Appendix—Details on Cups, Error Variance, and Grid Spacing
The following cups (table 8) and lids (table 9) were used and their weight recorded.
| Weight range (grams) | Average (grams) | Total cups |
|---|---|---|
| 26.85 to 26.95 | 26.90 | 5,000 |
| 26.75 to 26.85 | 26.80 | 4,000 |
| 26.65 to 26.75 | 26.70 | 3,000 |
| Weight range (grams) | Average (grams) | Total lids |
|---|---|---|
| 16.45 to 16.55 | 16.50 | 2,500 |
| 16.35 to 16.45 | 16.40 | 8,500 |
- Average cup weight = [(26.9*5) + (26.8*4) + (26.7*3)]/12 = 26.816667 grams
- Standard deviation = 0.07993 grams. Variance = 0.0063888049 grams
- Average lid weight = [(16.4*8.5) + (16.5*2.5)]/11 = 16.422727 grams
- Standard deviation = 0.04191 grams. Variance = 0.0017564481 grams
- Tare (average weight of cup and lid) = 43.23939 grams
Combined standard deviation:
The lowest possible cup and lid weight was 43.00 grams and the highest was 43.50. If a cup with retardant in it weighed less than 43.23939 grams, the computer program automatically switched to a tare weight of 43.00 to avoid negative gpc.
At a 99-percent confidence level (CI), the margin of error for the tare weight of 43.2393 grams is ± 0.23249 (2.576*0.09025 = 0.23249)
At a 95 percent CI, the margin of error for the tare weight is ± 0.17689 grams. (1.960*0.09025 = 0.17689 grams)
Error Variance Estimate for GPC
The error variance estimator for triangulated gpc values is:
![Equation: V(tgpc)=[V(triangulation) + V(cups)/nc + V(lids)/nl]*0.124087^2](images/Equation03.gif)
Where V (triangulation) is the triangulation variance. V (cups) is the variance for empty cups, and V (lids) is the variance for empty lids. nc and nl are the number of cups and the number of lids, respectively. 0.124087 is a constant that converts grams of retardant with density 1.095 grams per milliliter into gpc.
Mean square error (MSE) is an estimate of the triangulation variance. The three MSEs are 0.215, 0.262, and 0.256, which is an average MSE of 0.244.
0.003804 = [0.244 + 0.000000532 + 0.0000001597] * 0.1240872
Variance around triangulated gpc = 0.0038. Standard deviation around triangulated gpc = 0.0616.
An example of an analysis of variance (ANOVA) model (figure 21).
*** Analysis of Variance Model ***
Short Output:
Cell:
- aov(formula – Continuous ~ Ret + FlowRate, data – LineLengths05, qr – T, n.action – na.exclude)
Terms:
| Ret | FlowRate | Residuals | |
| Sum of Squares | 2790.75 | 7140.25 | 26.50 |
| Deg. of Freedom | 1 | 1 | 3 |
Residual standard error: 2.972092
Estimated effects may be unbalanced
| Df | Sum of Sq | Mean Sq | F Value | Pr(F) | |
| Ret | 1 | 2790.75 | 2790.750 | 315.9340 | 0.0003882841 |
| FlowRate | 1 | 7140.25 | 7140.250 | 808.3302 | 0.0000955337 |
| Residuals | 3 | 26.50 | 8.833 |
Tables of means
Grand mean
- 569.5
Ret
| GTSR | Water | |
| 554.25 | 600.00 | |
| rep | 4.00 | 2.00 |
FlowRate
| High | Low | |
| 527.25 | 590.63 | |
| rep | 2.00 | 4.00 |
Figure 21—Analysis of variance results.
