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The authors of the paper "Adolescents and MP3 Players: Too Many Risks, Too Few Precautions" (Pediatrics [2009]:e953-e958) concluded that more boys than girls listen to music at high volumes. This conclusion was based on data from independent random samples of 764 Dutch boys and 748 Dutch girls ages 12 to \(19 .\) Of the boys, 397 reported that they almost always listen to music at a high volume setting. Of the girls, 331 reported listening to music at a high volume setting. Do the sample data support the authors' conclusion that the proportion of Dutch boys who listen to music at high volume is greater than this proportion for Dutch girls? Test the relevant hypotheses using a 0.01 significance level.

Short Answer

Expert verified
The sample data supports the authors' conclusion if the calculated p-value is less than or equal to the significance level of 0.01.

Step by step solution

01

State the Null Hypothesis and Alternative Hypothesis

The Null Hypothesis (\(H_0\)) is that the proportion of boys who listen to music at high volume is equal to the proportion of girls. The Alternative Hypothesis (\(H_1\)) is that the proportion of boys who listen to music at high volume is greater than the proportion of girls.
02

Calculate Sample Proportions

The sample proportion of boys (\(p_1\)) who listen to music at high volume is 397 out of 764 and that of girls (\(p_2\)) is 331 out of 748. So, \(p_1 = \frac{397}{764}\) and \(p_2 = \frac{331}{748}\). Calculate these values.
03

Calculate Combined Sample Proportion

The combined sample proportion (\(p\)) is the total number of successes (listening to high volume music) divided by the total number of trials (total number of boys and girls). So, \( p = \frac{(397 + 331)}{(764 + 748)}\). Calculate this value.
04

Test Statistic Calculation

The test statistic (\(Z\)) for testing proportion is calculated using the formula \( Z = \frac{(p_1 - p_2)}{\sqrt{p(1 - p)(\frac{1}{n_1} + \frac{1}{n_2})}}\), where \(n_1\) and \(n_2\) are the sample sizes for boys and girls respectively. Plug in the values and calculate the \(Z\) score.
05

Determine the P-Value

To determine if the sample data supports the authors' conclusion, compare the p-value to the significance level. If the p-value ≤ 0.01 significance level, then reject the null hypothesis.

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Key Concepts

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

Statistical Significance
When conducting hypothesis testing in statistics, researchers seek to determine whether their findings reflect a true effect or whether they are due to chance. This is where the concept of statistical significance comes into play. A result is statistically significant if it is unlikely to have occurred by chance alone, according to a certain threshold, known as the significance level. Common significance levels are 0.05, 0.01, or 0.001, which correspond to a 5%, 1%, or 0.1% chance of rejecting the true null hypothesis (a type I error).

In the context of the exercise, using a 0.01 significance level means that there is only a 1% risk that the conclusion reached is incorrect if the null hypothesis is actually true. This stringent level of significance is chosen to ensure that if there is a claim that more boys than girls listen to music at high volumes, it is backed by strong evidence from the data.
Sample Proportions
Sample proportion represents a fraction of the sample with a particular characteristic. It's especially relevant in studies comparing attributes between two different groups. In the exercise, two sample proportions are calculated, one for Dutch boys (\( p_1 \) for boys) and one for Dutch girls (\( p_2 \) for girls), to compare the rates at which they listen to music at high volumes.

To compute each sample proportion, divide the number of individuals with the characteristic by the total number of individuals in that sample. Here, \( p_1 = \frac{397}{764} \) and \( p_2 = \frac{331}{748} \) represent the proportions of boys and girls, respectively, who listen to music at high volumes. These proportions are crucial in hypothesis testing as they serve as estimators for the true population proportions we are interested in comparing.
P-Value
The p-value is a fundamental concept in the field of hypothesis testing designed to measure the strength of evidence against the null hypothesis. After calculating the test statistic, the p-value tells us the probability of observing a result as extreme as the one obtained (or more extreme), assuming that the null hypothesis is true.

