Question 1 (4 points) The American College Health Association produced the National College Health Assessment (Andy Gardiner, "Surfacing from Depression," February 6, 2006). The assessment indicates that the percentage of U.S. college students who report having been diagnosed with depression has risen from 2000 . The assessment surveyed 47,202 students at 74 campuses. It discovered that \( 10.3 \% \) and \( 14.9 \% \) of students indicated that they had been diagnosed with depression in 2000 and 2004 , respectively. Assume that half of the students surveyed were surveyed in 2004. Indicate the margin of error for estimating \( p_{1}-p_{2} \) with \( \bar{p} 1-\bar{p} 2 \).

To calculate the margin of error for estimating the difference in proportions \( p_{1} - p_{2} \), we first need to determine the sample proportions. We have: - \( \bar{p}_1 = 0.103 \) (proportion of students diagnosed with depression in 2000) - \( \bar{p}_2 = 0.149 \) (proportion of students diagnosed with depression in 2004) For the sample sizes, since half of the 47,202 students were surveyed in each year, we have: - \( n_1 = n_2 = \frac{47,202}{2} = 23,601 \) Next, we can calculate the standard error (SE) of the difference in proportions: \[ SE = \sqrt{\frac{\bar{p}_1(1 - \bar{p}_1)}{n_1} + \frac{\bar{p}_2(1 - \bar{p}_2)}{n_2}} \] Substituting the values: \[ SE = \sqrt{\frac{0.103(1 - 0.103)}{23601} + \frac{0.149(1 - 0.149)}{23601}} \] \[ = \sqrt{\frac{0.103 \cdot 0.897}{23601} + \frac{0.149 \cdot 0.851}{23601}} \] \[ = \sqrt{\frac{0.09208}{23601} + \frac{0.126799}{23601}} \] \[ = \sqrt{\frac{0.218879}{23601}} \approx \sqrt{0.00000927} \approx 0.00305 \] To find the margin of error (ME), we typically use a z-score. For a 95% confidence level, the z-score is approximately 1.96. Thus: \[ ME = z \cdot SE \approx 1.96 \cdot 0.00305 \approx 0.00598 \] So, the margin of error for estimating \( p_{1} - p_{2} \) is approximately \( 0.00598 \) or about \( 0.6\% \).