One Management professor adopted a crossword puzzle exercise in his introductory management information systems (MIS) class, because he believes that the most important effects of engaging in a crossword exercise is the great workout it gives one’s brain—solving crossword puzzles requires several skills, such as spelling, reasoning, making inferences, evaluating choices, and drawing conclusions. That is, the exercise may enhance one’s ability to memorize words. He wants to know whether or not the crossword puzzle exercise can accelerate the learning of MIS vocabulary and terms. A total of 109 students in his two sections of a MIS class in fall 2009 participated in an experiment involving student learning and retention of MIS terms using the crossword puzzle exercise. One section (56 students) was defined as a control group, while the other section (53 students) was defined as an experimental group. One day about two weeks before the end of semester, students in the control group were given a list of terms taken from the text book used in the course; students in the experimental group were given a crossword puzzle exercise containing terms and phrases taken from the same text book and based on the same list of terms given to the control group. Before an in-class quiz, both groups had forty minutes to use their handouts to prepare; after forty minutes, both groups were given the same quiz. To ensure that experimental subjects worked the crossword puzzle, they had to submit a complete puzzle to the professor prior to the quiz. After the quiz, the professor computed the proportion of students who received A and B grades of the quiz for these two groups. If the proportion of A and B grades is significantly high, it implies that the crossword puzzle exercise may accelerate student learning. The proportion of the experimental group is 0.7547, while the proportion of the control group is 0.6429. He then developed a two-tailed test. The null hypothesis is: the proportion of A and B grades in the experimental group that used the crossword puzzle handout as an aid in studying for their quiz is not different from the proportion of A and B grades in the control group that did not have that handout. The alternative hypothesis is: the proportion of A and B grades in the experimental group that used the crossword puzzle handout as an aid in studying for their quiz is different from the proportion of A and B grades in the control group that did not have that handout. Since he did not have any further information about the population, he used t test statistic. Did the professor develop a correct hypothesis test and use a correct test statistic? If you think yes, explain why you think yes; if no, explain why not and how to develop a correct hypothesis test. Also, explain what it will happen (including the test statistic and p-value) if the professor developed a wrong hypothesis test? Discuss and explain your reasons. You must provide your statistical analysis and reasons.
Solution
The hypothesis formed is incorrect. Answer 1
‘That is, the exercise may enhance one’s ability to memorize words. He wants to know whether or not the crossword puzzle exercise can accelerate the learning of MIS vocabulary and terms.’
The above clearly suggest that the professor is expecting the experimental group to perform better. So, it should be a one-sided (right-tail, i.e., > type) test.
Thus, the correct hypotheses must be:
Null H0 : p1 = p2
Vs
Alternative HA : p1 > p2
where p1 and p2 are population proportion for experimental group and control group respectively.
The test statistic remains the same irrespective of the difference in hypotheses. Answer 2
But, the p-value would change and hence there is a possibility of conclusion being different. Answer 3
Now, to quantify the above reasoning,
Case 1 Alternative HA : p1 ≠ p2
|
n1 |
53 |
|
n2 |
56 |
|
p1hat |
0.7647 |
|
p2hat |
0.6429 |
|
phat |
0.702123853 |
|
Zcal |
1.389764648 |
|
α |
0.05 |
|
Zcrit |
1.959963985 |
|
p-value |
0.1646003564 |
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 α%, Critical Value = upper (α/2)% of N(0, 1), and
p-value = P(|Z| > Zcal)
Using Excel Functions Statistical NORMSINV and NORMSDIST,
Critical Value and p-value are found to be as shown in the above table.
Decision:
Since Zcal < Zcrit, or equivalently, since p-value > α, H0 is accepted.
Conclusion :
There is NOT enough evidence to conclude that the proportion of A and B grades in the experimental group that used the crossword puzzle handout as an aid in studying for their quiz is different from the proportion of A and B grades in the control group that did not have that handout..
Case 2 Alternative HA : p1 > p2
The test statistic remains the same at 1.39, but p-value becoming 0.08, the decision and hence the conclusion continues to be the same.
Thus, in the present scenario, using an incorrect Alternative did not change the conclusion. Answer 4
DONE
One Management professor adopted a crossword puzzle exercise in his introductory management information systems (MIS) class,...
Exercise. A researcher wants to test: : What’s wrong with this? Exercise 7.24. Although the Environmental Protection Agency (EPA) establishes the tests to determine the fuel economy of new cars, it often does not perform them. Instead, the test protocols are given to the car companies, and the companies perform the tests themselves. To keep the industry honest, the EPA runs some spot checks each year. Recently, the EPA announced that Hyundai and Kia must lower their fuel economy estimates...