Statistics

EXAMPLE: A law school is looking at the effectiveness of a particular Bar preparation class. They looked at 150 recent graduates and found that 100 failed the Bar exam on their first attempt. They also determined that 40 of the students who completed the Bar -prep c lass passed on their first attempt. The school completed a contingency table based on their analysis. Use the information in the table to discuss the following questions. Does knowing if a graduate has completed the class help the school predict how lik ely the graduate is to pass the Bar Exam on the first try? Why? What does you answer imply about the statistical relationship between completing the class and passing the Bar Exam? Use the relationship between Marginal, Joint and Conditional probability from this week's lesson to justify your answers. INITIAL POST Knowing that the graduate completed the Bar -prep class does help the school predict if graduates will pass the Bar exam on the first attempt. If completing the class affects the probability of passing the Bar Exam on the first attempt, then the Conditional Probability of Passing, given completing the class, P(Passing | Completed) will be different than the Marginal Probability of Passing P(Passing) .P(Passing) = 50/150 = 0.333 P(Passing | Comple ted) = P(Passing AND Completed) / P(Completed) P(Passing | Completed) = (40/150) / (69/150) = 0.580 Complete Bar-prep Class No Bar-prep Class Total Passed the Bar 40 10 50 Failed the Bar 29 71 100 Total 69 81 150 Knowing that a graduate completed the Bar -prep class increased their probability of passing the Bar Exam on the first attempt from 33.3% to 58.0%. Since t he outcome of one event (completing the class) affects the probability of the second event (passing the Bar on the first try) the two events are not independent. We can confirm this by using the Multiplication Rule, which states: “If A and B are independen t, the probability of A AND B is equal to the probability of A times the probability of B. P(Passing) = 50/150 = 0.333 P(Completed) = 69/150 = 0.460 P(Passing) * P(Completed) = 0.333 * 0.460 = 0.153 P(Passing AND Completed) = 40/150 =0.267 P(Passing AND Completed) ≠ P(Passing ) * P(Completed) so the two events are not independent