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For each of the following hypothesis testing scenarios, indicate whether or not the appropriate hypothesis test would be for a difference in population means. If not, explain why not. Scenario 1: The authors of the paper "Adolescents and MP3 Players: Too Many Risks, Too Few Precautions" (Pediatrics [2009]: e953-e958) studied 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. You would like to determine if there is convincing evidence that the proportion of Dutch boys who listen to music at high volume is greater than this proportion for Dutch girls. Scenario 2: The report "Highest Paying Jobs for \(2009-10\) Bachelor's Degree Graduates" (National Association of Colleges and Employers, February 2010 ) states that the mean yearly salary offer for students graduating with accounting degrees in 2010 is \(\$ 48,722\). A random sample of 50 accounting graduates at a large university resulted in a mean offer of \(\$ 49,850\) and a standard deviation of \(\$ 3,300\). You would like to determine if there is strong support for the claim that the mean salary offer for accounting graduates of this university is higher than the 2010 national average of \(\$ 48,722\). Scenario 3: Each person in a random sample of 228 male teenagers and a random sample of 306 female teenagers was asked how many hours he or she spent online in a typical week (Ipsos, January 25,2006 ). The sample mean and standard deviation were 15.1 hours and 11.4 hours for males and 14.1 and 11.8 for females. You would like to determine if there is convincing evidence that the mean number of hours spent online in a typical week is greater for male teenagers than for female teenagers.

Short Answer

Expert verified
Scenario 1 does not require a test for difference in population means, it needs difference in population proportions. Scenarios 2 and 3 require a test for difference in population means.

Step by step solution

01

Scenario 1 Analysis

In this scenario, the test is about the proportion of Dutch boys who listen to music at high volume s compared to Dutch girls. Because we are comparing proportions on a categorical variable - boys who listen to music at a high volume vs girls who listen to music at a high volume - we are looking at testing difference in population proportions, not means. So hypothesis test for difference in population means would not be appropriate for this.
02

Scenario 2 Analysis

In this scenario, the claim being tested is about the difference in the mean yearly salary offer for students graduating with accounting degrees in 2010. Since the problem is about determining whether the mean salary offer is different from the known national average, and salary being a numerical variable allows us to compute mean, this situation calls for a test of difference in population means.
03

Scenario 3 Analysis

In this situation, the goal is to see if there is evidence that the mean number of hours spent online in a typical week is greater for male teenagers than for female teenagers. Since the variable under study is number of hours spent online - a numerical value which we can compute a mean for - the appropriate hypothesis test would be a test for a difference in population means.

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Key Concepts

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

Difference in Population Means
Understanding the difference in population means is a foundational concept in inferential statistics, particularly when comparing two or more groups. When statisticians refer to the 'population,' they mean the entire group of individuals or items they are interested in studying. The 'mean' is the average measure.

For instance, in Scenario 2, the focus is on assessing whether the mean salary offer for graduates from a specific university differs from the national average. In such cases, a t-test for independent samples might be used if the variance between the two groups is assumed to be equal, and a Welch’s t-test if not. The null hypothesis (\( H_0 \textrm{: } \textmu_1 = \textmu_2 \)) in this scenario posits no difference in mean salary offers, while the alternative hypothesis (\( H_1 \textrm{: } \textmu_1 > \textmu_2 \)) suggests that there is a significant difference. If after conducting the appropriate test the results demonstrate statistical significance, we could reject the null hypothesis, supporting the claim that the mean salary offer at the university is different from the national average.

Scenario 3 also presents a case for examining the difference in means—specifically, the amount of time spent online between male and female teenagers. Here, the null hypothesis would imply no difference in average online hours (\( \textmu_{\text{males}} = \textmu_{\text{females}} \)), while the alternate hypothesis would argue that males spend more time online on average (\( \textmu_{\text{males}} > \textmu_{\text{females}} \)).
Difference in Population Proportions
When dealing with categorical data, particularly where we're interested in the proportion of a certain attribute within two or more groups, we analyze the difference in population proportions. A widely used hypothesis test for this purpose is the two-proportion z-test.

