In the 1996 regular baseball season, the World Series Champion New York Yankees

Question

In the 1996 regular baseball season, the World Series Champion New York Yankees played 80 games at home and 82 games away. They won 49 of their home games and 43 of the games played away. We can consider these games as samples from potentially large populations of games played at home and away. How much advantage does the Yankee home field provide?

a. Find the proportion of wins for the home games. Do the same for the away games.

b. Find the standard error needed to compute a confidence interval for the difference in the proportions.

c. Compute a 90% confidence interval for the difference between the probability that the Yankees win at home and the probability that they win when on the road. Are you convinced that the 1996 Yankees were more likely to win at home?

Explanation / Answer

a) Proportion of wins for the home games (p1) = 49/80 = 0.61

    Proportion of wins for for the games away(p2) = 43/82 = 0.52

b) The Pooled sample proportion (P) = (p1 * n1 + p2 * n2)/(n1 + n2)

                                                           = (0.61 * 80 + 0.52 * 82)/(80 + 82)

                                                           = 0.564

SE = sqrt(P * (1 - P) * (1/n1 + 1/n2))

      = sqrt(0.564 * (1 - 0.564) * (1/80 + 1/82))

      = 0.078

c) At 90% confidence interval the critical value is z0.05 = 1.645

The 90% confidence interval is

(p1 - p2) +/- z0.05 * SE

= (0.61 - 0.52) +/- 1.645 * 0.078

= 0.09 +/- 0.128

= -0.038, 0.218

As the the interval cointains the hypothised value 0, so there is not sufficient evidence to suopport that 1996 Yankees were more likely to win at home.

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