ArmCo Company initiated an employee empowerment program (in which employee feedback are collected and responded by...

ArmCo Company initiated an employee empowerment program (in which employee feedback are collected and responded by the management on a regular basis). The Company wishes to find out if employees at all five of its plants --- in the South, Midwest, Northeast, Southwest,and West --- have equally positive perceptions towards the empowerment program.Random samples of employees at the five plants were asked to rate the success of the empowerment program on a scale of 1 to 10, 10 being the most favorable rating. The collected data is shown to the right. Use Alpha = .05

1: What is the Null Hypothesis (select one)?

2. From the ANOVA table, what is the p-value?

3. From ANOVA table, what is F-statistics for the data?

4. From ANOVA table, what is F-critical?

5. Based on the ANOVA table, what is your decision?

6. Is there any indication of differences in mean ratings across the plants?

Empowerment Ratings from Five Plants

South   Midwest   Northeast   Southwest   West
7       7       7               6       6
1       6       5               9       6
8       10       5               7       6
7       3       5               4       6
2       9       4               7       3
9       10       3               6       4
3       8       4               6       8
8       4       5               7       6
4       9       5               7       6
7       9       1               6       6
6       7       3               6       1
7       6       3               6       6
3       6       7               7       5
8       4       3               6       6
6       6       6               5       5
6       9       3               7       5
2       6       5               8       5
3       5       6               6       4
6       5       3               7       3
8       1       7               5       3
7       6       5               8       6
4       3       1               6       6
3       4       3               7       6
3       4       4               6       5
5       4       1               7       9
6       4       3               8       8
7       7       4               3       4
9       4       6               4       5
9       4       6               6       5
5       4       7               7       5
4       2       2               4       5
5       5       6               6       2
3       2       5               7       1
7       8       7               7       5
5       3       5               4       2
7       2       8               3       4
4       7       9               7       5
       7       10               8       6
       5       5               9       4
       10       5               4       7
       10       4               4
       6       2               3
       3       4               5
       5       6               4
       2                       7
       6                       6
       4                       4
       5
       2
       7
       8
       7

Expert Solution

Using Excel, go to Data, select Data Analysis, choose Anova: Single Factor.

Anova: Single Factor

SUMMARY
Groups	Count	Sum	Average	Variance
South	37	204	5.513514	4.812312
Midwest	52	290	5.576923	5.817496
Northeast	44	208	4.727273	4.109937
Southwest	47	282	6	2.391304
West	40	200	5	3.025641


ANOVA
Source of Variation	SS	df	MS	F	P-value	F crit
Between Groups	45.26445	4	11.31611	2.781602	0.027732	2.413639
Within Groups	874.6628	215	4.068199

Total	919.9273	219

1. H0: µ1 = µ2 = µ3 = µ4 = µ5, Mean rate of success of the empowerment program for all 5 plants is same

H1: Mean rate of success of the empowerment program for at least one of the 5 plants is different

2. p-value = 0.0278

3. F-statistic = 2,782

4. F critical = 2.414

5. Since p-value is less than 0.05, we reject the null hypothesis.

6. Based on p-value, mean rate of success of the empowerment program for at least one of the 5 plants is different.

orchestra answered 2 years ago

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