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, 201 0). 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. In a report, use the sample information to:

1.) Provide a descriptive analysis of student achievement on Advanced Placement before and after the monetary incentive is offered.

2.) 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.

3.) At the 5% significance level, has the monetary incentive decreased the proportion of failing scores of 1 and 2?

4.) Assess the effectiveness of monetary incentives in improving student achievement.

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

6.) Define a Type II Error for this problem

Explanation / Answer

Solution

Part (1)

The given data is summarized in the table below:

Score

1 & 2

(Failing Scores)

3 & 4

5

(Top Score)

Total

Historical

54 (54%)

38 (38%)

8 (8%)

100

Current (incetivised)

61 (50%)

49 (40%)

12 (10%)

122 (100%)

Difference (C - H)

- 4

2

2

-

Apparently, with announcement of incentives, performance in test seems to have improved, albeit marginally. This is the descriptive part of the analysis. How much is the improvement and if that is significant will be revealed only by statistical tests. ANSWER

Part (2)

Let p1 and p2 be the proportion of top scores historically and with incentives respectively.

We want to test

Null Hypothesis: H0 p1 = p2   Vs Alternative: H0p1 < p2

Test Statistic:

Z = (p2cap – p1cap)/[pcap(1 - pcap){(1/n1) + (1/n2)} where p1cap and p2cap are sample proportions, n1, n2 are sample sizes and pcap = {(n1 x p1cap) + (n2 x p2cap)}/(n1 + n2).

Calculations:

p2cap = 12/122 = 0.0984; p1cap = 8/100 = 0.08;

pcap = (12 + 8)/(122 + 100) = 0.0901

(1 - pcap) = 0.9099

pcap(1 - pcap) = 0.08198199

{(1/n1) + (1/n2)} = 0.0082 + 0.01 = 0.0182

Z = 0.4763

Distribution, Critical Value and p-value:

Under H0, distribution of Z can be approximated by Standard Normal Distribution

So, given a level of significance of %, Critical Value = upper % of N(0, 1), and

p-value = P(Z > Zcal)

Using Excel Functions of N(0, 1), Critical Value (for = 5%) = 1.645 and p-value = 0.3169

Decision Criterion (Rejection Region):

Reject H0, if Zcal > Zcrit or if p-value < .

Decision:

Since Zcal < Zcrit, p-value > , H0 is accepted.

Conclusion :

There is not enough evidence to suggest that the incentive announcement has improved the top score.

DONE ANSWER    

Part (3)

Let p1 and p2 be the proportion of Failing Scores historically and with incentives respectively.

We want to test

Null Hypothesis: H0 p1 = p2   Vs Alternative: H0p1 > p2

Test Statistic:

Z = (p1cap – p2cap)/[pcap(1 - pcap){(1/n1) + (1/n2)} where p1cap and p2cap are sample proportions, n1, n2 are sample sizes and pcap = {(n1 x p1cap) + (n2 x p2cap)}/(n1 + n2).

Calculations:

p2cap = 61/122 = 0.5; p1cap = 54/100 = 0.54;

pcap = (61 + 54)/(122 + 100) = 0.5180

(1 - pcap) = 0.4920

pcap(1 - pcap) = 0.254856

{(1/n1) + (1/n2)} = 0.0082 + 0.01 = 0.0182

Z = 0.5873

Distribution, Critical Value and p-value:

Under H0, distribution of Z can be approximated by Standard Normal Distribution

So, given a level of significance of %, Critical Value = upper % of N(0, 1), and

p-value = P(Z > Zcal)

Using Excel Functions of N(0, 1), (for = 5%)

Critical Value = 1.645 and p-value = 0.2785

Decision Criterion (Rejection Region):

Reject H0, if Zcal > Zcrit or if p-value < .

Decision:

Since Zcal < Zcrit, p-value > , H0 is accepted.

Conclusion :

There is not enough evidence to suggest that the incentive announcement has brought down the failing scores.

DONE ANSWER    

Part (4)

As seen in the above parts, the incentive has neither brought down the failing scores nor enhanced the top score. Thus, the incentive does not seem to have any impact on the performance. ANSWER

Score

1 & 2

(Failing Scores)

3 & 4

5

(Top Score)

Total

Historical

54 (54%)

38 (38%)

8 (8%)

100

Current (incetivised)

61 (50%)

49 (40%)

12 (10%)

122 (100%)

Difference (C - H)

- 4

2

2

-

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