Management training programs are often conducted in order to teach supervisory s
ID: 3171054 • Letter: M
Question
Management training programs are often conducted in order to teach supervisory skills and thereby increase productivity. An industrial psychologist at a large company administers a similar test to six supervisors before and after a training program. The tests are designed to measure whether the training program significantly increased supervisory skills. Use =0.01.
X1 =Before Training
X2 =After Training
D=X2 – X1
D2=(X2 – X1)2
1
63
78
15
225
2
84
85
1
1
3
91
99
8
64
4
71
82
11
121
5
80
78
-2
4
6
83
84
1
1
D =34
D2 = 416
D =
D=After Training – Before Training = D=X2 – X1
17. State the Alternative Hypothesis
A
B
C
D
E None of the above
Answer for 17: A
18. Determine the mean D of the differences D=X2 – X1.
A -5.67
B 5.67
C 3.67
D -8.43
E None of the above
19. Compute the standard deviation of the differences
A 6.01
B 5.78
C -6.68
D 6.68
E None of the above
20. What critical value should be used at = 0.01?
A 4.032
B 3.707
C 3.365
D 3.143
E None of the above
21. Determine the test statistic
A2.08
B 1.34
C 3.39
D -2.00
E None of the above
22. What is the p-value?
A 0.0460
B 0.4603
C 0.0921
D -0.0460
E None of the above
23. At =0.01 the correct decision and conclusion are
A Reject H0, the training program significantly increased supervisory skills.
B Do not reject H0, the training program did not significantly increase supervisory skills.
C Reject H0, the training program did not significantly increase supervisory skills.
D Do not reject H0, the training program significantly increased supervisory skills.
E Insufficient information to make a decision
X1 =Before Training
X2 =After Training
D=X2 – X1
D2=(X2 – X1)2
1
63
78
15
225
2
84
85
1
1
3
91
99
8
64
4
71
82
11
121
5
80
78
-2
4
6
83
84
1
1
D =34
D2 = 416
D =
Explanation / Answer
18. Determine the mean D of the differences D=X2 – X1.
A -5.67
B 5.67
C 3.67
D -8.43
E None of the above
Solution:
We are given
D = 34 and n = 6
The formula for Dbar is given as below:
Dbar = D/n = 34/6 = 5.666667
Correct Answer: B. 5.67
19. Compute the standard deviation of the differences
A 6.01
B 5.78
C -6.68
D 6.68
E None of the above
Solution:
Standard deviation = sqrt[(D - Dbar)^2 / (n – 1)]
Calculation table is given as below:
D
(D - Dbar)^2
15
87.0489
1
21.8089
8
5.4289
11
28.4089
-2
58.8289
1
21.8089
Total
223.3334
(D - Dbar)^2 = 223.3334
(n – 1) = 6 – 1 = 5
[(D - Dbar)^2 / (n – 1)] = 223.3334/5 = 44.66668
Standard deviation = sqrt(44.66668) = 6.683313549
Correct Answer: D. 6.68
20. What critical value should be used at = 0.01?
A 4.032
B 3.707
C 3.365
D 3.143
E None of the above
Solution:
We are given sample size n = 6
Degrees of freedom = n – 1 = 6 – 1 = 5
Level of significance = alpha = 0.01
Test is one tailed.
So, by using t-table or excel
Critical value = 3.365
Correct answer: C. 3.365
21. Determine the test statistic
A2.08
B 1.34
C 3.39
D -2.00
E None of the above
Solution:
Test statistic formula is given as below:
Test statistic = t = Dbar / [Sd/sqrt(n)]
Test statistic = t = 5.6667/[6.6833/sqrt(6)]
Test statistic = t = 2.076896672
Correct Answer: A. 2.08
D
(D - Dbar)^2
15
87.0489
1
21.8089
8
5.4289
11
28.4089
-2
58.8289
1
21.8089
Total
223.3334
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