In: Statistics and Probability
The value of a sports franchise is directly related to the amount of revenue that a franchise can generate. Below is the data that represents the value (in $millions) and the annual revenue (in $millions) for 30 Major League Baseball franchises. Suppose you want to develop a simple linear regression model to predict franchise value based on annual revenue generated.
| 
 Team  | 
 Revenue  | 
 Value  | 
| 
 Baltimore  | 
 179  | 
 460  | 
| 
 Boston  | 
 310  | 
 1000  | 
| 
 Chicago White Sox  | 
 214  | 
 600  | 
| 
 Cleveland  | 
 178  | 
 410  | 
| 
 Detroit  | 
 217  | 
 478  | 
| 
 Kansas City  | 
 161  | 
 354  | 
| 
 Los Angeles Angels  | 
 226  | 
 656  | 
| 
 Minnesota  | 
 213  | 
 510  | 
| 
 New York Yankees  | 
 439  | 
 1850  | 
| 
 Oakland  | 
 160  | 
 321  | 
| 
 Seattle  | 
 210  | 
 585  | 
| 
 Tampa Bay  | 
 161  | 
 323  | 
| 
 Texas  | 
 233  | 
 674  | 
| 
 Toronto  | 
 188  | 
 413  | 
| 
 Arizona  | 
 186  | 
 447  | 
| 
 Atlanta  | 
 203  | 
 508  | 
| 
 Chicago Cubs  | 
 266  | 
 879  | 
| 
 Cincinnati  | 
 185  | 
 424  | 
| 
 Colorado  | 
 193  | 
 464  | 
| 
 Houston  | 
 196  | 
 549  | 
| 
 Los Angeles  | 
 230  | 
 1400  | 
| 
 Miami  | 
 148  | 
 450  | 
| 
 Milwaukee  | 
 195  | 
 448  | 
| 
 New York Mets  | 
 225  | 
 719  | 
| 
 Philadelphia  | 
 249  | 
 723  | 
| 
 Pittsburgh  | 
 168  | 
 336  | 
| 
 St. Louis  | 
 233  | 
 591  | 
| 
 San Diego  | 
 163  | 
 458  | 
| 
 San Francisco  | 
 230  | 
 643  | 
| 
 Washington  | 
 200  | 
 480  | 
(a ) Use the least-squares method to determine the regression coefficients (intercept and slope).
(b) Interpret the meaning of the intercept and slope in this problem.
(c) Predict the value of a baseball franchise that generates $150 million of annual revenue.
Using Excel<data<megastat<correlation/regression<regression
| Regression Analysis | ||||||
| r² | 0.790 | |||||
| r | 0.889 | |||||
| Std. Error | 150.955 | |||||
| n | 30 | |||||
| k | 1 | |||||
| Dep. Var. | Value y | |||||
| ANOVA table | ||||||
| Source | SS | df | MS | F | p-value | |
| Regression | 24,04,121.3683 | 1 | 24,04,121.3683 | 105.50 | 5.31E-11 | |
| Residual | 6,38,045.3317 | 28 | 22,787.3333 | |||
| Total | 30,42,166.7000 | 29 | ||||
| Regression output | confidence interval | |||||
| variables | coefficients | std. error | t (df=28) | p-value | 95% lower | 95% upper | 
| Intercept | -496.3022 | 110.71 | -4.48 | 0.00 | -723.09 | -269.51 | 
| Revenue x | 5.1961 | 0.5059 | 10.271 | 5.31E-11 | 4.1599 | 6.2324 | 

a)
Intercept (b0) -496.3022
Slope (b1) 5.1961
b)
A practical interpretation of the Y-intercept,b0, is not meaningful because no value is going to have a revenue of zero. The slope,b1, implies that for each increase of 1 million dollars in annual revenue, the value is expected to increase by the value of b1, in millions of dollars.
c)
Y=5.196 x-496.302 = 5.196*150-496.302=283.098
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