NCP manufactures printers and fax machines at plants located in City 1, City 2, and City 3. To measure how much employees at these plants know about quality management, a random sample of 6 employees was selected from each plant and the employees selected were given a quality awareness examination. The examination scores for these 18 employees are shown in table given below. The sample means, sample variances, and sample standard deviations for each group are also provided. Managers want to use these data to test the hypothesis that the mean examination score is the same for all three plants.
Plant 1 | Plant 2 | Plant 3 | |
City 1 | City 2 | City 3 | |
---|---|---|---|
74 | 61 | 82 | |
76 | 71 | 82 | |
74 | 63 | 84 | |
86 | 74 | 81 | |
81 | 68 | 82 | |
71 | 65 | 93 | |
Sample mean | 77 | 67 | 84 |
Sample variance | 30.4 | 24.4 | 20.4 |
Sample standard deviation | 5.51 | 4.94 | 4.52 |
Set up the ANOVA table and test for any significant difference in the mean examination score for the three plants. Use a=0.5.
Source of Variation |
Sum of Squares |
Degrees of Freedom |
Mean Square (to 2 decimals) |
(to 2 decimals) |
-value (to 4 decimals) |
Treatments | |||||
Error | |||||
Total |
The p-value is - Select your answer -less than .01between .01 and .025between .025 and .05between .05 and .10greater than .10Item 11 .
What is your conclusion?
excel data analysis tool for one factor anova,steps are:
write data>menu>data>data analysis>anova :one factor>enter required labels>ok, and following o/p Is obtained,
Anova: Single Factor | |||||
SUMMARY | |||||
Groups | Count | Sum | Average | Variance | |
City 1 | 6 | 462 | 77 | 30.4 | |
City 2 | 6 | 402 | 67 | 24.4 | |
City 3 | 6 | 504 | 84.00 | 20.40 | |
ANOVA | |||||
Source of Variation | SS | df | MS | F | P-value |
Between Groups | 876.000 | 2 | 438.00 | 17.47 | 0.000 |
Within Groups | 376.00 | 15 | 25.07 | ||
Total | 1252.00 | 17 |
the p value is less than 0.01
since,p-value <α=0.05, reject Ho
hence, there is enough confidence that there is a significant difference in the mean examination score for the three plants at α=0.05
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