The smaller the p-value, the stronger the evidence against the null hypothesis since it indicates that such an extreme result is less likely to occur simply by chance. If the p-value is less or equal to the significance level (in this case, 0.01), it suggests that the observed data are inconsistent with the null hypothesis, and thus, the null hypothesis should be rejected. In our exercise example, comparing the calculated p-value to the 0.01 threshold will help us draw conclusions about music listening habits among boys and girls.
Null and Alternative Hypotheses
In hypothesis testing, we set up two opposing statements—the null hypothesis (\( H_0 \)) and the alternative hypothesis (\( H_1 \) or \( H_a \)). The null hypothesis represents a position of no effect or no difference, serving as the default assumption. It is what we seek to test against the evidence. The alternative hypothesis, on the other hand, embodies the new claim we're testing for—an effect or a difference in a specific direction (greater than, less than, or not equal).

In our study example, the null hypothesis states that there's no difference in the music-listening habits at high volume between boys and girls (\( p_1 = p_2 \)). The alternative hypothesis suggests that there is indeed a difference, specifically that more boys listen to high volumes than girls (\( p_1 > p_2 \)). The testing process involves using data to determine which hypothesis is more likely to be true.

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Most popular questions from this chapter

Common Sense Media surveyed 1,000 teens and 1,000 parents of teens to learn about how teens are using social networking sites such as Facebook and MySpace ("Teens Show, Tell Too Much Online," San Francisco Chronicle, August 10,2009 ). The two samples were independently selected and were chosen to be representative of American teens and parents of American teens. When asked if they check online social networking sites more than 10 times a day, 220 of the teens surveyed said yes. When parents of teens were asked if their teen checked networking sites more than 10 times a day, 40 said yes. Use a significance level of 0.01 to determine if there is convincing evidence that the proportion of all parents who think their teen checks social networking sites more than 10 times a day is less than the proportion of all teens who check more than 10 times a day.

The article "College Graduates Break Even by Age 33" (USA Today, September 21,2010 ) reported that \(2.6 \%\) of college graduates were unemployed in 2008 and \(4.6 \%\) of college graduates were unemployed in \(2009 .\) Suppose that the reported percentages were based on independently selected representative samples of 500 college graduates in each of these two years. Construct and interpret a \(95 \%\) confidence interval for the difference in the proportions of college graduates who were unemployed in these 2 years.

The report "Audience Insights: Communicating to Teens (Aged \(12-17)^{\prime \prime}\) (www.cdc.gov, 2009 ) described teens' attitudes about traditional media, such as TV, movies, and newspapers. In a representative sample of American teenage girls, \(41 \%\) said newspapers were boring. In a representative sample of American teenage boys, \(44 \%\) said newspapers were boring. Sample sizes were not given in the report. a. Suppose that the percentages reported were based on samples of 58 girls and 41 boys. Is there convincing evidence that the proportion of those who think that newspapers are boring is different for teenage girls and boys? Carry out a hypothesis test using \(\alpha=.05\). b. Suppose that the percentages reported were based on samples of 2,000 girls and 2,500 boys. Is there convincing evidence that the proportion of those who think that newspapers are boring is different for teenage girls and boys? Carry out a hypothesis test using \(\alpha=0.05\). c. Explain why the hypothesis tests in Parts (a) and resulted in different conclusions.

A hotel chain is interested in evaluating reservation processes. Guests can reserve a room by using either a telephone system or an online system that is accessed through the hotel's web site. Independent random samples of 80 guests who reserved a room by phone and 60 guests who reserved a room online were selected. Of those who reserved by phone, 57 reported that they were satisfied with the reservation process. Of those who reserved online, 50 reported that they were satisfied. Based on these data, is it reasonable to conclude that the proportion who are satisfied is greater for those who reserve a room online? Test the appropriate hypotheses using a significance level of \(0.05 .\)

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