In Scenario 1, researchers are examining whether a higher proportion of Dutch boys listen to music at a high volume compared to Dutch girls, which is a categorical variable (high volume vs not). Therefore, the suitable hypothesis test examines the difference in proportions, not means. The null hypothesis would state that the proportions are equal (\( p_1 = p_2 \textrm{, where } p \textrm{ is the population proportion} \)), and the alternative hypothesis suggests a higher proportion in boys (\( p_{\text{boys}} > p_{\text{girls}} \)). To analyze this, the proportion of each group exhibiting the behavior is computed and then compared using the z-test. If the p-value obtained is lower than the chosen alpha level (often 0.05), then it suggests a statistically significant difference in proportions.

This approach is pivotal in many real-world contexts, such as clinical trials, marketing research, or opinion polling, where understanding differences in proportions can guide decisions and strategies.
Statistical Significance
Statistical significance is a determination about the strength of the evidence against the null hypothesis provided by the data. In simple terms, it evaluates whether the observed differences or effects in a study are likely due to genuine effects rather than random chance.

In hypothesis testing, we calculate a p-value, which represents the probability of obtaining the observed results, or more extreme, assuming that the null hypothesis is true. If this p-value is less than the predetermined significance level (commonly set at 0.05), then we may declare the results statistically significant and reject the null hypothesis.

In Scenarios 2 and 3, determining statistical significance is crucial for making inferences about the population means. For Scenario 1, statistical significance would help establish if the difference in proportions between boys and girls listening to music at a high volume is meaningful and not a product of random variation. It’s important to remember, however, that statistical significance does not imply practical significance; even if the test indicates a significant difference, it must still be evaluated in context to determine its real-world relevance and impact.
Numerical Variable Analysis
Numerical variable analysis deals with variables that take on quantifiable values. In statistical testing, particularly for differences in means, these variables are crucial since they allow for calculations of averages and variances among populations or samples.

Both Scenario 2 and Scenario 3 utilize numerical variable analysis to interpret data. In Scenario 2, the numerical variable is the salary offer extended to graduates, and in Scenario 3, it is the hours spent online by teenagers. Descriptive statistics, like mean and standard deviation, assist in summarizing these numerical variables to draw comparisons. The analysis applies tests such as the t-test to determine if the observed differences in means are unlikely to have occurred under the null hypothesis.

Deep knowledge of numerical variables also helps in assessing assumptions relevant to the chosen statistical tests, such as normality, homogeneity of variance, and independence. Ensuring these conditions are met reinforces the reliability of the hypothesis test conducted on numerical data.

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Most popular questions from this chapter

The paper "Ladies First?" A Field Study of Discrimination in Coffee Shops" (Applied Economics [2008]: 1-19) describes a study in which researchers observed wait times in coffee shops in Boston. Both wait time and gender of the customer were observed. The mean wait time for a sample of 145 male customers was 85.2 seconds. The mean wait time for a sample of 141 female customers was 113.7 seconds. The sample standard deviations (estimated from graphs in the paper) were 50 seconds for the sample of males and 75 seconds for the sample of females. Suppose that these two samples are representative of the populations of wait times for female coffee shop customers and for male coffee shop customers. Is there convincing evidence that the mean wait time differs for males and females? Test the relevant hypotheses using a significance level of 0.05

The article "More Students Taking AP Tests" (San Luis Obispo Tribune, January 10,2003 ) provided the following information on the percentage of students in grades 11 and 12 taking one or more AP exams and the percentage of exams that earned credit in 1997 and 2002 for seven high schools on the central coast of California. $$ \begin{array}{cccccc} & {\begin{array}{c} \text { Percentage of } \\ \text { Students Taking } \\ \text { One or More } \\ \text { AP Exams } \end{array}} & & {\begin{array}{c} \text { Percentage of } \\ \text { Exams That } \\ \text { Earned College } \end{array}} \\ { 2 - 3 } { 5 - 6 } & & & & {\text { Credit }} \\ { 2 - 3 } { 5 - 6 } \text { School } & 1997 & 2002 & & 1997 & 2002 \\ & 1 & 13.6 & 18.4 & & 61.4 & 52.8 \\ 2 & 20.7 & 25.9 & & 65.3 & 74.5 \\ 3 & 8.9 & 13.7 & & 65.1 & 72.4 \\ 4 & 17.2 & 22.4 & & 65.9 & 61.9 \\ 5 & 18.3 & 43.5 & & 42.3 & 62.7 \\ 6 & 9.8 & 11.4 & & 60.4 & 53.5 \\ 7 & 15.7 & 17.2 & & 42.9 & 62.2 \\ \hline \end{array} $$ a. Assuming that it is reasonable to regard these seven schools as a random sample of high schools located on the central coast of California, carry out an appropriate test to determine if there is convincing evidence that the mean percentage of exams earning college credit at central coast high schools in 1997 and in 2002 were different. b. Do you think it is reasonable to generalize the conclusion of the test in Part (a) to all California high schools? Explain. c. Would it be appropriate to use the paired-samples \(t\) test with the data on percentage of students taking one or more AP exams? Explain.

