In: Statistics and Probability
National Bearings manufactures bearings at plants located in Portland Oregon, Houston Texas, and Jacksonville Florida. To measure employee knowledge of Total Quality Management (TQM), six employees were randomly selected at each plant and tested. The test scores for these employees are given in DATA. Managers want to know if, on average, knowledge of TQM is equal across the 3 plants. Test equality of mean scores at x = 0.05
Ovservation | Portland | Houston | Jacksonville |
1 | 85 | 71 | 64 |
2 | 75 | 75 | 69 |
3 | 82 | 73 | 67 |
4 | 76 | 74 | 74 |
5 | 71 | 69 | 80 |
6 | 85 | 82 | 72 |
A. none of the answers are correct
B. Equality of means is not rejected with pvalue 0.295. Knowledge of TQM is equal across the 3 plants.
C. Equality of means in not rejected with pvalue of 0.060. Knowledge of TQM is equal across the 3 plants.
D. Equality of means is rejected with pvalue 0.010. Knowledge of TQM is not equal across the 3 plants.
E. Equality of means is rejected with pvalue 0.003. Knowledge of TQM is not equal across the 3 plants
Consider
: Average Knowledge of TQM at Portland plant.
: Average Knowledge of TQM at Houston plant.
: Average Knowledge of TQM at Jacksonville plant.
We have to testing the hypothesis
Whether or not Average Knowledge of TQM is equal across the plants ?
i.e Null Hypothesis -
against
Alternative Hypothesis- Ha : Knowledge of TQM is not equal across the 3 plants.
By using Excel
Observation | Portland | Houston | Jacksonville | ||||||||
1 | 85 | 71 | 64 | Anova: Single Factor | |||||||
2 | 75 | 75 | 69 | ||||||||
3 | 82 | 73 | 67 | SUMMARY | |||||||
4 | 76 | 74 | 74 | Groups | Count | Sum | Average | Variance | |||
5 | 81 | 69 | 80 | Column 1 | 6 | 484 | 80.66667 | 18.66667 | |||
6 | 85 | 82 | 72 | Column 2 | 6 | 444 | 74 | 20 | |||
Column 3 | 6 | 426 | 71 | 32 | |||||||
ANOVA | |||||||||||
Source of Variation | SS | df | MS | F | P-value | F crit | |||||
Between Groups | 293.7778 | 2 | 146.8889 | 6.235849 | 0.010692 | 3.68232 | |||||
Within Groups | 353.3333 | 15 | 23.55556 | ||||||||
Total | 647.1111 | 17 |
Alpha : level of significance = 0.05
From excel output p-value = 0.01062
Since p-value < level of significance alpha.
We reject the null Hypothesis at 5% level of significance.
Correct Answer : D - Equality of means is rejected with p-value 0.010. Knowledge of TQM is not equal across the 3 plants.