91Ó°ÊÓ

Binge drinking The PACE project (pace.uhs.wisc.edu) at the University of Wisconsin in Madison deals with problems associated with high-risk drinking on college campuses. Based on random samples, the study states that the percentage of UW students who reported bingeing at least three times within the past two weeks was \(42.2 \%\) in \(1999(n=334)\) and \(21.2 \%\) in \(2009(n=843)\) a. Estimate the difference between the proportions in 1999 and \(2009,\) and interpret. b. Find the standard error for this difference. Interpret it. c. Construct and interpret a \(95 \%\) confidence interval to estimate the true change, explaining how your interpretation reflects whether the interval contains \(0 .\) d. State and check the assumptions for the confidence interval in part \(\mathrm{c}\) to be valid.

Step by step solution

01

Calculate the Estimated Difference in Proportions

The difference between the proportions for the years 1999 and 2009 is calculated by subtracting the two percentages. The proportion in 1999 is 42.2% and in 2009 is 21.2%. The estimated difference in these proportions is: \[ \hat{p}_{1999} - \hat{p}_{2009} = 0.422 - 0.212 = 0.21 \] This means there was a decrease of 21% in the percentage of students binge drinking three times in the past two weeks from 1999 to 2009.

02

Calculate the Standard Error (SE) for the Difference

The formula for the standard error of the difference between two independent proportions is given by: \[ SE = \sqrt{ \frac{\hat{p}_{1999} \times (1 - \hat{p}_{1999})}{n_{1999}} + \frac{\hat{p}_{2009} \times (1 - \hat{p}_{2009})}{n_{2009}} } \] Substituting the values: \[ SE = \sqrt{ \frac{0.422 \times 0.578}{334} + \frac{0.212 \times 0.788}{843} } = \sqrt{\frac{0.243516}{334} + \frac{0.167056}{843}} \approx \sqrt{0.000729 + 0.000198} = \sqrt{0.000927} \approx 0.0304 \] This standard error indicates the extent to which the sampling distribution of the difference between proportions would vary.

03

Construct the 95% Confidence Interval

To construct a 95% confidence interval for the difference in proportions, we use the formula: \[ \text{Confidence Interval} = (\hat{p}_{1999} - \hat{p}_{2009}) \pm Z \times SE \] For a 95% confidence level, the Z-score is 1.96. Thus, \[ CI = 0.21 \pm 1.96 \times 0.0304 \] Calculating the interval gives: \[ CI = 0.21 \pm 0.0596 \] Which results in the interval: \[ (0.1504, 0.2694) \] This confidence interval implies we are 95% confident that the true difference in proportions falls between 15.04% and 26.94%.

04

Interpret Whether the Confidence Interval Contains Zero

Since the interval \((0.1504, 0.2694)\) does not contain zero, we can conclude that there is a statistically significant decrease in the proportion of students who reported binge drinking from 1999 to 2009.

05

Check Assumptions for Confidence Interval

The main assumption for this confidence interval is that the samples are independent random samples. Additionally, the sample sizes must be large enough for the normal approximation to be valid, which can be checked using the rule that \(n\hat{p} \) and \(n(1-\hat{p})\) for both years should be larger than 5. For 1999: - \(334 \times 0.422 = 140.91\) and \(334 \times 0.578 = 193.09\)For 2009: - \(843 \times 0.212 = 178.596\) and \(843 \times 0.788 = 664.404\)All values are greater than 5, thus validating the assumptions needed for the interval calculation.

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Key Concepts

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

Difference of Proportions

The difference of proportions is a crucial concept when comparing two groups over time. In the context of the PACE project study, this refers to the change in binge drinking rates between 1999 and 2009 among UW students. To find this difference, we subtract the proportion of students who binge drank in 2009 (21.2%) from those in 1999 (42.2%). The calculation is straightforward: 0.422 - 0.212 = 0.21.
This result tells us there was a 21% reduction in binge drinking over the decade. Understanding this helps in grasping how student behaviors change over time and can aid in evaluating the effectiveness of interventions. A positive difference would indicate a decrease, as seen here, while a negative difference would suggest an increase in the behavior being studied.

Standard Error

Standard error (SE) is a measure that tells us about the variability of a statistic. In the case of the difference in proportions, it quantifies the extent to which the difference between two proportions may vary from sample to sample.
In mathematical terms, the standard error for the difference of two independent proportions is given by: \[ SE = \sqrt{ \frac{\hat{p}_{1999} \times (1 - \hat{p}_{1999})}{n_{1999}} + \frac{\hat{p}_{2009} \times (1 - \hat{p}_{2009})}{n_{2009}} } \] Using our data, this calculates to approximately 0.0304.
By assessing the SE, we understand the degree of uncertainty or "noise" we might have around the estimated difference. A smaller SE indicates more precision in our estimates.

Statistical Significance

Statistical significance tells us whether the observed difference is likely to be genuine or due to random chance. In the context of the confidence interval constructed for this study, any interval not containing zero suggests statistical significance.
For the PACE project, the confidence interval for the difference in binge drinking rates was (15.04%, 26.94%). Since zero is not within this range, we infer that there's a statistically significant decrease in binge drinking rates.
This significance implies we can confidently state that the drinking behavior among students has indeed changed between 1999 and 2009, rather than the observed difference being a result of sampling variability or randomness.

Assumptions for Confidence Intervals

For the confidence interval about the difference of proportions to be valid, certain assumptions must be satisfied. Primarily, these include the independence of samples and the condition that sample sizes are adequate.
Independence requires that the selection of participants for one sample doesn’t influence the selection of the other. Large enough sample sizes ensure that the sampling distribution approximates normality, making confidence interval calculations more reliable.
In this study, both sample sizes were checked: \( n \hat{p} \) and \( n(1-\hat{p}) \) were greater than 5 for both years, meeting the sufficiency requirement. This checks ensure that the confidence interval is applicable and trustworthy for making inferences about the population from the sample data.