Let p^ = the sample statistic (23/79) and p(o) = the expected population statistic (1/6). Let the α =.05 be the threshold. If the discovered p-value is < .05 we reject the null hypothesis. The null hypothesis is that p^ = p(o). For this let's use a One-proportion z-test.
At this point in the tournament there had been 29 games, so let n = 29. p^ = p(hat) (I am unsure how to type hats or subscripts), p(o) = p and z-score = z. Plug the values into the formula to get the z-score... the z-score = 1.79, making the p-value < .05 so we reject the null hypothesis.
Does it make sense that such a high percentage of goals would be scored in the final 15 minutes? It brings to mind an analysis I read about the US presidential elections. The author asserts that the trailing candidate should pursue the "pull the goalie strategy" used in hockey. The trailing hockey team is so desperate to score, the coach pulls the goalie and puts a better goal score into the game. By pulling the goalie the team expects to have more chances to score but will certainly be easier to score upon. The team that is trailing wants to utilize a strategy that increases the overall variance at the expense of the optimal strategy. Over the long-term this high variance strategy is not as effective as the optimal strategy, but the trailing team is not playing for the long term. It is playing for the short term. The trailing team will use a high variance strategy, resulting in it and the opposition scoring more goals.
This was evident today in the Euro 2008 finals. Germany pursued the high variance strategy in the final 15 minutes at the expense of the optimal strategy. The German team had more chances to score, but at the same time allowed the Spanish side some great opportunities.
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