It is common in sports to predict outcomes using logistic regression or if you want to predict a continuous response linear regression. In logistic regression the response Y has 2 values (win or lose in the case of a sporting event outcome) So the outcome is usually coded as 0 for loss and 1 for win. The model is y=a1 X1 + a2 X2 +a3 X3 +....+ e. The response y=log(p/1-p) where p is the unknown probability of success and X1, X2, X2,... are the vsrisble that are used to predict y. In linear regression you have the same form as with logistic regression except that y is the response rather than a function of it. e is the random component or stochastic element. It is there to express error (sometimes measurement error in determining y). In these models the error term is assumed to be additive. Also the Xi are assumed to be determined without error. Using the model and data for the Xis and the corresponding ys the regression parameters a1, a2,a3, ... are estimated (usually by least squares). Often the error term e is assumed to be normally distributed with me an 0. They are independent of the Xis and independent of each other. The variability of the error term is defined by the standard deviation sigma which is assumed to be constant (i. e. the same for each sample). For the normal distribution the mean and standard deviation uniquely determine the distribution. These models are general and the technique can be applied in any sport. The covariate and response variables can differ from sport to sport. It is important to keep the modeling assumptions in mind and if they don't seem to apply alternative approaches mey need to be considered.

It is common in sports to predict outcomes using logistic regression or if you want to predict a continuous response linear regression. In logistic regression the response Y has 2 values (win or lose in the case of a sporting event outcome) So the outcome is usually coded as 0 for loss and 1 for win. The model is y=a1 X1 + a2 X2 +a3 X3 +....+ e. The response y=log(p/1-p) where p is the unknown probability of success and X1, X2, X2,... are the vsrisble that are used to predict y. In linear regression you have the same form as with logistic regression except that y is the response rather than a function of it. e is the random component or stochastic element. It is there to express error (sometimes measurement error in determining y). In these models the error term is assumed to be additive. Also the Xi are assumed to be determined without error. Using the model and data for the Xis and the corresponding ys the regression parameters a1, a2,a3, ... are estimated (usually by least squares). Often the error term e is assumed to be normally distributed with me an 0. They are independent of the Xis and independent of each other. The variability of the error term is defined by the standard deviation sigma which is assumed to be constant (i. e. the same for each sample). For the normal distribution the mean and standard deviation uniquely determine the distribution. These models are general and the technique can be applied in any sport. The covariate and response variables can differ from sport to sport. It is important to keep the modeling assumptions in mind and if they don't seem to apply alternative approaches may need to be considered.

The American Statistical Association has a section on Statistics in Sports. Hal Stern whose article you cited is a member of that section. He has written substantially on sports. other members Jim Albert and Michael Schell have written books specifically on baseball. You might want to consider joining the association and the section. On the ASA website there is an eGroup for the section where questions like yours are discussed. I am a member but haven't actively done research in it. Many members of the section have though.

Other prominent members that I have not yet mentioned are Carl Morris and Scott Berry.

What is the sample size and do you have bins with fewer than 5 observations? I ask these questions because very small departures from normality can be detect in large samples. Also the chi square test is approximate and doesn't work well when there are bins with fewer than 5 observations. The median seems a lot larger than the mean which would indicate high skewness. But that doesn't show up in the skewness statistic.The standard deviation is large but the standard error is small which indicate a large sample size.

Since the ratio of the standard deviation tio the standard error should be the sqaure root of n, your sample size must be close to 600. I calculated 576 and looking at our data again I see count =567 is probably your sample size and that would make sense. I would like to see a histogram and a Q-Q plot. The conclusion is non-normal but it is not yet clear to me that the departure from normality is large enough to worry about.