Homework SoursePlease explain solution and resultsSolutionUsing technology
Please explain solution and results.
Solution
Using technology, (In Excel, you can use Data-->Data Analysis-->Anova: Single Factor), we obtain these results:
As we can see, the P value is < 0.01.
Thus, we Reject the null hypothesis that the means are equal.
Thus, there is significant differences in conductivity due to coating type in at least 2 coating types. [CONCLUSION]
| Anova: Single Factor | ||||||
| SUMMARY | ||||||
| Groups | Count | Sum | Average | Variance | ||
| Column 1 | 4 | 580 | 145 | 15.33333 | ||
| Column 2 | 4 | 581 | 145.25 | 44.25 | ||
| Column 3 | 4 | 526 | 131.5 | 9.666667 | ||
| Column 4 | 4 | 517 | 129.25 | 4.25 | ||
| Column 5 | 4 | 581 | 145.25 | 7.583333 | ||
| ANOVA | ||||||
| Source of Variation | SS | df | MS | F | P-value | F crit |
| Between Groups | 1060.5 | 4 | 265.125 | 16.34892 | 2.41E-05 | 3.055568 |
| Within Groups | 243.25 | 15 | 16.21667 | |||
| Total | 1303.75 | 19 |
