Homework Answers
|
ANOVA |
|||||
|
observation |
|||||
|
Sum of Squares |
df |
Mean Square |
F |
Sig. |
|
|
Between Groups |
658.667 |
2 |
329.333 |
67.364 |
.000 |
|
Within Groups |
44.000 |
9 |
4.889 |
||
|
Total |
702.667 |
11 |
|||
|
Multiple Comparisons |
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Dependent Variable: observation LSD |
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|
(I) Manufacturer |
(J) Manufacturer |
Mean Difference (I-J) |
Std. Error |
Sig. |
95% Confidence Interval |
|
|
Lower Bound |
Upper Bound |
|||||
|
1.00 |
2.00 |
-11.00000* |
1.56347 |
.000 |
-14.5368 |
-7.4632 |
|
3.00 |
7.00000* |
1.56347 |
.002 |
3.4632 |
10.5368 |
|
|
2.00 |
1.00 |
11.00000* |
1.56347 |
.000 |
7.4632 |
14.5368 |
|
3.00 |
18.00000* |
1.56347 |
.000 |
14.4632 |
21.5368 |
|
|
3.00 |
1.00 |
-7.00000* |
1.56347 |
.002 |
-10.5368 |
-3.4632 |
|
2.00 |
-18.00000* |
1.56347 |
.000 |
-21.5368 |
-14.4632 |
|
|
*. The mean difference is significant at the 0.05 level. |
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Part a
Sum of squares ,treatment = 658.667
Sum of squares, Error = 44.00
Mean Square, Treatment = 329.333
Mean Squares, Error = 4.889
Test statistic value = 67.364
p value = 0.0000..
Conclusion : We reject H0 and conclude that there is significant difference between treatments(Manufacturer).
Part b
Fisher's LSD value = 2.61
Mean significant difference between the time for manufacturer 1 and 3 is 7.000 with p value 0.000.. < 0.05
Hence we conclude that there is significant difference between the time for manufacturer 1 and 3
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