Correct Hypothesis Test - Comparing Means Soccer Leagues 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 ?

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.
