AP Statistics probability question - Sampling Distributions

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? A study revealed that 67% of students who enrolled in an orientation course returned from a second year, while only 55% of students who did not enroll in an orientation course returned for a second year. A random sample of 20 from students who took the course and 25 students who did not participate in an orientation course was taken. What is the probability that the difference in between the sample that did that the orientation has at least a 10% higher return rate than the sample of students who did not take the orientation class? Use the 4 step process. (state, plan, do, conclude).

Answer 1

Hey Shilpa!

Great to see your interest in AP Statistics! Let’s walk through your probability question using the four-step process: State, Plan, Do, and Conclude.

First, stating the problem, we want to determine whether college students who take a freshman orientation course are more likely to stay in college compared to those who don’t. According to the study, 67% of students who enrolled in the orientation course returned after their second year, while only 55% of those who didn’t take the course returned. You’ve taken a sample of 20 students from the orientation group and 25 from the non-orientation group. The goal is to find the probability that the difference in return rates between these two samples is at least ten percentage points higher for the orientation group.

Next, in the planning phase, we’ll use the concept of the sampling distribution of the difference between two proportions. This involves calculating the sample proportions, determining the variability or standard error of the difference, finding how many standard deviations the desired difference is from the observed difference (which gives us the z-score), and then using this z-score to find the corresponding probability.

Moving on to the calculation step, we start by noting the sample proportions: 67% for the orientation group and 55% for the non-orientation group, which gives us a difference of twelve percentage points. To understand how much this difference might vary just by chance, we calculate the standard error, which takes into account the sample sizes and the proportions from both groups. After performing these calculations, we find that the standard error is approximately 0.1446. Using this, we determine the z-score for a ten percentage point difference. The z-score tells us how many standard errors the desired difference is away from our observed difference. In this case, the z-score is approximately negative 0.138.

Finally, in the conclusion, we interpret the z-score using the standard normal distribution (often referred to as the z-table). A z-score of negative 0.138 corresponds to a probability of about 44.5% to the left. Since we’re interested in the probability that the difference is at least ten percentage points in favor of the orientation group, we look at the right side of the distribution. Subtracting the left-side probability from one gives us approximately a 55.5% probability. This means there is about a 55.5% chance that the orientation group’s return rate is at least ten percentage points higher than the non-orientation group’s in your samples. While this suggests a positive impact of the orientation course on student retention, the probability isn’t extremely high, indicating that further research with larger samples or additional factors might be necessary to draw more definitive conclusions.

Hope this helps!