Trying to see if using this regression calculation is an acceptable process to arrive at support labor.
Example by multiplying the total material times the X Variable 1 Coefficient and adding the Intercept.
Is this a good calculation or is there a better process ?
Material $300,000,000
Intercept 56.13
X Variable 1 0.000000098
Support labor = ($300,000,000 * 0.000000098) + 56.13 = 86
Actual Material $ | Actual support for Regression |
$ 338,073,262 | 91 |
$ 343,425,344 | 94 |
$ 396,713,846 | 94 |
$ 475,330,113 | 98 |
$ 530,814,611 | 104 |
$ 748,507,682 | 130 |
$ 823,531,326 | 133 |
$ 785,887,674 | 135 |
$ 778,807,225 | 144 |
$ 763,340,407 | 126 |
Regression Statistics | ||||||||
Multiple R | 0.97 | |||||||
R Square | 0.93 | |||||||
Adjusted R Square | 0.93 | |||||||
Standard Error | 5.532463 | |||||||
Observations | 10 | |||||||
ANOVA | ||||||||
df | SS | MS | F | Significance F | ||||
Regression | 1 | 3499.684674 | 3499.684674 | 114.3383191 | 5.14E-06 | |||
Residual | 8 | 244.8652175 | 30.60815219 | |||||
Total | 9 | 3744.549892 | ||||||
Coefficients | Standard Error | t Stat | P-value | Lower 95% | Upper 95% | Lower 95.0% | Upper 95.0% | |
Intercept | 56.13 | 5.7794 | 9.7123 | 1.05510E-05 | 42.8036 | 69.4582 | 42.8036 | 69.4582 |
X Variable 1 | 0.000000098 | 9.20E-09 | 10.6929 | 5.13637E-06 | 7.71952E-08 | 1.19645E-07 | 7.71952E-08 | 1.19645E-07 |
Material $ | Support |
$ 300,000,000 | 86 |
$ 335,000,000 | 89 |
$ 370,000,000 | 93 |
$ 405,000,000 | 96 |
$ 440,000,000 | 99 |
$ 475,000,000 | 103 |
$ 510,000,000 | 106 |
$ 545,000,000 | 110 |
$ 580,000,000 | 113 |
$ 615,000,000 | 117 |
$ 650,000,000 | 120 |
$ 685,000,000 | 124 |
$ 720,000,000 | 127 |
$ 755,000,000 | 130 |
$ 790,000,000 | 134 |
$ 825,000,000 | 137 |
$ 860,000,000 | 141 |
$ 895,000,000 | 144 |
$ 930,000,000 | 148 |
$ 965,000,000 | 151 |
$ 1,000,000,000 | 155 |
$ 1,035,000,000 | 158 |