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I am studying performance of football teams. I want to prove one team is significantly more successful at home then other teams.

I calculate the ratio between points per game at home and points per game away. I want to prove that this ratio for one team is significantly different from the population.

I am familiar with t-tests. But I believe I cannot use them here. The ratios that I am testing is continuous, but the point per game is derived from the number of points a team gains in a game which can be 0,1 or 3. I do not have and IV, I am just comparing different populations.

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I calculate the ratio between points per game at home and points per game away. I want to prove that this ratio for one team is significantly different from the population.

Taking this as what you want to do. That is, 1) use the total points from all games, and not use each game as a separate observation for that team. 2) Use Home points / Total points as the relevant metric. And, 3) compare this for one team to the population value.

A simple way would be to calculate a confidence interval for the binomial proportion of Home points / Total points. If this confidence interval doesn't contain the proportion for the population, then this team is significantly different than the population.

Here, 0.56 is outside the confidence interval, suggest the team is different from the population.

Home.points = 100

Total.points = 150

Population.prop = 0.56

binom.test(Home.points, Total.points)

   ### 95 percent confidence interval:
   ###  0.5851570 0.7414436

Another way to do essentially the same thing is use a binomial test using the the population proportion as the theoretical proportion. A significant p-value suggests that the team is different than the population.

Here, the p-value < 0.05, suggest the team is different from the population.

binom.test(Home.points, Total.points, Population.prop)

   ### number of successes = 100, number of trials = 150, p-value = 0.008429
   ### alternative hypothesis: true probability of success is not equal to 0.56
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  • $\begingroup$ Thank you for the answer. There is one problem though, since I am measuring points not observations, the test gives less significance to worse teams. For example if one team has 15 home point and 30 all together the test would say the difference is less significant then for a team with 40 and 80 point, even though the sample size is the same. $\endgroup$ – Borut Flis Aug 6 '18 at 6:47
  • $\begingroup$ @BorutFlis , Yes, that's a good point. $\endgroup$ – Sal Mangiafico Aug 6 '18 at 14:01
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I think it won't be a mistake to use the tests applied to a contingency table (e.g. Pearson's chi-squared test or Fisher's exact test).

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    $\begingroup$ Maybe you could elaborate on how to apply such tests in this case? $\endgroup$ – The Laconic Aug 5 '18 at 15:06
  • $\begingroup$ @TheLaconic Make a contingency table where cols are games (away or home) and rows are teams. Then follow instructions for these tests in your statistical software. $\endgroup$ – Maksim Terpilowski Aug 5 '18 at 15:30

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