To test the hypothesis, we shall make a number of simplifying assumptions. First of all, we shall ignore the fact that some of t
he games were played between Stanford and Cal: we shall pretend that all the games were played against other teams in the conference. One strong version of the hypothesis that the two teams have equal skill is that the outcomes of the games would have been the same had the two teams swapped schedules. That is, suppose that when Washington played Stanford on a particular day, Stanford won. Under this strong hypothesis, had Washington played Cal that day instead of Stanford, Cal would have won. A weaker version of the hypothesis is that the outcome of Stanford's games is determined by independent draws from a 0-1 box that has a fraction pC of tickets labeled "1" (Stanford wins the game if the ticket drawn is labeled "1"), that the outcome of Berkeley's games is determined similarly, by independent draws from a 0-1 box with a fraction pS of tickets labeled "1," and that pS = pC. This model has some shortcomings. (For instance, when Berkeley and Stanford play each other, the independence assumption breaks down, and the fraction of tickets labeled "1" would need to be 50%. Also, it seems unreasonable to think that the chance of winning does not depend on the opponent. We could refine the model, but that would require knowing more details about who played whom, and the outcome.) Nonetheless, this model does shed some light on how surprising the records would be if the teams were, in some sense, equally skilled. This box model version allows us to use Fisher's Exact test for independent samples, considering "treatment" to be playing against Stanford, and "control" to be playing against Cal, and conditioning on the total number of wins by both teams (26). Problem 2. The test statistic is (Q3) If the null hypothesis is true, the test statistic has a (Q4) distribution with parameter(s) (Q5) Problem 3. The P-value for a one-sided test against the alternative hypothesis that the Stanford team is better than the Cal team is (Q6) .0277777778 At significance level 5%, we should reject the null hypothesis. (Q7) Problem 4. If the null hypothesis is true, the expected value of the test statistic is (Q8) 8.5 and the standard error of the test statistic is (Q9) 1.518928194 The z-score of the test statistic is (Q10) The normal approximation to the P-value for Fisher's exact test against the alternative that the Stanford team is better than the Berkeley team is (Q11) 10.50655275 Now consider the z test using independent samples. We pretend that each team's wins and losses are independent random samples with replacement from 0-1 boxes in which the fraction of ones represents the probability that that team wins each game it plays. The number of tickets labeled "1" in the sample is the number of games the team wins. Problem 5. The sample percentage of games won by Stanford is (Q12) On the assumption that the null hypothesis is true, the bootstrap estimate of the standard error of the sample percentage of games won by Stanford is (Q13) (Hint: if the null hypothesis is true, then with the simplifications we made it is as if the teams independently draw at random with replacement from the same box of tickets. What percentage of the tickets in that box would you estimate to be labeled "1?") The sample percentage of games won by Cal is (Q14) On the assumption that the null hypothesis is true, the bootstrap estimate of the standard error of the sample percentage of games won by Cal is (Q15) The difference in sample percentages of games won by Stanford and Cal is (Q16) On the assumption that the null hypothesis is true, the bootstrap estimate of the standard error of the difference in sample percentages is (Q17) The z-score for the difference in sample percentages is (Q18) The approximate P-value for z test against the two-sided alternative that the Stanford and Berkeley teams have different skills is (Q19) At significance level 1%, we should reject the null hypothesis
