/*! 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 46 The report "Young People Living ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The report "Young People Living on the Edge" (Greenberg Quinlan Rosner Research, 2008 ) summarizes a survey of people in two independent random samples. One sample consisted of 600 young adults (age 19 to 35 ) and the other sample consisted of 300 parents of children age 19 to \(35 .\) The young adults were presented with a variety of situations (such as getting married or buying a house) and were asked if they thought that their parents were likely to provide financial support in that situation. The parents of young adults were presented with the same situations and asked if they would be likely to provide financial support to their child in that situation. a. When asked about getting married, \(41 \%\) of the young adults said they thought parents would provide financial support and \(43 \%\) of the parents said they would provide support. Carry out a hypothesis test to determine if there is convincing evidence that the proportion of young adults who think parents would provide financial support and the proportion of parents who say they would provide support are different. b. The report stated that the proportion of young adults who thought parents would help with buying a house or apartment was \(37 .\) For the sample of parents, the proportion who said they would help with buying a house or an apartment was . \(27 .\) Based on these data, can you conclude that the proportion of parents who say they would help with buying a house or an apartment is significantly less than the proportion of young adults who think that their parents would help?

Short Answer

Expert verified
The results of the hypothesis tests will depend on the respective p-values calculated. If p-value is less than 0.05, there would be enough evidence to suggest that the proportions are significantly different for marriage support. Likewise, if p-value is less than 0.05 for house buying support, we can conclude that parents are less likely to help than what the young adults think. Otherwise, we fail to reject the null hypotheses.

Step by step solution

01

Setup hypotheses for marriage support

Understand the hypotheses. Null hypothesis (H0): The proportion of young adults who think their parents would provide marriage support and the proportion of parents who say they would provide marriage support are equal. This can be written as \(p1 = p2\). Alternative hypothesis (H_A): The two proportions are not equal, written as \(p1 ≠ p2\).
02

Calculate test statistic for marriage support

Compute the pooled proportion which is the total number of successful responses divided by the total number of responses, and then calculate the test statistic (Z). For this study: The pooled proportion \(p = (0.41*600 + 0.43*300) / (600 + 300)\) The test statistic is calculated by \[ Z = \frac{p1 - p2}{\sqrt{p * (1- p) * [ 1/n1 + 1/n2 ]}} = \frac{0.41 - 0.43}{\sqrt{p * (1- p) * [ 1/600 + 1/300 ]}} \]
03

Conduct hypothesis test for marriage support

Get the p-value associated with the test statistic from a standard normal (Z) distribution table or use a statistical software to do so. If the p-value is less than or equal to the significance level (usually 0.05), reject the null hypothesis.
04

Setup hypotheses for house buying support

Set up the hypotheses. Null hypothesis (H0): The proportion of parents who would help their children to buy a house or an apartment are equal to or greater than the proportion of young adults who think their parents would provide such support. This can be written as \(p2 ≥ p1\). Alternative hypothesis (H_A): The proportion of parents who would help their children to buy a house or an apartment is less than the proportion of young adults who think their parents would provide such support, written as \(p2 < p1\).
05

Calculate test statistic for house buying support

Same step like the previous test statistic calculation but this time with data on house buying support: \[ Z = \frac{p1 - p2}{\sqrt{p * (1- p) * [ 1/n1 + 1/n2 ]}} = \frac{0.37 - 0.27}{\sqrt{p * (1- p) * [ 1/600 + 1/300 ]}} \]
06

Conduct hypothesis test for house buying support

Get the p-value associated with the test statistic. Since it is a one-tailed test, if the p-value is less than or equal to the significance level (usually 0.05), reject the null hypothesis.

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.

Independent Random Samples
In statistics, independent random samples are crucial for conducting unbiased hypothesis tests. They refer to two or more samples that are selected from their respective populations in such a way that the selection of a data point in one sample has no influence on the selection of data points in another sample. Moreover, within each sample, the individual observations must be randomly selected and each member of the population must have an equal chance of being included. This ensures that our sample represents the population well and that the comparative analysis between the groups, as in the case of the survey on young adults and parents, is valid and free of confounding variables.
Null Hypothesis
The null hypothesis is a fundamental concept in hypothesis testing, serving as the default or 'status quo' assumption. It asserts that there is no effect or no difference between populations or that any observed effect is due to chance. For instance, in our study concerning financial support for marriage, the null hypothesis posits that the proportion of young adults who believe their parents would provide financial support is the same as the proportion of parents who say they would provide support. In formal terms, we state it as H0: p1 = p2. The null hypothesis is what we attempt to reject based on our sample data.
Alternative Hypothesis
Contrasting with the null hypothesis is the alternative hypothesis, which suggests there is a statistically significant effect or difference between the populations. It represents the outcome that the study aims to support. From the aforementioned survey, the alternative hypothesis for financial support for marriage proposes that the proportions of young adults and parents differ in their readiness to provide support. Mathematically, this is expressed as HA: p1 ≠ p2. If we achieve sufficient evidence against the null, we accept the alternative hypothesis.
Pooled Proportion
When comparing two proportions from independent samples, the pooled proportion represents the combined rate of success across both samples. It is calculated by adding the number of successes in both groups and dividing by the total number of observations in both groups. The pooled proportion is used to calculate the standard error of the difference between two sample proportions under the assumption that the null hypothesis is true, as it provides a common estimate of the proportion that applies to both populations. For example, our study's pooled proportion for the belief in financial support for marriage is derived from both the young adults' and parents' responses.
Test Statistics
Test statistics are the standardized values used to determine whether to reject the null hypothesis within the context of a significance level. In the case of comparing two proportions, we often use the Z-test statistic. It quantifies the degree to which the observed data differ from what the null hypothesis would expect. This is calculated by subtracting the difference between the sample proportions and dividing by the standard error. The resulting Z-value allows us to compare our result against a standard normal distribution to find the p-value, which indicates the probability of obtaining a test statistic at least as extreme as ours if the null hypothesis were true.

