Question

er’s license exam. An incoming class of 100 students is randomly assigned to two groups, each of size 50. One group is taught by Instructor A; the other is taught by Instructor B. At the end of the course, 30 of Instructor A’s students and 22 of Instructor B’s students pass the state exam. Do these results give convincing evidence that Instructor A is more effective?A driving school wants to find out which of its two instructors is more effective at preparing students to pass the state’s driver’s license exam. An incoming class of 100 students is randomly assigned to two groups, each of size 50. One group is taught by Instructor A; the other is taught by Instructor B. At the end of the course, 30 of Instructor A’s students and 22 of Instructor B’s students pass the state exam. Do these results give convincing evidence that Instructor A is more effective?

Answer 1

Answer:

Yes, there is evidence to support that claim that instructor 1 is more effective than instructor 2

Step-by-step explanation:

We can conduct a hypothesis test for the difference of 2 proportions.  If there is no difference in instructor quality, then the difference in proportions will be zero.  That makes the null hypothesis

H0:  p1 - p2 = 0

The question is asking whether instructor 1 is more effective, so if he is, his proportion will be larger than instructor 2, so the difference would result in a positive number.  This makes the alternate hypothesis

Ha:  p1 - p2 > 0

This is a right tailed test (the > or < sign always point to the critical region like an arrowhead)

We will use a significance level of 95% to conduct our test.  This makes the critical values for our test statistic: z > 1.645.  

If our test statistic falls in this region, we will reject the null hypothesis.

See the attached photo for the hypothesis test and conclusion