In: Statistics and Probability
5. Suppose a rehabilitation psychologist has developed a new job skills training program for people who have not been able to hold a job. Of the 14 people who agree to be in the study, the researcher randomly picks seven of these volunteers to be in a experimental group who will go through the special training program. The other seven volunteers are put in a control group who will go through an ordinary job skills training program. After finishing the training program (of whichever type), all 14 are placed in similar jobs. A month later, each volunteer’s employer is asked to rate how well the new employee is doing using a 9 point scale where a score of 1 indicates “very poor” and 9 indicates “Excellent.” The following ratings for the 14 employees are given below. New Training Program: 6 4 9 7 7 3 6
Ordinary Training Program: 6 1 5 3 1 1 4
A.Identify the independent variable and its levels; also identify the dependent variable and its measurement scale.
Independent variable is the new job skills training program for people who have not been able to hold a job. Dependent variable is how well the new employees are doing after the training.
B. State the null hypothesis for the study described. The average of new job skills training program for people who have not been able to hold a job will be the same as the average ordinary training program. Any difference observed between the average of training from both groups is solely due to random error. C. Use Excel to conduct the appropriate statistical test for the study described (p=.05). Be sure to properly state your statistical conclusion regarding the null hypothesis
D. Provide an interpretation of your answer in part C.
E. What type of statistical error might you have made in part C? A Type error may have been made ( rejecting the Ho when it is true.
Your answers for (A) and (B) are correct
(C)
Data:
n1 = 7
n2 = 7
x1-bar = 6
x2-bar = 3
s1 = 2
s2 = 2.081665999
Hypotheses:
Ho: μ1 = μ2
Ha: μ1 ≠ μ2
Decision Rule:
α = 0.05
Degrees of freedom = 7 + 7 - 2 = 12
Lower Critical t- score = -2.178812827
Upper Critical t- score = 2.178812827
Reject Ho if |t| > 2.178812827
Test Statistic:
Pooled SD, s = √[{(n1 - 1) s1^2 + (n2 - 1) s2^2} / (n1 + n2 - 2)] = √(((7 - 1) * 2^2 + (7 - 1) * 2.08166599946613^2)/(7 + 7 -2)) = 2.041
SE = s * √{(1 /n1) + (1 /n2)} = 2.04124145231932 * √((1/7) + (1/7)) = 1.091089451
t = (x1-bar -x2-bar)/SE = 2.749545417
p- value = 0.017617108
Decision (in terms of the hypotheses):
Since 2.749545417 > 2.178812827 we reject Ho
(D)
Conclusion (in terms of the problem):
There is sufficient evidence of a significant difference between the two group means. The average of new job skills training program for people who have not been able to hold a job is significantly different from the average ordinary training program.
(E)
In (C), we rejected Ho. If Ho were true, then we would be making a type I error.
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