Homework SourseDetermine the estimated regression equation that can be used
Determine the estimated regression equation that can be used to predict the overall score given the scores for Itineraries/Schedule, Shore Excursions, and Food/Dining.
Use the F test to determine the overall significance of the relationship. What is your conclusion at the .05 level of significance?
Use the t test to determine the significance of each independent variable. What is your conclusion at the .05 level of significance?
Remove any independent variable that is not significant from the estimated regression equation. What is your recommended estimated regression equation?
Please show in Excel form! thank you!!
| Ships | Overall | Itineraries/Schedule | Shore Excursions | Food/Dining |
| Seabourn Odyssey | 94.4 | 94.6 | 90.9 | 97.8 |
| Seabourn Pride | 93.0 | 96.7 | 84.2 | 96.7 |
| National Geographic | 92.9 | 100.0 | 100.0 | 88.5 |
| Seabourn Sojourn | 91.3 | 88.6 | 94.8 | 97.1 |
| Paul Gauguin | 90.5 | 95.1 | 87.9 | 91.2 |
| Seabourn Ledgend | 90.3 | 92.5 | 82.1 | 98.8 |
| Seabourn Spirit | 90.2 | 96.0 | 86.3 | 92.0 |
| Silver Explorer | 89.9 | 92.6 | 92.6 | 88.9 |
| Silver Spirit | 89.4 | 94.7 | 95.9 | 90.8 |
| Seven Seas Navigator | 89.2 | 90.6 | 83.3 | 90.5 |
| Silver Whisperer | 89.2 | 90.9 | 82.0 | 88.6 |
| National Geographic E. | 89.1 | 93.1 | 93.1 | 89.7 |
| Silver Cloud | 88.7 | 92.6 | 78.3 | 91.3 |
| Celebrity Xpedition | 87.2 | 93.1 | 91.7 | 73.6 |
| Silver Shadow | 87.2 | 91.0 | 75.0 | 89.7 |
| Silver Wind | 86.6 | 94.4 | 78.1 | 91.6 |
| SeaDream II | 86.2 | 95.5 | 77.4 | 90.9 |
| Wind Star | 86.1 | 94.9 | 76.5 | 91.5 |
| Wind Surf | 86.1 | 92.1 | 72.3 | 89.3 |
| Wind Spirit | 85.2 | 93.5 | 77.4 | 91.9 |
Solution
Determine the estimated regression equation that can be used to predict the overall score given the scores for Itineraries/Schedule, Shore Excursions, and Food/Dining.
Overall score=38.0838 + 0.1063*Itineraries/Schedule + 0.2215*Shore Excursions+0.2447*Food/Dining
----------------------------------------------------------------------------------------------------------------
Use the F test to determine the overall significance of the relationship. What is your conclusion at the .05 level of significance?
Null hypothesis: there is no overall significance of the relationship
Alternative hypothesis: there is the overall significance of the relationship
Since the p-value of F test is less than 0.05, we reject the null hypothesis. So we can conclude that there is the overall significance of the relationship
----------------------------------------------------------------------------------------------------------------
Use the t test to determine the significance of each independent variable. What is your conclusion at the .05 level of significance?
For Itineraries/Schedule, since the p-value of t test is larger than 0.05, we can not conclude that Itineraries/Schedule variable is significant.
For Shore Excursions, since the p-value of t test is less than 0.05, we can conclude that Shore Excursions variable is significant.
For Food/Dining, since the p-value of t test is less than 0.05, we can conclude that Food/Dining variable is significant.
----------------------------------------------------------------------------------------------------------------
Remove any independent variable that is not significant from the estimated regression equation. What is your recommended estimated regression equation?
Overall score=47.2640 + 0.2297*Shore Excursions + 0.2455*Food/Dining
| Regression output | confidence interval | |||||
| variables | coefficients | std. error | t (df=16) | p-value | 95% lower | 95% upper |
| Intercept | 38.0838 | 14.3664 | 2.651 | .0174 | 7.6285 | 68.5391 |
| Itineraries/Schedule | 0.1063 | 0.1418 | 0.750 | .4643 | -0.1943 | 0.4069 |
| Shore Excursions | 0.2215 | 0.0449 | 4.931 | .0002 | 0.1263 | 0.3168 |
| Food/Dining | 0.2447 | 0.0676 | 3.618 | .0023 | 0.1013 | 0.3880 |
