Recent research has shown that creative people are more likely to cheat than the
ID: 3365988 • Letter: R
Question
Recent research has shown that creative people are more likely to cheat than their less-creative counterparts (Gino & Ariely, 2011). Participants in the study first completed creativity assessment questionnaires and then returned to the lab several days later for a series of tasks. One task was a multiple-choice general knowledge test for which the participants circled their answers on the test sheet. Afterward, they were asked to transfer their answers to a bubble sheet for computer scoring. However, the experimenter admitted that the wrong bubble sheet had been copied so that the correct answers were still faintly visible. Thus, the participants had an opportunity to cheat and inflate their test scores. Higher scores were valuable because participants were paid based on the number of correct answers. However, the researchers had secretly coded the original tests and the bubble sheets so that they could measure the degree of cheating for each participant. Assuming that the participants were divided into two groups based on their creativity scores, the following data are similar to the cheating scores obtained in the study.
High-Creativity Low-Creativity
Participants Participants___
n = 27 n = 27
M = 7.41 M = 4.78
SS = 749.5 SS = 830
a.Use a one-tailed test with = .05 to determine whether these data are sufficient to conclude that high-creativity people are more likely to cheat than people with lower levels of creativity.
b.Compute Cohen’s d to measure the size of the effect
Explanation / Answer
(a)
Data:
n1 = 27
n2 = 27
x1-bar = 7.41
x2-bar = 4.78
s1 = 5.37
s2 = 5.65
Hypotheses:
Ho: 1 2
Ha: 1 > 2
Decision Rule:
= 0.05
Degrees of freedom = 27 + 27 - 2 = 52
Critical t- score = 1.67468915
Reject Ho if t > 1.67468915
Test Statistic:
Pooled SD, s = [{(n1 - 1) s1^2 + (n2 - 1) s2^2} / (n1 + n2 - 2)] = (((27 - 1) * 5.37^2 + (27 - 1) * 5.65^2)/(27 + 27 - 2)) = 5.511778297
SE = s * {(1 /n1) + (1 /n2)} = 5.51177829742815 * ((1/27) + (1/27)) = 1.500116045
t = (x1-bar -x2-bar)/SE = (7.41 - 4.78)/1.50011604489391 = 1.7531977
p- value = 0.04273075
Decision (in terms of the hypotheses):
Since 1.7531977 > 1.674689154 we reject Ho and accept Ha
Conclusion (in terms of the problem):
It appears that M1 > M2
(b)
Cohen's d = (M2 - M1)/Pooled standard deviation = (4.78 - 7.41)/5.511 = -0.477
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