The effort to reward city students for passing Advanced Placement tests is part

Question

The effort to reward city students for passing Advanced Placement tests is part of a growing trend nationally and internationally. Financial incentives are offered in order to lift attendance and achievement rates. One such program in Dallas, Texas, offers $100 for every Advanced Placement test on which a student scores a three or higher (Reuters, September 20, 2010). A wealthy entrepreneur decides to experiment with the same idea of rewarding students to enhance performance, but in Chicago. He offers monetary incentives to students at an inner-city high school. Due to this incentive, 122 students take the Advancement Placement tests. Twelve tests are scored at 5, the highest possible score. There are 49 tests with scores of 3 and 4, and 61 tests with failing scores of 1 and 2. Historically, about 100 of these tests are taken at this school each year, where 8% score 5, 38% score 3 and 4, and the remaining are failing scores of 1 and 2.

1) Conduct a hypothesis test that determines, at the 5% significance level, whether the monetary incentive has resulted in a higher proportion of scores of 5, the highest possible score. (Show work)

2)  Define a Type I Error for this problem. What is the probability of making a Type I Error?

3) Define a Type II Error for this problem.

Explanation / Answer

Part (1)

Let X = number of score 5 out of 100 (historical figure) and

Y = number of score 5 out of 122 (under incentive). Then,

Proportion of scores of 5 historical, p1 = 8/100 = 0.08 and

Proportion of scores of 5 under incentive, p2 = 12/122 = 0.0984

To test if monetary incentive has resulted in a higher proportion of scores of 5, we test the null hypothesis: p1 = p2 versus the alternative: p1 < p2, (i.e., the claim) in the population, with the test statistic:

Z = (p2 – p1)/[p(1 - p){(1/100) + (1/122)}

= 0.0184/0.0386 [p = (12 + 8)/(122 + 100)]

   = 0.4763

The critical value for this test is upper 5% point of N(0,1) which is 1.645 as per the standard table.

Since the calculated value of Z is less than the critical value, the null hypothesis: p1 = p2 is accepted implying that there is no evidence to suggest that monetary incentive has resulted in a higher proportion of scores of 5. ANSWER

Part (2)

Type I Error for the above problem is the error of rejecting the null hypothesis: p1 = p2, when in fact, p1 = p2.

By definition of Type I Error and Level of Significance for any test, the probability of making a Type I Error = Level of Significance.

Thus, the answer is 5% ANSWER

Part (3)

Type II Error for this problem is the error of accepting the null hypothesis:

p1 = p2, when in fact, p1 < p2 (i.e., when alternative is true in reality). ANSWER

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.