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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

"Smartest People Often Dumbest About Sunburns" is the headline of an article that appeared in the San Luis Obispo Tribune (July 19,2006 ). The article states that "those with a college degree reported a higher incidence of sunburn that those without a high school degree \(-43 \%\) versus \(25 \% . "\) Suppose that these percentages were based on random samples of size 200 from each of the two groups of interest (college graduates and those without a high school degree). Is there convincing evidence that the proportion experiencing a sunburn is greater for college graduates than it is for those without a high school degree? Test the appropriate hypotheses using a significance level of 0.01 .

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