Test of Signficiance for Win Percentage I am looking at the win percentage of a teams using 3 defenders versus teams using 4 defenders and would like to see if the difference in means is statistically signifcant and test within R. However, I first want to know if I am setting it up properly.
Let

*

*$p_1=0.30\ $ be the probability of winning with a Back 3 and

*$p_2=0.35\ $ be the probability of winning with a Back 4.

*$n_1=100\ $ be the number of games with a Back 3, and

*$n_2=123\ $ be the number of games with a Back 4

So my hypothesis tests are as follows:
$$H_0:p_1\le p_2\ \text{ and } \ H_A:p_1\gt p_2$$
I want to test if it is statistically greater rather than simply statistically different. Would this be advised or is it best practice to simply test the means? Since win percentage is a binomial probaility, I need to calculate the test statistic which is:
$$z=\frac{\hat p_1-\hat p_2}{\sqrt{\hat p(1-\hat p)\left(\frac{1}{n_1}+\frac{1}{n_2}\right)}}$$
where $\hat p=\frac{n_1\hat p_1+n_2\hat p_2}{n_1+n_2}$.
In calcuating this, I have the following:
$$\hat p=\frac{n_1\hat p_1+n_2\hat p_2}{n_1+n_2}$$
$$\hat p=\frac{100\times 0.3+123\times 0.35}{100+123}$$
$$\hat p=0.355157$$
So when I plug in,
$$z=\frac{0.3-0.35}{\sqrt{0.355157\cdot (1-0.355157)\cdot\left(\frac{1}{100}+\frac{1}{123}\right)}}$$
Which is simply $z=-1.55$ which is less than the signicant value of $-1.96$, i.e. we fail to reject the null.
Is this process correct?
 A: The procedure is correct, but several things need to be said.


*

*This is only valid if the two samples are independent, so make sure this working assumption holds in your problem.

*It's not quite correct to say "Let $p_1=0.30$ be the probability of winning with a Back 3 and $p_2=0.35$..." because the probabilities of winning are unknown, or else there will be no need to test this hypothesis. What you are referring to in here is the point estimates (proportions) based on your sample. 

*You did not specify the significance level of this test. This is a one-side, test, so the critical value -1.96 corresponds to size $\alpha=.025$. If this is correct, then you're good. If $\alpha=0.05$ then the appropriate critical value should be $-1.64$.
A: I don't know what this game is, but if a team may independently select the number of defenders (whatever that means) and the team 1 selected 3 an the team 2 selected 4, the probability of the 1st and 2nd winning can be $0.3$ and $0.35$ correspondingly at the same time only if this game can end in a draw with the probability of $1.0-0.3-0.35=0.35$.
Another consideration is that the teams 1 and 2 may have 4 defenders at the same time. Should it affect the probability of winning (is it somewhere between $0.3$ and $0.35$ in this case, or maybe the probability of ending in a draw will rise from $0.35$ to, let's say, $0.7$)?
