In a graduate college, the average length of registration time during a semester is 120 minutes with a standard deviation of 25 minutes. With the introduction of a new registration procedure a random sample of 50 students took an average of 80 minutes with a standard deviation of 12 minutes.

**Step 1. State the Hypotheses** We wish to test if the new registration procedure reduces the average registration time. Define the population mean registration time under the new procedure as \( \mu \). The hypotheses are: - \( H_0: \mu = 120 \) minutes (the new procedure does not change the time) - \( H_1: \mu < 120 \) minutes (the new procedure reduces the time) **Step 2. Choose the Appropriate Test Statistic** Since the historical average registration time and its standard deviation are provided and we assume them to represent the population parameters, we can employ the \( z \)-test. The test statistic is given by \[ z = \frac{\bar{x} - \mu_0}{\sigma / \sqrt{n}}, \] where - \( \bar{x} = 80 \) minutes is the sample mean, - \( \mu_0 = 120 \) minutes is the claimed population mean, - \( \sigma = 25 \) minutes is the population standard deviation, - \( n = 50 \) is the sample size. **Step 3. Compute the Test Statistic** First, compute the standard error (SE): \[ \text{SE} = \frac{\sigma}{\sqrt{n}} = \frac{25}{\sqrt{50}}. \] Since \( \sqrt{50} \approx 7.071 \), \[ \text{SE} \approx \frac{25}{7.071} \approx 3.536. \] Next, calculate the \( z \)-score: \[ z = \frac{80 - 120}{3.536} = \frac{-40}{3.536} \approx -11.31. \] **Step 4. Determine the p-value and Decision** For a one-tailed \( z \)-test, a \( z \)-score of approximately \(-11.31\) is extremely low. The corresponding p-value is essentially zero. Since the p-value is far below any common significance level (e.g., \( \alpha = 0.05 \) or \( 0.01 \)), we reject the null hypothesis. **Step 5. Conclusion** There is overwhelming statistical evidence that the new registration procedure decreases the registration time.