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The authors of the paper "Adolescents and MP3 Players: Too Many Risks, Too Few Precautions" \((P \mathrm{e}-\) diatrics [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 age 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 .01 significance level.

Short Answer

Expert verified
Based on the hypothesis test, there is strong evidence to reject the null hypothesis that there is no difference in proportions and conclude that the proportion of boys who listen to music at high volume is significantly greater than that of girls.

Step by step solution

01

State the Null and Alternative Hypotheses

The null hypothesis is that there is no difference in the proportions of boys and girls listening to music at high volumes, so \(p1 = p2\). The alternative hypothesis is that the proportion of boys (p1) listening to music at high volumes is greater than the proportion of girls (p2), so \(p1 > p2\).
02

Calculate the sample proportions

The sample proportion for boys (p1) is 397 out of 764 or 0.52 and for girls (p2) it is 331 out of 748 or 0.44.
03

Calculate the combined proportion

The combined proportion (p) is calculated as (x1+x2) / (n1+n2). Here, x1=397, x2=331, n1=764, n2=748. So, p = (397+331) / (764+748) = 0.48.
04

Calculate the Standard Error

We compute the Standard Error (SE) with the formula \(\sqrt{ p(1-p)(1/n1 + 1/n2)}\). Plugging in p=0.48, n1=764, n2=748 gives us SE = 0.028.
05

Compute the test statistic

The test statistic (z) is calculated as (p1-p2) / SE. Plugging in p1=0.52, p2=0.44, and SE=0.028, we get z=2.86.
06

Obtain the p-value

Using a standard Normal table or a software, we find the one-tailed p-value corresponding to z=2.86. The p-value here is 0.002.
07

Compare the p-value with the significance level

The p-value (0.002) is less than our significance level (0.01). So, we reject the null hypothesis.

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Key Concepts

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

Null Hypothesis
In hypothesis testing, the null hypothesis, often denoted as \(H_0\), is a statement that there is no effect or no difference. It serves as the starting assumption for statistical testing. In the context of our exercise, the null hypothesis is that the proportion of boys and girls who listen to music at high volumes is the same: \(p_1 = p_2\). This means we initially presume there is no difference in behavior between boys and girls, similar to a baseline or status quo.

If our calculated evidence (through data analysis) significantly contradicts the null hypothesis, we have reason to reject it. However, unless proven by such evidence, the null hypothesis remains in effect. Think of it as 'innocent until proven guilty'—the null hypothesis is "assumed true" unless the data convincingly suggests otherwise.
  • Serves as a baseline or a default position.
  • Assured to be true until statistically disproven.
  • Challenges biases in comparative studies.
Alternative Hypothesis
The alternative hypothesis, denoted as \(H_a\), is the statement we want to test and potentially accept if the null hypothesis is rejected. This hypothesis reflects the researchers' actual expectations. In our case, the alternative hypothesis is that the proportion of boys who listen to music at high volumes is greater than the proportion for girls: \(p_1 > p_2\).

This hypothesis represents a specific claim or effect we believe to be true or wish to provide evidence for. Deciding on an alternative hypothesis is a critical step in hypothesis testing because it guides the direction of the study and the interpretation of results.
  • Represents the expected research outcome.
  • Contradicts the null hypothesis.
  • Drives study objectives and decision-making.
Standard Error
Standard Error (SE) measures the variability or dispersion of a sample statistic from its population mean estimation. In hypothesis testing, the SE indicates how much the sample proportion might vary from the true population proportion. It forms the basis for constructing confidence intervals or computing test statistics.

In this exercise, the SE is computed using the formula:\[ SE = \sqrt{ p(1-p) \left( \frac{1}{n_1} + \frac{1}{n_2} \right)} \]where \(p\) is the combined sample proportion, and \(n_1\) and \(n_2\) are the sample sizes for boys and girls, respectively.

Having a smaller SE typically suggests more reliable estimates, as it means less variability in the sample data. Calculating SE allows researchers to gauge the accuracy of their sample and helps in assessing statistical significance.
  • Quantifies the precision of statistical estimates.
  • Aids in hypothesis testing and confidence interval estimation.
  • Depends inversely on sample size; larger samples usually reduce SE.
Significance Level
The significance level, often denoted as \(\alpha\), is the threshold or "cut-off" point used to determine whether a hypothesis test result is statistically significant. In this exercise, a significance level of 0.01 means there is a 1% risk of concluding that a difference exists when there isn't one (Type I error).

Choosing a significance level involves balancing the risks of Type I and Type II errors. A smaller \(\alpha\) reduces the likelihood of a Type I error but may increase the chance of a Type II error (failing to detect a true effect).

To decide the outcome of the hypothesis test, the p-value obtained from the test statistic is compared to the significance level. A p-value less than \(\alpha\) indicates that the difference is statistically significant, justifying rejecting the null hypothesis.
  • Helps determine statistical significance.
  • A lower level reduces risk of false positives.
  • Critical in validating research findings.

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Most popular questions from this chapter

The paper "The Psychological Consequences of Money" (Science [2006]: \(1154-1156\) ) describes several experiments designed to investigate the way in which money can change behavior. In one experiment, participants completed one of two versions of a task in which they were given lists of five words and were asked to rearrange four of the words to create a sensible phrase. For one group, half of the 30 unscrambled phrases related to money, whereas the other half were phrases that were unrelated to money. For the second group (the control group), none of the 30 unscrambled phrases related to money. Participants were 44 students at Florida State University. Participants received course credit and \(\$ 2\) for their participation. The following description of the experiment is from the paper: Participants were randomly assigned to one of two conditions, in which they descrambled phrases that primed money or neutral concepts. Then participants completed some filler questionnaires, after which the experimenter told them that the experiment was finished and gave them a false debriefing. This step was done so that participants would not connect the donation opportunity to the experiment. As the experimenter exited the room, she mentioned that the lab was taking donations for the University Student Fund and that there was a box by the door if the participant wished to donate. Amount of money donated was the measure of helping. We found that participants primed with money donated significantly less money to the student fund than participants not primed with money \([t(38)=2.13, P<0.05]\) The paper also gave the following information on amount donated for the two experimental groups. a. Explain why the random assignment of participants to experimental groups is important in this experiment. b. Use the given information to verify the values of the test statistic and degrees of freedom (38, given in parentheses just after the \(t\) in the quote from the paper) and the statement about the \(P\) -value. Assume that both sample sizes are 22 . c. Do you think that use of the two-sample \(t\) test was appropriate in this situation? Hint: Are the assumptions required for the two-sample \(t\) test reasonable?

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