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The article "More Teen Drivers See Marijuana as OK; It's a Dangerous Trend" (USA TODAY, February 23,2012 ) describes two surveys of U.S. high school students. One survey was conducted in 2009 and the other was conducted in \(2011 .\) In \(2009,78 \%\) of the people in a representative sample of 2300 students said marijuana use is very distracting or extremely distracting to their driving. In \(2011,70 \%\) of the people in a representative sample of 2294 students answered this way. Use the five-step process for estimation problems \(\left(\mathrm{EMC}^{3}\right)\) to construct and interpret a \(99 \%\) large-sample confidence interval for the difference in the proportion of high school students who believed marijuana was very distracting or extremely distracting in 2009 and this proportion in 2011 .

Short Answer

Expert verified
The 99% confidence interval for the difference in the proportion of high school students who believed marijuana usage is distracting to their driving between 2009 and 2011 is (0.0523, 0.1077). This implies a significant decrease in the proportion of students who considered marijuana distracting to their driving, with a decrease ranging from 5.23% to 10.77% during this period. The trend indicates that more high school students in 2011 regarded marijuana as less problematic for their driving as compared to those in 2009.

Step by step solution

01

State the estimation problem

We want to estimate the difference between the proportion of students who believed marijuana usage is distracting to their driving in 2009 and this proportion in 2011.
02

Plan the method

We will use the large-sample confidence interval formula for comparing two proportions to construct a 99% confidence interval. The formula is given by: \[CI = (\hat{p_1} - \hat{p_2}) \pm z^* \sqrt{\frac{\hat{p_1}(1-\hat{p_1})}{n_1}+\frac{\hat{p_2}(1-\hat{p_2})}{n_2}}\] Where \(\hat{p_1}\) and \(n_1\) are the proportion and sample size from 2009, \(\hat{p_2}\) and \(n_2\) are the proportion and sample size from 2011, and \(z^*\) is the critical value for a 99% confidence level (which is 2.576).
03

Identify the data and perform the calculations

Given data: \(\hat{p_1} = 0.78, n_1 = 2300, \hat{p_2} = 0.70, n_2 = 2294\) Now, we can plug these values into the large-sample confidence interval formula: \[ CI = (0.78 - 0.70) \pm 2.576 \sqrt{\frac{0.78(1-0.78)}{2300}+\frac{0.70(1-0.70)}{2294}} \] Calculating the confidence interval: \[ CI = 0.08 \pm 2.576 \times \sqrt{0.00006883 + 0.00007333} \] \[ CI = 0.08 \pm 2.576 \times 0.01078 \] \[ CI = 0.08 \pm 0.0277 \] So the 99% confidence interval is (0.0523, 0.1077).
04

Make a conclusion

We can conclude with 99% confidence that the difference in the proportion of high school students who believed marijuana usage is distracting to their driving (2009 proportion minus 2011 proportion) lies between 5.23% and 10.77%.
05

Interpret the result

This result implies that there is a significant decrease in the proportion of students who believed marijuana is distracting to their driving between 2009 and 2011. The proportion of students who believed marijuana was very distracting or extremely distracting to their driving decreased by at least 5.23% and possibly as much as 10.77% during this period. This trend suggests that more high school students in 2011 considered marijuana as not as problematic for their driving as compared to those in 2009.

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Key Concepts

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

Large-Sample Confidence Interval
When we talk about large-sample confidence intervals, particularly when comparing proportions, we're dealing with a statistical method that allows us to estimate the range within which we expect the true difference in proportions to fall. Confidence intervals for large samples are useful because they provide insight into how much difference exists between two samples, such as survey results from different years.

In the context of our exercise, we're examining two groups of high school students from 2009 and 2011. We use their responses to estimate the range in which the true difference in opinions on marijuana being distracting when driving lies. Because we have large samples (2300 and 2294 students respectively), our calculations are more reliable, leading to a more accurate interval estimate. This interval helps us understand the change in attitude over time, by giving us a clear indication of how much this perspective shifted, accounting for sample size and variability.
Difference in Proportions
The difference in proportions is a concept utilized when comparing two distinct categorical datasets to find out how much one proportion differs from the other. It's especially useful in cases where surveys or experiments are conducted at different times or with different groups.

In our exercise, we've observed two surveys, each involving high school students. The 2009 survey showed that 78% of students felt marijuana use was distracting, while in 2011, only 70% felt the same way. The difference in proportions is simply the 2009 proportion minus the 2011 proportion, which is 0.08 or 8%.
  • This difference shows us an observed trend in changing opinions between the two years.
  • By examining this difference, we can interpret whether such changes are random fluctuations or signify a real shift in attitude that warrants attention.
Estimation Process
The estimation process is a structured approach to deriving values that summarize sample data, offering insights about the population from which the data were drawn. It is crucial in making informed decisions based on survey results, like those discussed here.

For this exercise, a five-step estimation process was used:
  • Firstly, we defined the estimation problem clearly, helping us focus on what exactly needed to be measured—in this case, the difference in student opinions over two years.
  • The next step involved planning how to collect and analyze the relevant data, employing the large-sample confidence interval formula.
  • Then, we gathered the given data from both surveys and incorporated these into our calculations to estimate the confidence interval.
  • After computing the necessary values, we drew conclusions about the estimated range of the difference in proportions.
  • Finally, these results were interpreted to understand their real-world implications—such as whether students' perceptions of marijuana and driving had indeed changed significantly.
Critical Value for Confidence Level
The critical value is a fundamental component in calculating confidence intervals. It represents the number of standard errors to move away from the sample statistic to create a specified level of confidence. The higher the critical value, the wider the confidence interval, reflecting greater uncertainty in the estimate.

In this exercise, the designated confidence level was 99%, which means we want a high degree of certainty in our results. To achieve this, we used a critical value of 2.576, derived from the standard normal distribution.
  • This value indicates that we expect 99% of similar sample differences to fall within this interval, ensuring robustness in our estimation.
  • The choice of this critical value affects the width of the confidence interval—higher confidence levels yield wider intervals, giving a more cautious range but one that is more likely to encompass the true population difference.
By understanding and selecting the appropriate critical value, we improve the accuracy and reliability of our confidence interval estimation, making our conclusions more dependable.

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Most popular questions from this chapter

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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