Question

Are college students who take a freshman orientation course more or less likely to stay in college than those who do not take such a course? The article "A Longitudinal Study of the Retention and Academic Performance of Participants in Freshmen Orientation Courses" (Journal of College Student Development \([1994]: 444-\) 449) reported that 50 of 94 randomly selected students who did not participate in an orientation course returned for a second year. Of 94 randomly selected students who did take the orientation course, 56 returned for a second year. Construct a \(95 \%\) confidence interval for \(\pi_{1}-\pi_{2}\), the difference in the proportion returning for students who do not take an orientation course and those who do. Give an interpretation of this interval.

Short answer

The \(95\%\) confidence interval for \(\pi_{1}-\pi_{2}\) will provide a range of values that includes the true population parameter \(95\%\) of the time. The interpretation will depend on the computed interval. If 0 is included in the interval, then there is no significant difference in the return rates for the two groups.

Step-by-step solution

Identify Given Data

From the problem, we know that 50 of 94 students who did not participate in the orientation course returned for a second year, therefore \(\pi_{1} = \frac{50}{94}\). Similarly, 56 of 94 students who did take the course, returned for a second year, so \(\pi_{2} = \frac{56}{94}\). We are asked to compute a \(95\%\) confidence interval.

Calculate the Standard Error

The standard error (SE) for \(\pi_{1} - \pi_{2}\) can be calculated using the formula: \[ \text{SE}_{\pi_{1} - \pi_{2}} = \sqrt{\frac{\pi_{1} (1 - \pi_{1})}{n_1} + \frac{\pi_{2} (1 - \pi_{2})}{n_2}} \] Where \(n_1\) and \(n_2\) represent the number of individuals in each group, both of which are 94 in this case.

Calculate the Confidence Interval

A \(95\%\) confidence interval can be calculated using the following formula: \[ CI = (\pi_{1} - \pi_{2}) \pm Z_{0.025} \times \text{SE}_{\pi_{1} - \pi_{2}} \] Here \(Z_{0.025}\) is the z-value for a \(95\%\) interval, which is 1.96.

Interpret the Confidence Interval

The confidence interval will give the range of values within which we are \(95\%\) confident that the difference between the two proportions lies. If the interval includes 0, it suggests no significant difference between the likelihood of returning to college based on whether or not they attended the orientation. If all values in the interval are positive (or negative), it would suggest a significant difference.