confidence intervals have a confidence level of 95%, their widths depend on the sample proportion and sample size, which are different in both cases.

Step by step solution

01

Calculate the sample proportions

We are given that for the first proportion (football fans), 1216 out of 2096 are football fans. So, the sample proportion is: \[\hat{p}_1 = \frac{1216}{2096}\] Similarly, for the second proportion (football fans who think new rules are effective), 692 out of 1216 think the new rules are effective. So, the sample proportion is: \[\hat{p}_2 = \frac{692}{1216}\]
02

Calculate the 95% confidence intervals

We will now use the formula mentioned in the Analysis section for both proportions: For football fans (case a): \[CI_1 = \hat{p}_1 \pm 1.96 \sqrt{\frac{\hat{p}_1(1 - \hat{p}_1)}{2096}}\] For football fans who think new rules are effective (case b): \[CI_2 = \hat{p}_2 \pm 1.96 \sqrt{\frac{\hat{p}_2(1 - \hat{p}_2)}{1216}}\]
03

Interpret the confidence intervals

We will now interpret the confidence intervals obtained in Step 2. a. The 95% confidence interval for the proportion of U.S. adults who consider themselves to be football fans is \(CI_1\). This means that we are 95% confident that the true proportion of adults in the U.S. who are football fans lies within this range. b. The 95% confidence interval for the proportion of football fans who think that the new rules have been effective in limiting head injuries is \(CI_2\). This means that we are 95% confident that the true proportion of football fans who think the new rules are effective lies within this range. c. The confidence intervals in Parts (a) and (b) are not the same width because they have different sample proportions and sample sizes. Even though both confidence intervals have a confidence level of 95%, their widths depend on the sample proportion and sample size, which are different in both cases.

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.

Hypothesis Testing
Hypothesis testing is a fundamental procedure in statistics used to determine whether there is enough evidence in a sample of data to infer that a certain condition is true for the entire population.

In the context of the exercise, let's say you want to test the hypothesis that the majority of U.S. adults are football fans. You would set this is as your null hypothesis, typically denoted as H0, and the opposite (that the majority are not fans) as your alternative hypothesis, denoted as H1. The sample proportion of 1216 fans out of 2096 adults provides the data needed to test the null hypothesis.

Hypothesis testing involves calculating a test statistic, which in the case of proportions is typically the z-score. The z-score measures how many standard deviations the observed proportion is from a hypothesized population proportion under H0. If this z-score is large enough, you would reject the null hypothesis in favor of the alternative hypothesis. This is essentially what we're doing when we calculate a confidence interval: we're checking if the hypothesized proportion (like exactly 50% for majority) falls within that interval. If it doesn't, we have evidence to reject H0.
Proportion Estimation
Proportion estimation is about calculating the fraction of a population that contains a specific characteristic, based on sample data. For instance, in our NFL example, we estimated two proportions: the proportion of U.S. adults who are football fans (1216 out of 2096), and the proportion of these fans who believe new rules are effectively limiting head injuries (692 out of 1216).

When we calculate a confidence interval for these proportions, we're saying, 'We are th{95} percent confident that the true proportion in the entire population falls within this range.' It's crucial to remember that the confidence level (like 95%) reflects the degree of certainty we have in this estimation process—it does not mean that the actual proportion lies within the interval with a probability of 95%.

Estimating a proportion with high precision requires a sufficiently large sample size, and the width of the confidence interval is inversely related to this size. The more data you have, the narrower, and thus more precise, your confidence interval becomes.
Statistical Significance
Statistical significance is a term used to describe the likelihood that the difference in outcomes between two groups is not due to random chance. In terms of hypothesis testing, a result is statistically significant if it is unlikely to have occurred by chance, given that the null hypothesis is true.

The p-value is the probability of obtaining an effect at least as extreme as the one in your sample data, assuming the null hypothesis is true. If this p-value is smaller than the threshold significance level (often set at 0.05), you declare that the results are statistically significant.

