I'm trying to work out if the home advantage for two soccer leagues are different, where teams try to score more goals than their opponent to win, otherwise drawing or losing.

Home advantage is the added benefit of playing on your own turf, and I've defined it as

$$ \gamma=\frac{\sum x_i}{\sum y_i}, $$ where $\gamma $ is the advantage from playing at home, $x_i$ is the total number of goals scored a home team in game ${i}$ across a season and $y_i$ the number of goals scored by away teams.

$x$ and $y$ are the sum of poisson distributions with low means that cannot be approximated by the normal distribution.

The data gets to a 2 x 2 table:

\begin{array}{|c|c|} \hline & League A & League B \\ \hline Home & 853& 741\\ \hline Away & 579 & 50 8 \\ \hline \gamma & 1.473 & 1.459 \\ \hline \end{array}

Now a Chi test screams out here but I'm not sure whether any significance difference found would the right one: there is a definitely a difference between home and away, but I want to check if there is a difference between leagues.

Would the correct way of doing to calculate the $y$ of the two league's combined, use this to work out the expected ratios of home to away and then apply the normal chi squared test ?

  • $\begingroup$ I batted my eye and thought of a Poisson regression. What do you think? $\endgroup$
    – jassis
    Feb 13, 2022 at 1:26
  • 1
    $\begingroup$ Table shows no significance using chisq.test in R. P-value is $0.99.$ Unclear why you say it 'screams out'. // The two $\gamma$s are a bit different, but is that difference of practical importance? $\endgroup$
    – BruceET
    Feb 13, 2022 at 1:32

1 Answer 1


One could define 'home advantage' in various ways, but it seems natural to look at proportions of wins, and it easy to test whether they are significantly different.

Here, the proportions of wins are $0.5957$ and $0.5947,$ respectively--very nearly the same. So, it should be no surprise that prop.test in R does not find them significantly different.

prop.test(c(853, 741), c(1432, 1246))

        2-sample test for equality of proportions 
        with continuity correction

data:  c(853, 741) out of c(1432, 1246)
X-squared = 0.00013009, df = 1, p-value = 0.9909
alternative hypothesis: two.sided
95 percent confidence interval:
 -0.03705508  0.03898976
sample estimates:
   prop 1    prop 2 
0.5956704 0.5947030

This test is essentially the same as the chi-squared test in my comment and gives the same P-value.


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