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## Addendum on Pythagorean Expectation May 20, 2010

Posted by tomflesher in Baseball, Economics.
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I noted below that the sample size of 13 games is too small to make a determination as to whether the proportions of conditions expected to predict the winning team – the home team, the team with the higher Pythagorean expectation, the team with more runs scored, and the team with the higher run differential – is significantly different from chance. If chance were the only determinant of the winner, then we would expect each proportion to be .5, since you’d expect a randomly-selected home team to win half the games, a randomly-selected team with higher run differential to win half the games, and so on.

Making the standard statistical assumptions, the margin of error using proportions is $\sqrt{\frac{p(1-p)}{n}}$ . Three of the proportions were .46, meaning that the margin of error would be $\sqrt{\frac{.46(.54)}{13}} = \sqrt{\frac{.2484}{13}}$ which simplifies to $\sqrt{.0191} = {.1382}$. Using 12 degrees of freedom, a t-table shows that the critical value for 95% confidence  is 2.18. Thus, the binomial confidence interval method, tells us we can be 95% sure that the true value of the proportion lies within the range .46 ± 2.18*.1382 = .46 ± .30 = .16 … .76. Clearly, this range is far too large to reject the conclusion that the proportion is significantly different from .5.

For the simple measure of more runs, the proportion was .31, meaning that the margin of error is $\sqrt{\frac{.31(.69)}{13}} = \sqrt{\frac{.2139}{13}}$ or $\sqrt{.0165} = {.1283}$. The 95% confidence interval around .31 is .31 ± 2.18*.1283 = .31 ± .2797 = .03 … .59. Again, .5 is included in this range.

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## Comments»

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