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The Insurance Institute for Highway Safety issued a news release titled "Teen Drivers Often Ignoring Bans on Using Cell Phones" (June 9,2008 ). The following quote is from the news release: Just \(1-2\) months prior to the ban's Dec. \(1,2006,\) start, \(11 \%\) of teen drivers were observed using cell phones as they left school in the afternoon. About 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) are representative of these populations of teen drivers. Suppose also that 200 teen drivers were observed before the ban (so \(n_{1}=200\) and \(\hat{p}_{1}=0.11\) ) and that 150 teen drivers were observed after the ban. a. Construct and interpret a \(95 \%\) large-sample 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 (a)? What does this imply about the difference in the population proportions?

Short Answer

Expert verified
95% confidence interval for the difference in proportion is obtained from step 4, and a conclusion about the difference in population proportions before and after the ban is reached in step 5.

Step by step solution

01

Identify the given data

Before the ban, the sample size \(n_{1}\) is 200 and proportion \(\hat{p}_{1}\) equals 0.11. After the ban, the sample size was not given but the proportion \(\hat{p}_{2}\) is 0.12. The desired confidence level is 95%.
02

Calculate the standard error of the difference

The formula for standard error (SE) of the difference in two proportions is \(\sqrt{ \frac{\hat{p}_{1}(1-\hat{p}_{1})}{n_{1}} + \frac{\hat{p}_{2}(1-\hat{p}_{2})}{n_{2}}}\) . Substituting the given values into the formula, we obtain \(SE = \sqrt{ \frac{0.11(1-0.11)}{200} + \frac{0.12(1-0.12)}{150}}\) . Calculate the numerical value of the standard error.
03

Determine the critical value

For a 95% confidence level, the critical value (z*) from the standard normal distribution is 1.96.
04

Construct the confidence interval

The confidence interval is given by the formula: \((\hat{p}_{1} - \hat{p}_{2}) \pm (z* \cdot SE)\) . Substitute the calculated and given values into the formula to obtain the confidence interval.
05

Interpret the confidence interval

The confidence interval gives a range of plausible values for the difference in population proportions. If the interval includes zero, this indicates no significant difference in population proportions before and after the ban.

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Key Concepts

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

Statistics Education
In the realm of statistics education, it's essential to master the interpretation of various statistical measures. One such measure that students encounter is the confidence interval, which provides a range of values to estimate population parameters based on sample data. A critical aspect here is understanding that the confidence interval is not about the probability of a parameter falling within that range but instead concerns how often the interval, if calculated repeatedly from multiple samples, would contain the parameter. This foundational concept underpins inferential statistics and helps in comprehending more complex statistical procedures such as hypothesis testing and error calculation.
Inferential Statistics
The heart of inferential statistics is in making educated guesses about population parameters based on sample statistics. With such methodologies, we can make assertions about a population without the need to observe every member of it. The exercise at hand, where we seek to determine the confidence interval for the difference in proportions of teen drivers using cell phones before and after a ban, exemplifies the power of inferential statistics. It allows the conclusions drawn from the observed samples to potentially extend to the overall population of teen drivers, highlighting the essential ability of inferential methods to inform policy and decision-making.
Proportion Hypothesis Testing
When discussing proportion hypothesis testing, we're dealing with the testing of claims or hypotheses regarding population proportions. Generally, the null hypothesis suggests no effect or no difference, while the alternative hypothesis posits a definite effect or difference. For our specific scenario, where we explore teen drivers' cellphone usage before and after a ban, we are effectively probing whether there is a significant change in behaviour. The constructed confidence interval is a tool for this hypothesis test — if the interval does not include zero, it provides evidence against the null hypothesis of no difference.
Standard Error Calculation
The concept of standard error calculation is a cornerstone of statistical analysis, quantifying the variability of an estimated parameter across different samples. In the context of a confidence interval for difference in proportions, standard error measures the extent to which the sample proportion might vary from the true population proportion. The lower the standard error, the more precise the estimate is. In the exercise provided, the calculation of standard error incorporates both sample proportions and sizes, crucial in determining the width of the confidence interval. Understanding this calculation is vital as it influences the precision and reliability of conclusions drawn in inferential statistics.

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Most popular questions from this chapter

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 (ages 19 to 35 ), and the other sample consisted of 300 parents of young adults ages 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 in that situation. 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 their parents would provide financial support and the proportion of parents who say they would provide support are different.

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 study by the Kaiser Family Foundation provided one of the first looks at media use among the very youngest children - those from 6 months to 6 years of age (Kaiser Family Foundation, \(2003,\) www.kff.org). The director of the Kaiser Family Foundation's Program for the Study of Entertainment Media and Health said, "It's not just teenagers who are wired up and tuned in, its babies in diapers as well." Because previous research indicated that children who have a TV in their bedroom spend less time reading than other children, the authors of the Foundation study were interested in learning about the proportion of kids who have a TV in their bedroom. They collected data from two independent random samples of parents. One sample consisted of parents of children ages 6 months to 3 years old. The second sample consisted of parents of children ages 3 to 6 years old. They found that the proportion of children who had a TV in their bedroom was 0.30 for the sample of children ages 6 months to 3 years and 0.43 for the sample of children ages 3 to 6 years old. Suppose that each of the two sample sizes was 100 . a. Construct and interpret a \(95 \%\) large-sample confidence interval for the proportion of children ages 6 months to 3 years who have a TV in their bedroom. Hint: This is a one-sample confidence interval. b. Construct and interpret a \(95 \%\) large-sample confidence interval for the proportion of children ages 3 to 6 years who have a TV in their bedroom. c. Do the confidence intervals from Parts (a) and (b) overlap? What does this suggest about the two population proportions? d. Construct and interpret a \(95 \%\) large-sample confidence interval for the difference in the proportions for children ages 6 months to 3 years and for children ages 3 to 6 years. e. Is the interval in Part (d) consistent with your answer in Part (c)? Explain.

The report referenced in the previous exercise also stated that the proportion who thought their parents would help with buying a house or renting an apartment for the sample of young adults was 0.37 . For the sample of parents, the proportion who said they would help with buying a house or renting an apartment was 0.27 . Based on these data, can you conclude that the proportion of parents who say they would help with buying a house or renting an apartment is significantly less than the proportion of young adults who think that their parents would help?

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.

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