Question
In: Statistics and Probability
Using the analysis containing all four independent variables, calculate the 99% confidence internal for the coefficient...
Using the analysis containing all four independent variables, calculate the 99% confidence internal for the coefficient of V2.
| V1 | V2 | V3 | V4 | Y |
| 10 | 2 | 5 | 2 | 24.4 |
| 11 | 2 | 4 | 2 | 22 |
| 12 | 2 | 3 | 2 | 22.4 |
| 13 | 4 | 2 | 3 | 24.5 |
| 14 | 3 | 1 | 2 | 21 |
| 15 | 5 | 5 | 2 | 37.4 |
| 16 | 5 | 4 | 2 | 37.8 |
| 17 | 4 | 2 | 1 | 25.5 |
| 18 | 5 | 1 | 2 | 24 |
| 19 | 5 | 5 | 1 | 25 |
| 20 | 2 | 3 | 1 | 22.1 |
| 21 | 4 | 4 | 3 | 28.8 |
| 22 | 2 | 3 | 2 | 23 |
| 23 | 3 | 2 | 2 | 28.3 |
| 24 | 3 | 1 | 1 | 32.7 |
| 25 | 2 | 1 | 3 | 19.1 |
| 26 | 2 | 4 | 2 | 26.1 |
| 27 | 5 | 5 | 2 | 41.5 |
| 28 | 3 | 3 | 2 | 29.6 |
| 29 | 5 | 2 | 1 | 33.6 |
Solutions
Expert Solution
| Coefficients | Standard Error | t Stat | P-value | Lower 95% | Upper 95% | Lower 99.0% | Upper 99.0% | ||
| Intercept | 6.994048297 | 5.964970426 | 1.172520197 | 0.259274 | -5.71999 | 19.70808 | -10.583 | 24.5711 | |
| V1 | 0.352776111 | 0.171117793 | 2.061598065 | 0.057019 | -0.01195 | 0.717505 | -0.15146 | 0.857011 | |
| V2 | 2.745443148 | 0.789977715 | 3.475342525 | 0.003391 | 1.061646 | 4.429241 | 0.417606 | 5.073281 | |
| V3 | 1.548762884 | 0.693337078 | 2.233780557 | 0.041149 | 0.07095 | 3.026576 | -0.4943 | 3.591828 | |
| V4 | -0.217883059 | 1.555740492 | -0.14005103 | 0.890483 | -3.53387 | 3.098099 | -4.8022 | 4.366437 | |
| 99% confidence interval for V2= Beta(V2)+-t(n-(k+1),probability)(Standard error of V2) | |||||||||
| 99% ci for V2=2.74+-F28t((20-(4+1),0.995)*0.7899 | |||||||||
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