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The news release referenced in the previous exercise also included data from independent samples of teenage drivers and parents of teenage drivers. In response to a question asking if they approved of laws banning the use of cell phones and texting while driving, \(74 \%\) of the teens surveyed and \(95 \%\) of the parents surveyed said they approved. The sample sizes were not given in the news release, but suppose that 600 teens and 400 parents of teens were surveyed and that these samples are representative of the two populations. Do the data provide convincing evidence that the proportion of teens who approve of banning cell phone and texting while driving is less than the proportion of parents of teens who approve? Test the relevant hypotheses using a significance level of 0.05 The report titled "Digital Democracy Survey" (2016, www .deloitte.com/us/tmttrends, retrieved December 16,2016\()\) stated that \(31 \%\) of the people in a representative sample of adult Americans age 33 to 49 rated a landline telephone among the three most important services that they purchase for their home. In a representative sample of adult Americans age 50 to \(68,48 \%\) rated a landline telephone as one of the top three services they purchase for their home. Suppose that the samples were independently selected and that the sample size was 600 for the 33 to 49 age group sample and 650 for the 50 to 68 age group sample. Does this data provide convincing evidence that the proportion of adult Americans age 33 to 49 who rate a landline phone in the top three is less than this proportion for adult Americans age 50 to 68 ? Test the relevant hypotheses using \(\alpha=0.05\)

Short Answer

Expert verified
In conclusion, the hypothesis test results in rejecting the null hypothesis (\(H_0\): \(p_1 = p_2\)) in favor of the alternative hypothesis (\(H_a\): \(p_1 < p_2\)), with a p-value of 0 at a 0.05 significance level. This suggests convincing evidence that the proportion of teens who approve of banning cell phone and texting while driving is less than the proportion of parents of teens who approve.

Step by step solution

01

Define the null and alternative hypotheses

Let \(p_1\) be the proportion of teens who approve of banning cell phone and texting while driving, and \(p_2\) be the proportion of parents of teens who approve. We want to test whether \(p_1\) is less than \(p_2\). Null hypothesis (\(H_0\)): \(p_1 = p_2\) Alternative hypothesis (\(H_a\)): \(p_1 < p_2\)
02

Calculate the sample proportions and their difference

Let \(\hat{p}_1\) and \(\hat{p}_2\) be the sample proportions. We have: \(\hat{p}_1 = \frac{0.74 * 600}{600} = 0.74\) \(\hat{p}_2 = \frac{0.95 * 400}{400} = 0.95\) Difference: \(\hat{p}_1 - \hat{p}_2 = 0.74 - 0.95 = -0.21\)
03

Calculate the standard error

Standard error (SE) is calculated as: \[ SE = \sqrt{\frac{\hat{p}_1 (1-\hat{p}_1)}{n_1} + \frac{\hat{p}_2 (1-\hat{p}_2)}{n_2}} \] Plugging in the values, we get: \[ SE = \sqrt{\frac{0.74 * (1-0.74)}{600} + \frac{0.95 * (1-0.95)}{400}} = 0.0267 \]
04

Calculate the test statistic (z-score)

The test statistic (z) is calculated using the difference of sample proportions (\(\hat{p}_1 - \hat{p}_2\)) and the standard error (SE): \[ z = \frac{(\hat{p}_1 - \hat{p}_2) - (p_1 - p_2)}{SE} \] Under the null hypothesis, \(p_1 - p_2 = 0\): \[ z = \frac{-0.21}{0.0267} = -7.87 \]
05

Find the p-value

We have a left-tailed test, so we need to find the area to the left of z in the standard normal distribution. Using a z-table or calculator, we find that the p-value is approximately 0.
06

Compare p-value with significance level and make a conclusion

We set a significance level of 0.05. Since the p-value (0) is smaller than the significance level (0.05), we reject the null hypothesis. Therefore, the data provides convincing evidence that the proportion of teens who approve of banning cell phone and texting while driving is less than the proportion of parents of teens who approve.

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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** (denoted as \(H_0\)) is the statement that there is no effect or difference, and it serves as a starting point for statistical testing. It is a statement of equality, often reflecting a baseline or the status quo. For example, if we write \(H_0: p_1 = p_2\), we claim that the proportion of teens who approve of banning cell phone use while driving is equal to that of the parents.

Here are some key points about the null hypothesis:
  • It is considered true until evidence suggests otherwise.
  • In many tests, the null hypothesis is a hypothesis of no difference or no effect.
  • We only look for evidence to reject the null hypothesis, not to prove it.
  • In practice, rejecting the null hypothesis implies that the alternative hypothesis may be true.
Understanding the null hypothesis is crucial because it lays the foundation for the statistical test and helps establish whether we'll be focusing on finding evidence for its rejection.
Alternative Hypothesis
The **alternative hypothesis** (denoted as \(H_a\)) is the statement we aim to find evidence for in a hypothesis test. It is formulated as the opposite of the null hypothesis, proposing that there is an effect or a difference. For example, \(H_a: p_1 < p_2\) suggests that the proportion of teens approving bans on cell phone use while driving is less than that of parents.

