In: Math
Let be the proportion of away matches won in even year and let be the same for odd year. Then the null and alternate hypothesis would be the following:
We will use the following test statistic:
We will use 5% level of significance. Then from the data given,
Therefore the value of the test statistic from the sample would be 0.498. Now observe that this is a right tailed test. Hence the p-value for this test would be
As the p-value is greater than the level of significance hence we fail to reject the null hypothesis. Hence he is wrong.