Analysis Methodology
- Two-factor ANOVA with replication for each alternative and the three levels of simultaneous clients
- Difference in means for 1, 5, 10 clients
Aggregated Results
| Two-Factor ANOVA with Replication: 99% Confidence | ||||||
| Source of Variation | SS | df | MS | F | P-value | F crit |
| Sample | 4240.85 | 2 | 2120.42 | 105.26 | 0 | 4.61 |
| Columns | 1793.07 | 1 | 1793.07 | 89.01 | 0 | 6.64 |
| Interaction | 604.63 | 2 | 302.32 | 15.01 | 0 | 4.61 |
| Within | 120743.78 | 5994 | 20.14 | |||
| Total | 127832.32 | 5999 | ||||
With 99% confidence, we find the following intervals for our difference in means.
| 99% Confidence Interval for Difference in Means | ||
| Number of Simultaneous Connections | c1 | c2 |
| 1 | 0.262 | 0.336 |
| 5 | 1.090 | 1.166 |
| 10 | 1.809 | 1.897 |
Interpretation
Based on the confidence intervals obtained from ANOVA, we can tell that, with 99% confidence, there is a statistically significant difference between the two alternatives, as our computed F values exceed the critical F values. According to the difference in means, this is true regardless of the number of simultaneous connections (based on our measurements). None of these confidence intervals include zero, allowing us to make this statement. In fact, the significance of the difference widens as the number of connections increases, with the non-serialized alternative being between 1.809 and 1.897 ms slower than the serialized alternative.