The authors of the paper "Ultrasound Techniques Applied to Body Fat Measurement in Male and Female Athletes" (Journal of Athletic Training [2009]: \(142-147\) ) compared two different methods for measuring body fat percentage. One method uses ultrasound and the other method uses X-ray technology. The accompanying table gives body fat percentages for 16 athletes using each of these methods (a subset of the data given in a graph that appeared in the paper). For purposes of this exercise, you can assume that the 16 athletes who participated in this study are representative of the population of athletes. Do these data provide convincing evidence that the mean body fat percentage measurement differs for the two methods? Test the appropriate hypotheses using \(\alpha=0.05\). $$ \begin{array}{crr} \text { Athlete } & \text { X-ray } & \text { Ultrasound } \\ \hline 1 & 5.00 & 4.75 \\ 2 & 7.00 & 3.75 \\ 3 & 9.25 & 9.00 \\ 4 & 12.00 & 11.75 \\ 5 & 17.25 & 17.00 \\ 6 & 29.50 & 27.50 \\ 7 & 5.50 & 6.50 \\ 8 & 6.00 & 6.75 \\ 9 & 8.00 & 8.75 \\ 10 & 8.50 & 9.50 \\ 11 & 9.25 & 9.50 \\ 12 & 11.00 & 12.00 \\ 13 & 12.00 & 12.25 \\ 14 & 14.00 & 15.50 \\ 15 & 17.00 & 18.00 \\ 16 & 18.00 & 18.25 \end{array} $$

For each of the following hypothesis testing scenarios, indicate whether or not the appropriate hypothesis test would be about a difference in population means. If not, explain why not. Scenario 1: The international polling organization Ipsos reported data from a survey of 2,000 randomly selected Canadians who carry debit cards (Canadian Account Habits Survey, July 24,2006 ). Participants in this survey were asked what they considered the minimum purchase amount for which it would be acceptable to use a debit card. You would like to determine if there is convincing evidence that the mean minimum purchase amount for which Canadians consider the use of a debit card to be acceptable is less than \(\$ 10\). Scenario 2: Each person in a random sample of 247 male working adults and a random sample of 253 female working adults living in Calgary, Canada, was asked how long, in minutes, his or her typical daily commute was ("Calgary Herald Traffic Study," Ipsos, September 17,2005 ). You would like to determine if there is convincing evidence that the mean commute times differ for male workers and female workers. Scenario 3: A hotel chain is interested in evaluating reservation processes. Guests can reserve a room using either a telephone system or an online system. 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. You would like to determine if it reasonable to conclude that the proportion who are satisfied is higher for those who reserve a room online.

In a study of the effect of college student employment on academic performance, the following summary statistics for GPA were reported for a sample of students who worked and for a sample of students who did not work (University of Central Florida Undergraduate Research Journal, Spring 2005\()\) : $$ \begin{array}{cccc} & \begin{array}{c} \text { Sample } \\ \text { Size } \end{array} & \begin{array}{c} \text { Mean } \\ \text { GPA } \end{array} & \begin{array}{c} \text { Sandard } \\ \text { Deviation } \end{array} \\ \begin{array}{c} \text { Students Who Are } \\ \text { Employed } \end{array} & 184 & 3.12 & 0.485 \\ \begin{array}{c} \text { Students Who Are } \\ \text { Not Employed } \end{array} & 114 & 3.23 & 0.524 \\ \hline \end{array} $$ The samples were selected at random from working and nonworking students at the University of Central Florida. Does this information support the hypothesis that for students at this university, those who are not employed have a higher mean GPA than those who are employed?

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