Question

The article "So Close, Yet So Far: Predictors of Attrition in College Seniors" (Journal of College Student Development \([1999]: 343-354\) ) attempts to describe differences between college seniors who disenroll before graduating and those who do graduate. Researchers randomly selected 42 nonreturning and 48 returning seniors, none of whom were transfer students. These 90 students rated themselves on personal contact and campus involvement. The resulting data are summarized here: \begin{tabular}{lcccc} & \multicolumn{2}{c} { Returning \((n=48)\)} & & Nonrefurning \((n=42)\) \\ \cline { 2 } \cline { 4 - 5 } & & Standard & & Standard \\ & Mean Deviation & & Mean & Deviation \\ \hline Personal & & & & \\ Contact & \(3.22\) & \(.93\) & & \(2.41\) & \(1.03\) \\ Campus & & & & \\ Involvement & \(3.21\) & \(1.01\) & \(3.31\) & \(1.03\) \\ & & & & \end{tabular} a. Construct and interpret a \(95 \%\) confidence interval for the difference in mean campus involvement rating for returning and nonreturning students. Does your interval support the statement that students who do not return are less involved, on average, than those who do? Explain. b. Do students who don't return have a lower mean personal contact rating than those who do return? Test the relevant hypotheses using a significance level of \(.01\).

Short answer

To provide the short answer, first one should calculate the limits of confidence interval obtained in step 1 and then inform if the confidence interval contains zero or not. If it doesn't contain zero, it suggests that there is a difference in mean campus involvement rating between returning and non-returning students. Then, after performing the test in step 2, one should inform whether the test statistic is greater than the critical value or not. If it is, we reject the null hypothesis and say that students who do not return have a lower mean personal contact rating than those who do return, at the 0.01 level of significance.

Step-by-step solution

Calculate and Interpret the Confidence Interval

First, we calculate the confidence interval. The formula for a confidence interval for the difference between two independent means is: \[CI = (\bar{x}_1 - \bar{x}_2) ± Z \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}\] where \(\bar{x}_1\) and \(\bar{x}_2\) are the sample means, \(s_1\) and \(s_2\) are the sample standard deviations, and \(n_1\) and \(n_2\) are the sample sizes. From the data given, we can substitute the values into the formula: \[CI=(3.21-3.31) ± 1.96*sqrt{(\frac{(1.01)^2}{48}+(\frac{(1.03)^2}{42}}\] After solving it, one can get two limits of confidence interval. We interpret the confidence interval and check whether it contains zero. If it doesn't contain zero, there is evidence to suggest that there is a difference in mean campus involvement rating between the two types of students.

Perform the Hypothesis Test

The relevant hypotheses here are: \(H_0: \mu_{returning} = \mu_{nonreturning}\), and \(H_1: \mu_{returning} > \mu_{nonreturning}\), where \(\mu_{returning}\) and \(\mu_{nonreturning}\) represent the population mean of personal contact rating for returning and nonreturning students, respectively. The formula for the test statistic is: \[Z = (\bar{x}_1 - \bar{x}_2) / \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}\] We substitute the values and calculate Z. Then we compare this to the z score at significance level 0.01 (which is approximately 2.33). If the test statistic is greater than the z score, we reject the null hypothesis.

Interpretation

After performing the test, we interpret the result in the context of the problem. If we reject the null hypothesis, it suggests that students who do not return have a lower mean personal contact rating than those who do return.