Confidence intervals relate to statistical significance in that if a confidence interval for a difference between two groups does not contain zero (for differences) or one (for ratios), we can say there's a statistically significant difference. For example, when constructing confidence intervals for proportions in our exercise, if the interval does not contain the value corresponding to 'no effect' (which might be 50% if testing for a majority), then we can say the observed effect is statistically significant.

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

A random sample will be selected from the population of all adult residents of a particular city. The sample proportion \(\hat{p}\) will be used to estimate \(p,\) the proportion of all adult residents who are employed full time. For which of the following situations will the estimate tend to be closest to the actual value of \(p\) ? i. \(n=500, p=0.6\) ii. \(n=450, p=0.7\) iii. \(n=400, p=0.8\)

In mid-2016 the United Kingdom (UK) withdrew from the European Union (an event known as "Brexit"), causing economic concerns throughout the world. One indicator that economists use to monitor the health of the economy is the proportion of residential properties offered for sale at auction that are successfully sold. An article titled "Going, going, gone through the roof-sky's the limit at auction" (October \(22,2016,\) www.estateagenttoday .co.uk/features/2016/10/going-going-gone-through-theroof-the-skys-the-limit- at-auction, retrieved May 4,2017 ) reported the success rate of a sample of 26 residential properties offered for sale at auctions in the UK in the summer of \(2016 .\) For this sample of properties, 14 of the 26 residential properties were successfully sold. Suppose it is reasonable to consider these 26 properties as representative of residential properties offered at auction in the post-Brexit UK. a. Would it be appropriate to use the large-sample confidence interval for a population proportion to estimate the proportion of residential properties successfully sold at auction in the post-Brexit UK? Explain. b. Would it be appropriate to use a bootstrap confidence interval for a population proportion to estimate the proportion of residential properties successfully sold at auction in the post-Brexit UK? Explain. c. Use the accompanying output from the "Bootstrap Confidence Interval for One Proportion" Shiny app to report a \(95 \%\) bootstrap confidence interval for the population proportion of residential properties successfully

Consider taking a random sample from a population with \(p=0.70\) a. What is the standard error of \(\hat{p}\) for random samples of size \(100 ?\) b. Would the standard error of \(\hat{p}\) be smaller for samples of size 100 or samples of size \(400 ?\) c. Does decreasing the sample size by a factor of \(4,\) from 400 to \(100,\) result in a standard error of \(\hat{p}\) that is four times as large?

In a survey of 800 college students in the United States, 576 indicated that they believe that a student or faculty member on campus who uses language considered racist, sexist, homophobic, or offensive should be subject to disciplinary action ("Listening to Dissenting Views Part of Civil Debate," USA TODAY, November 17,2015 ). Assuming that the sample is representative of college students in the United States, construct and interpret a \(95 \%\) confidence interval for the proportion of college students who have this belief.

It probably wouldn't surprise you to know that Valentine's Day means big business for florists, jewelry stores, and restaurants. But did you know that it is also a big day for pet stores? In January \(2015,\) the National Retail Federation conducted a survey of consumers in a representative sample of adult Americans ("Survey of Online Shopping for Valentine's Day 2015," nrf.com/news/delivering-customer-delight-valentines-day, retrieved November 14,2016)\(.\) One of the questions in the survey asked if the respondent planned to spend money on a Valentine's Day gift for his or her pet. a. The proportion who responded that they did plan to purchase a gift for their pet was 0.212 . Suppose that the sample size for this survey was \(n=200 .\) Construct and interpret a \(95 \%\) confidence interval for the proportion of all adult Americans who planned to purchase a Valentine's Day gift for their pet. b. The actual sample size for the survey was much larger than \(200 .\) Would a \(95 \%\) confidence interval calculated using the actual sample size have been narrower or wider than the confidence interval calculated in Part (a)?

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.