11.13 A management consulting company presents a three-day seminar on project ma
ID: 3182718 • Letter: 1
Question
11.13 A management consulting company presents a three-day seminar on project management to various clients. The seminar is basically the same each time it is given. However, sometimes it is presented to high-level managers, sometimes to midlevel managers, and sometimes to low-level managers. The seminar facilitators believe evaluations of the seminar may vary with the audience. Suppose the following data are randomly selected evaluation scores from different levels of managers who attend the seminar. The ratings on a scale from 1 to 10, with 10 being the highest. Use a one-way ANOVA to determine whether there is a significant difference in the evaluations according to manager level. Assume alpha= 0.05.
High Level Mid Level Low Level
7 8 5
7 9 6
8 8 5
7 10 7
9 9 4
10 8
8
1) please set up ANOVA test
(If you do not have ANOVA table for 11.13, use
SSC=53.37, SSE=22.657 and rests are easily obtained.)
2) write down conclusion using CV and p-value approach
3) write down implication
4) set up Tukey table
(Note: Last column of Tukey table should be one of two, same or not same)
Explanation / Answer
Solution:
Part 1
From the given data, the ANOVA table is calculated as below:
No
X1
X2
X3
X1^2
X2^2
X3^2
1
7
8
5
49
64
25
2
7
9
6
49
81
36
3
8
8
5
64
64
25
4
7
10
7
49
100
49
5
9
9
4
81
81
16
6
10
8
100
64
7
8
64
Total
38
62
35
292
554
215
ANOVA
Source of Variation
SS
df
MS
F
P-value
F critical
Between Groups
29.61
2
29.61/2 = 14.80
14.80/1.26 = 11.76
0.001
3.68
Within Groups
18.89
15
18.89/15 = 1.26
Total
48.5
17
Part 2
The critical value is given as below:
Critical value = 3.68 (By using excel)
Test statistic F = 11.76
Test statistic F > Critical value
So, we reject the null hypothesis that the average values for high level, mid level and low level are same. Now, we have to see p-value approach given as below:
P-value = 0.001
Alpha value = 0.05
P-value < Alpha value
So, according to decision rule, we reject the null hypothesis that the average values for high level, mid level and low level are same.
Part 3
This one way ANOVA implies that there is a significant difference in the average ratings for the given three levels.
Part 4
The Tukey table for comparison of means is given as below:
Multiple Comparisons
(I) Level
(J) Level
Mean Difference (I-J)
Std. Error
Sig.
High Level
Mid Level
-1.25714
.65710
.169
Low Level
1.76667*
.67953
.050
Mid Level
High Level
1.25714
.65710
.169
Low Level
3.02381*
.62434
.001
Low Level
High Level
-1.76667*
.67953
.050
Mid Level
-3.02381*
.62434
.001
No
X1
X2
X3
X1^2
X2^2
X3^2
1
7
8
5
49
64
25
2
7
9
6
49
81
36
3
8
8
5
64
64
25
4
7
10
7
49
100
49
5
9
9
4
81
81
16
6
10
8
100
64
7
8
64
Total
38
62
35
292
554
215
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