A quality-control manager at a large manufacturing company will like to measure
ID: 3392294 • Letter: A
Question
A quality-control manager at a large manufacturing company will like to measure his employees’ knowledge of total quality management. A quality awareness test is administered to a random sample of 6 employees at each of 3 plants in Atlanta, Dallas and Seattle. The test scores are shown below:
Atlanta
Dallas
Seattle
85
71
59
75
75
64
82
73
62
76
74
69
71
69
75
85
82
67
At = 0.05, is there sufficient evidence that the employees’ knowledge of total quality management differs among the three plants?
Question: True or false: Ho: µ(atl) = µ(dallas)= µ(seattle)
True
False
4 points
Question 2
True or false: SSG (between groups) = 520
True
False
4 points
Question 3
True or false: SSTotal = 946
True
False
4 points
Question 4
DFE (degrees of freedom within groups) = 17
True
False
4 points
Question 5
True or false: MSG (between groups) = 258
True
False
4 points
Question 6
True or false: MSE (within groups) = 30.67
True
False
4 points
Question 7
True or false: F test statistic = 9.0
True
False
4 points
Question 8
True or false: p-value = 0.0027
True
False
4 points
Question 9
True or false: At = 0.05, reject Ho. Employees’ knowledge of total quality management differs among the three plants.
True
False
Atlanta
Dallas
Seattle
85
71
59
75
75
64
82
73
62
76
74
69
71
69
75
85
82
67
Explanation / Answer
Using technology (like Excel below, in Data-->Data Analysis-->Anova Single Factor), we have the summary:
Thus,
1. TRUE
2. FALSE (it is 516)
3. TRUE
4. FALSE
5. TRUE
6. FALSE. It is 28.666667.
7. TRUE.
8. TRUE.
9. TRUE. it is because P < 0.05.
Anova: Single Factor SUMMARY Groups Count Sum Average Variance Column 1 6 474 79 34 Column 2 6 444 74 20 Column 3 6 396 66 32 ANOVA Source of Variation SS df MS F P-value F crit Between Groups 516 2 258 9 0.002703 3.68232 Within Groups 430 15 28.66667 Total 946 17Related Questions
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