Essentials of Statistics for the B...

8th Edition
Frederick J Gravetter + 1 other
ISBN: 9781133956570



Essentials of Statistics for the B...

8th Edition
Frederick J Gravetter + 1 other
ISBN: 9781133956570
Textbook Problem

Recent research has shown that creative people are more likely to cheat than their less-creative counter parts (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.


  1. a. Use a one-tailed test with α — 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.
  2. b. Compute Cohen’s d to measure the size of the effect.
  3. c. Write a sentence demonstrating how the results from the hypothesis test and the measure of effect size would appear in a research report.


To determine

To Find: Whether the data is sufficient to conclude that the high-creativity people are most likely to cheat then the low creativity people using one-tailed test at α=0.05 for the given question


Given info:

Two samples each with size 27 have means 7.41 and 4.78 respectively. Similarly, both of them have sum of squared difference (SS) equal to 749.5 and 830 respectively.


Step 1: Null Hypothesis is H0:μ1μ20 and Alternate Hypothesis is H1:μ1μ2>0

Step 2: For the given sample, the t-statistics will have degrees of freedom equals:


For a one-tailed test with α=0.05 and df=52 , the critical value (CV)  is obtained from the t-table as t=1.671

Step 3: The t-statistics is calculated using formula:


But before that calculate pooled variance, sp2 , using formula:


Substitute 749.5 for SS1 , 830 for SS2 and 26 for both df1 and df2 in the above formula



To determine

To Find: The value of Cohen’s d to measure effect size.

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