Here are some insights into the alternative hypothesis:
  • It is what you aim to support in a research study.
  • Comes into play once the null hypothesis has been rejected.
  • May represent a new effect, or difference detected in the sample data.
  • Shapes the direction of the test: left-tailed, right-tailed, or two-tailed.
A strong grasp of the alternative hypothesis allows for designing impactful tests and interpreting outcomes meaningfully. It conveys what the study is exploring beyond the initial assumption.
Sample Proportion
The **sample proportion** provides a snapshot of how a segment of the population behaves or thinks. It's computed by dividing the number of "successes" or instances of a trait by the total sample size. In our example, * e have two sample proportions: \(\hat{p}_1 = 0.74\) which represents that 74% of teens approve of a ban, and \(\hat{p}_2 = 0.95\), indicating that 95% of parents approve.

Key points to understand about sample proportions:
  • It reflects the estimate of a true population proportion.
  • Essential for comparing different populations or groups.
  • Comparison often involves calculating the difference or ratio between groups, i.e., \(\hat{p}_1 - \hat{p}_2\).
  • Sample size influences the reliability and stability of the estimate.
Sample proportions help us make inferences about larger groups based on smaller, manageable samples.
P-Value
The **p-value** in hypothesis testing is a measure that helps us determine the strength of evidence against the null hypothesis. Specifically, it represents the probability of obtaining test results at least as extreme as the observed data, under the assumption that the null hypothesis is true. In our scenario, a left-tailed test gave us a p-value of 0, meaning there is a significant evidence to reject the null hypothesis.

Some important aspects of p-values:
  • A low p-value (usually less than 0.05) indicates strong evidence against the null hypothesis, prompting its rejection.
  • The smaller the p-value, the stronger the evidence suggesting the alternative hypothesis is true.
  • It provides a threshold (significance level, \(\alpha\)) for making decisions whether to accept or reject \(H_0\).
  • P-value calculations are intrinsic to various statistical tests, leading to decisive research conclusions.
With a clear understanding of p-values, you can make informed decisions about hypotheses, strengthening the analysis carried out in your studies.

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Most popular questions from this chapter

The Interactive Advertising Bureau surveyed a representative sample of 1000 adult Americans and a representative sample of 1000 adults in China (“Majority of Digital Users in U.S. and China Regularly Shop and Purchase via E-Commerce," November \(10,2016,\) www.iab.com, retrieved December 15,2016 ). They reported that American shoppers are much more likely to use a credit or a debit card to make an online purchase. This conclusion was based on finding that \(63 \%\) of the people in the United States sample said they pay with a credit or a debit card, while only \(34 \%\) of those in the China sample said that they used a credit card or a debit card to pay for online purchases. To determine if the stated conclusion is justified, you want to carry out a test of hypotheses to determine if there is convincing evidence that the proportion who pay with a credit card or a debit card is greater for adult Americans than it is for adult Chinese. a. What hypotheses should be tested to answer the question of interest? b. Are the two samples large enough for the large-sample test for a difference in population proportions to be appropriate? Explain. c. Based on the following Minitab output, what is the value of the test statistic and what is the value of the associated \(P\) -value? If a significance level of 0.01 is selected for the test, will you reject or fail to reject the null hypothesis? d. Interpret the result of the hypothesis test in the context of this problem.

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 \(0.05 .\) (Hint: See Example \(11.4 .)\)

The Bureau of Labor Statistics report referenced in the previous exercise also reported that \(7.3 \%\) of high school graduates were unemployed in October 2013 and \(5.7 \%\) of high school graduates were unemployed in October \(2014 .\) Suppose that the reported percentages were based on independently selected representative samples of 400 high school graduates in each of these 2 years. a. Construct and interpret a 99\% large-sample confidence interval for the difference in the proportions of high school graduates who were unemployed in these 2 years. b. Is the confidence interval from Part (a) wider or narrower than the confidence interval calculated in the previous exercise? Give two reasons why it is wider or narrower.

The article "Fish Oil Staves Off Schizophrenia" (USA TODAY, February 2,2010 ) describes a study in which 81 patients age 13 to 25 who were considered at risk for mental illness were randomly assigned to one of two groups. Those in one group took four fish oil capsules daily. Those in the other group took a placebo. After 1 year, \(5 \%\) of those in the fish oil group and \(28 \%\) of those in the placebo group had become psychotic. Is it appropriate to use the largesample \(z\) test to test hypotheses about the difference in the proportions of patients receiving the fish oil and the placebo treatments who became psychotic? Explain why or why not.

In a test of hypotheses about a difference in treatment proportions, what does it mean when the null hypothesis is not rejected?

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