several variables on total winnings of 100 randomly selected

several variables on total winnings of 100 randomly selected PGA golfers in 2004. The are: Variable Age AvgDriveYds Description Player\'s age in years Average length of drive in yards % of drives landing in the fairway % of greens reached in regulation (ie, par-2) Average number of putts taken per round % of pars saved when off the green in regulation Number of golf tournaments played GreensReg AvgNumPutts SavePct NumEvents The dependent variable is: Dependent Var Description TotWinnings1000s Total winnings in thousands of $ results A multiple linear regression model was created using the above variables. Here are the main Regression Analysis R 0.435 Adjusted R2 0.392 R 0.659 n 100 Std. Eror 1058.283 Dep. Var. TotWinningst000s ANOVA table Source Regression Residual Total 79,221,681 103,036,680 182,258,361 11,317,383 10.11 255E-09 1.119.964 92 Regression output t (d-92) p-value 1.625 1075 VIF vanables Intercept Age AvgDriveYds DriveAcc std error 18,021.42 11,087.17 19.34 21.52 33.73 49.95 -13,745.68 4,816.89 21.63 24.25 350 7272 1297 -1.051 2962 2774 2709 0081 2.736 5813 8.78E-08 1775 2.8540053 1.198 1.481 14211201 1069 1.722 6.77 22.61 91.37 290.32 SavePct NumEvents 32.02 1.000 3199 mean VF a) (6 pts) Write down the fitted regression model. Only include variables that are statistically significant

Solution

We are given that the independent variables are,

X1 = Age

X2 = AvgDriveYds

X3 = DriveAcc

X4 = GreensReg

X5 = AvgNumputts

X6 = SavePct

X7 = NumEvents

And the dependent variable (Y) is TotWinnings1000s

Fitted regression model:

Y = 18021.42 + 6.77*X1 - 22.61*X2 - 91.37*X3 + 290.32*X4 - 13745.68*X5 + 32.02*X6 - 24.25*X7

The fitted regression model only include variables that are statistically significant.

Statistically significance means we have to check P-value < 0.05

Variables having P-value < 0.05 that variables are statistically significant.

From the output we can see that P-value for X3 is 0.0081 < 0.05

P-value for X4 is 8.78E-08 < 0.05

P-value for X5 is 0.0053 < 0.05

So the variables X3, X4 and X5 are statistically significant.

The new model will be,

Y = 18021.42 - 91.37*X3 + 290.32*X4 - 13745.68*X5

When using an alpha or type I error 0.05 since the P-value in the ANOVA table is so small, you can conclude that at least one of the independent variables is statistically significant.

This is the correct answer or statement because there are three variables which are statistically significant.

At least one of the independent variable means greator than one independent variable is statistically significant.

R² = 0.435= 43.5%

R² represents expresses the proportion of the variation in Y which is explained by variables X1,..........,X7.

The model explains 43.5% of the variability of the response data around its mean.

Adjusted R² = 0.392

The adjusted R-squared compares the explanatory power of regression models that contain different numbers of predictors.

The adjusted R-squared is a modified version of R-squared that has been adjusted for the number of predictors in the model.

The adjusted R-squared can be negative, but it’s usually not. It is always lower than the R-squared.

If we add independent variabel  the R-squared increases, even if due to chance alone. It never decreases. Consequently, a model with more terms may appear to have a better fit simply because it has more terms.

The adjusted R-squared increases only if the new term improves the model more than would be expected by chance.

The proportion of total variability in TotWinnings100s is explained by the linear relationship between TotWinnings1000s with the seven independent variables is 0.435 that is 0.435*100 = 43.5%