In: Economics
The following table lists a portion of Major League Baseball’s (MLB’s) leading pitchers, each pitcher’s salary (In $ millions), and earned run average (ERA) for 2008.
a-1. Estimate the model: Salary = β0 + β1ERA + ε. (Negative values should be indicated by a minus sign. Enter your answers, in millions, rounded to 2 decimal places.)Click here for the Excel Data File
Salary | ERA | |
J. Santana | 17.0 | 2.25 |
C. Lee | 2.0 | 2.37 |
T. Lincecum | 0.1 | 2.57 |
C. Sabathia | 10.0 | 2.11 |
R. Halladay | 7.0 | 2.52 |
J. Peavy | 5.7 | 2.77 |
D. Matsuzaka | 6.7 | 2.44 |
R. Dempster | 6.6 | 2.89 |
B. Sheets | 11.5 | 2.90 |
C. Hamels | 0.3 | 2.91 |
Salaryˆ=Salary^= + ERA
a-2. Interpret the coefficient of ERA.
A one-unit increase in ERA, predicted salary decreases by $6.37 million.
A one-unit increase in ERA, predicted salary increases by $6.37 million.
A one-unit increase in ERA, predicted salary decreases by $16.70 million.
A one-unit increase in ERA, predicted salary increases by $16.70 million.
b. Use the estimated model to predict salary for each player, given his ERA. For example, use the sample regression equation to predict the salary for J. Santana with ERA = 2.25. (Round coefficient estimates to at least 4 decimal places and final answers, in millions, to 2 decimal places.)
c. Derive the corresponding residuals. (Negative values should be indicated by a minus sign. Round coefficient estimates to at least 4 decimal places and final answers, in millions, to 2 decimal places.)
a-1. Salary = 23.0715-6.3667*ERA
we need to use regression function in excel data analysis to get the results as below:
SUMMARY OUTPUT |
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Regression Statistics |
||||||
Multiple R |
0.3480 |
|||||
R Square |
0.1211 |
|||||
Adjusted R Square |
0.0113 |
|||||
Standard Error |
5.2169 |
|||||
Observations |
10 |
|||||
ANOVA |
||||||
df |
SS |
MS |
F |
Significance F |
||
Regression |
1 |
30.0044 |
30.0044 |
1.1025 |
0.3244 |
|
Residual |
8 |
217.7246 |
27.2156 |
|||
Total |
9 |
247.7290 |
||||
Coefficients |
Standard Error |
t Stat |
P-value |
Lower 95% |
Upper 95% |
|
Intercept |
23.0715 |
15.6886 |
1.4706 |
0.1796 |
-13.1066 |
59.2496 |
ERA |
-6.3667 |
6.0636 |
-1.0500 |
0.3244 |
-20.3494 |
7.6160 |
a-2. Option 1
From the above table we can observe that the coefficient is
negative indicates about the negative relationship, hence one unit
increase in ERA decreases salary by $ 6.37 million
b.
Salary |
ERA |
Forecast |
17 |
2.25 |
23.0715-6.3667*2.25 = 8.75 |
2 |
2.37 |
23.0715-6.3667*2.37 = 7.98 |
0.1 |
2.57 |
23.0715-6.3667*2.57 = 6.71 |
10 |
2.11 |
23.0715-6.3667*2.11 = 9.64 |
7 |
2.52 |
23.0715-6.3667*2.52=7.03 |
5.7 |
2.77 |
23.0715-6.3667*2.77 =5.44 |
6.7 |
2.44 |
23.0715-6.3667*2.44=7.54 |
6.6 |
2.89 |
23.0715-6.3667*2.89=4.67 |
11.5 |
2.9 |
23.0715-6.3667*2.90=4.61 |
0.3 |
2.91 |
23.0715-6.3667*2.91=4.54 |
c.
Salary |
ERA |
Forecast |
Residuals = Actual-Forecast |
17 |
2.25 |
8.75 |
=17-8.75 = 8.25 |
2 |
2.37 |
7.98 |
=2-7.98 = -5.98 |
0.1 |
2.57 |
6.71 |
=0.1-6.71 =-6.61 |
10 |
2.11 |
9.64 |
=10-9.64 = 0.36 |
7 |
2.52 |
7.03 |
=7-7.03 = -0.03 |
5.7 |
2.77 |
5.44 |
=5.7-5.44 = 0.26 |
6.7 |
2.44 |
7.54 |
= 6.7-7.54 = -0.84 |
6.6 |
2.89 |
4.67 |
= 6.6-4.67 = 1.93 |
11.5 |
2.9 |
4.61 |
=11.5-4.61 = 6.89 |
0.3 |
2.91 |
4.54 |
=0.3-4.54 = -4.24 |