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

Are college students who take a freshman orientation course more or less likely to stay in college than those who do not take such a course? The article "A Longitudinal Study of the Retention and Academic Performance of Participants in Freshmen Orientation Courses" (journal of College Student Development [1994]: \(444-\) 449) reported that 50 of 94 randomly selected students who did not participate in an orientation course returned for a second year. Of 94 randomly selected students who did take the orientation course, 56 returned for a second year. Construct a \(95 \%\) confidence interval for \(p_{1}-p_{2}\), the difference in the proportion returning for students who do not take an orientation course and those who do. Give an interpretation of this interval.

An individual can take either a scenic route to work or a nonscenic route. She decides that use of the nonscenic route can be justified only if it reduces the mean travel time by more than 10 minutes. a. If \(\mu_{1}\) is the mean for the scenic route and \(\mu_{2}\) for the nonscenic route, what hypotheses should be tested? b. If \(\mu_{1}\) is the mean for the nonscenic route and \(\mu_{2}\) for the scenic route, what hypotheses should be tested?

The paper "Ready or Not? Criteria for Marriage Readiness among Emerging Adults" (journal of Adolescent Research [2009]: 349-375) surveyed emerging adults (defined as age 18 to 25 ) from five different colleges in the United States. Several questions on the survey were used to construct a scale designed to measure endorsement of cohabitation. The paper states that "on average, emerging adult men \((M=3.75, S D=1.21)\) reported higher levels of cohabitation cndorsement than cmerging adult women \((\mathrm{M}=3.39, \mathrm{SD}=1.17) .\) " The sample sizes were 481 for women and 307 for men. a. Carry out a hypothesis test to determine if the reported difference in sample means provides convincing evidence that the mean cohabitation endorsement for emerging adult women is significantly less than the mean for emerging adult men for students at these five colleges. b. What additional information would you want in order to determine whether it is reasonable to generalize the conclusion of the hypothesis test from Part (a) to all college students?

Some commercial airplanes recirculate approximately \(50 \%\) of the cabin air in order to increase fuel efficiency. The authors of the paper "Aircraft Cabin Air Redrculation and Symptoms of the Common Cold" (journal of the American Medical Association [2002]: 483-486) studied 1100 airline passengers who flew from San Francisco to Denver between January and April 1999\. Some passengers traveled on airplanes that recirculated air and others traveled on planes that did not recirculate air. Of the 517 passengers who flew on planes that did not recirculate air, 108 reported post-flight respiratory symptoms, while 111 of the 583 passengers on planes that did recirculate air reported such symptoms. Is there sufficient evidence to conclude that the proportion of passengers with post-flight respiratory symptoms differs for planes that do and do not recirculate air? Test the appropriate hypotheses using \(\alpha=.05\). You may assume that it is reasonable to regard these two samples as being independently selected and as representative of the two populations of interest.

The Insurance Institute for Highway Safety issucd a press release titled "Teen Drivers Often lgnoring Bans on Using Cell Phones" (June 9,2008 ). The following quote is from the press release: Just \(1-2\) months prior to the ban's \(\underline{\text { Dec. } 1,2006}\) start, 11 percent of teen drivers were observed using cell phones as they left school in the afternoon. Abour 5 months after the ban took effect, \(12 \%\) of teen drivers were observed using cell phones. Suppose that the two samples of teen drivers (before the ban, after the ban) can be regarded as representative of these populations of teen drivers. Suppose also that 200 teen drivers were observed before the ban (so \(n_{1}=200\) and \(\hat{p}_{1}=.11\) ) and 150 teen drivers were observed after the ban. a. Construct and interpret a \(95 \%\) confidence interval for the difference in the proportion using a cell phone while driving before the ban and the proportion after the ban. b. Is zero included in the confidence interval of Part (c)? What does this imply about the difference in the population proportions?

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.